An electron experiences a downward magnetic force of 7.00×10 −14 N when it is travelling at 1.8×10 5 m/s south through a magnetic field. Calculate the magnitude and direction of the magnetic field. a. 1.6⊤ down b. 4.3×10 11 T down C. 2.3×10 8 ⊤ down d. 2.4 T down A charged particle is travelling west through a downward magnetic field and it experiences a magnetic force directed to the north. Using the appropriate hand rule, determine if the charge is negative or positive. Explain all finger directions and the palm direction. Calculate the magnitude and the direction of the magnetic force acting on an alpha particle that is travelling upwards at a speed of 3.00×10 5 m/s through a 0.525 T west magnetic field. Explain all finger directions and the palm direction.

Answers

Answer 1

The magnitude of the magnetic field is approximately 2.430 T, and it is directed downward.The magnitude of the magnetic force acting on the alpha particle is approximately 3.15 × 10⁵N, and it is directed north, based on the right-hand rule.

To calculate the magnitude and direction of the magnetic field in the first scenario:

Force on the electron (F) = 7.00 × 10⁽⁻¹⁴⁾ N,

Velocity of the electron (v) = 1.8 × 10⁵ m/s.

The formula for the magnetic force on a charged particle moving through a magnetic field is given by:

F = qvB sin(θ),

where F is the force, q is the charge of the particle, v is the velocity, B is the magnetic field strength, and θ is the angle between the velocity vector and the magnetic field vector.

In this case, the force is downward, the velocity is south, and the angle is 90 degrees (because the velocity is perpendicular to the force). Therefore, sin(θ) = 1.

Rearranging the formula, we can solve for the magnetic field strength (B):

B = F / (qv).

Substituting the given values:

B = (7.00 × 10⁽⁻¹⁴⁾ N) / (1.6 × 10⁽⁻¹⁹⁾⁾ C × 1.8 × 10⁵ m/s).

B = 2.430 T.

For the second scenario, using the appropriate hand rule:

When a charged particle is moving in a magnetic field, the thumb points in the direction of the force, the index finger points in the direction of the magnetic field, and the middle finger points in the direction of the velocity.

If the magnetic force is directed to the north and the velocity of the particle is west, then the magnetic field must be directed upward. Since the force is directed opposite to the velocity, the charge of the particle must be negative.

Regarding the calculation of the magnitude and direction of the magnetic force acting on an alpha particle:

Velocity of the alpha particle (v) = 3.00 × 10⁵m/s,

Magnetic field strength (B) = 0.525 T.

Using the formula:

F = qvB sin(θ),

where F is the force, q is the charge of the particle, v is the velocity, B is the magnetic field strength, and θ is the angle between the velocity vector and the magnetic field vector.

Since the alpha particle is traveling upward, and the magnetic field is west, the angle θ is 90 degrees. Therefore, sin(θ) = 1.

Substituting the given values into the formula:

F = (2e)(3.00 × 10⁵ m/s)(0.525 T)(1).

F = 3.15 × 10⁵ N.

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Related Questions

A 28 g ball of clay traveling east at 3.2 m/s collides with a 32 g ball of clay traveling north at 2.8 m/s

Answers

The two balls will move together at a velocity of 2.987 m/s at an angle between east and north after the collision.

When the 28 g ball of clay traveling east at 3.2 m/s collides with the 32 g ball of clay traveling north at 2.8 m/s, the two balls will stick together due to the conservation of momentum.
To solve this problem, we can use the principle of conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision.
The momentum of an object is given by the product of its mass and velocity. Therefore, the momentum of the 28 g ball of clay before the collision is (28 g) * (3.2 m/s) = 89.6 g·m/s east, and the momentum of the 32 g ball of clay before the collision is (32 g) * (2.8 m/s) = 89.6 g·m/s north.


After the collision, the two balls stick together, so their total mass is 28 g + 32 g = 60 g. The momentum of the combined mass can be calculated by adding the momenta of the individual balls before the collision.
Therefore, the total momentum after the collision is 89.6 g·m/s east + 89.6 g·m/s north = 179.2 g·m/s at an angle between east and north.
To calculate the velocity of the combined balls after the collision, divide the total momentum by the total mass: (179.2 g·m/s) / (60 g) = 2.987 m/s.

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Suppose you wanted to levitate a person of mass 75.0 kg at 0.397 m above an equally charged plate on the ground below (near Earth) using electric force. What charge would the person and the charged plate have in microcoulombs (1,000,000 μC = 1 C) to three significant digits in order to balance the person's weight at that height?

Answers

To balance the person's weight at a height of 0.397 m, both the person and the charged plate should have charges of approximately 22.6 microcoulombs (μC).

The electric force between two charged objects can be calculated using Coulomb's law: F = (k * |q1 * q2|) / r²

Where F is the force, k is the electrostatic constant (approximately 9 × 10^9 N·m²/C²), q1 and q2 are the charges on the objects, and r is the distance between them. In this case, the electric force should be equal to the weight of the person: F = m * g

Where m is the mass of the person (75.0 kg) and g is the acceleration due to gravity (approximately 9.8 m/s²). Setting these two forces equal, we have: (m * g) = (k * |q1 * q2|) / r²

Now, since both the person and the plate have equal charges, we can rewrite the equation as: (m * g) = (k * q^2) / r²

Rearranging the equation to solve for q, we get: q = √((m * g * r²) / k)

Substituting the given values:
q = √((75.0 kg * 9.8 m/s² * (0.397 m)²) / (9 × 10^9 N·m²/C²))

Calculating the value: q ≈ 2.26 × 10^-5 C

Converting to microcoulombs: q ≈ 22.6 μC

Therefore, to balance the person's weight at a height of 0.397 m, both the person and the charged plate should have charges of approximately 22.6 microcoulombs (μC).

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1) If you add the vectors 12m South and 10m 35° N of E. the angle of the resultant is ____° S of E
2) A 125N box is pulled east along a horizontal surface with a force of 60.0N acting at an angle of 42.0°. if the force of frction is 25.0N, what is the acceleration of the box?

Answers

The acceleration of the box is 2.75 m/s².

1) If you add the vectors 12m South and 10m 35° N of E. the angle of the resultant is 25° S of E.

Consider the given vectors: The first vector is 12 m towards southThe second vector is 10 m towards the northeast which makes 35° with the east. We can represent both the vectors graphically and find their sum vector to determine the resultant vector.

When two vectors are added together, the resultant vector is obtained as shown below:

The angle of the resultant vector with the east is given by:

                          tanθ = (Ry/Rx)Where,Ry = 12 m - 10 sin 35°

                            Ry = 12 m - 5.7735 m

                           Ry = 6.2265 m

                            Rx = 10 cos 35°

                         Rx = 8.1773 m

Now, tanθ = (6.2265/8.1773)θ = tan-1(6.2265/8.1773)θ

                                    = 36.869898 mθ = 37°

The angle of the resultant vector is 37° S of E.

2) A 125N box is pulled east along a horizontal surface with a force of 60.0N acting at an angle of 42.0°. if the force of frction is 25.0N,

In this question, the force that acts on the box is 60 N at an angle of 42°.

The force of friction that acts on the box is 25 N.

The net force that acts on the box is given by:

                            Fnet = F - fWhere,F = 60 Nf = 25 NThe net force Fnet = 35 N.

The acceleration a of the box is given by:Fnet = ma35 = m × a

The mass of the box m = 125/9.81 m/s²m = 12.71 kgTherefore, a = 35/12.71a = 2.75 m/s²

The acceleration of the box is 2.75 m/s².

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Show that the first Covarient derivative of metric tensor th

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The first covariant derivative of the metric tensor is a mathematical operation that describes the change of the metric tensor along a given direction. It is denoted as ∇μgνρ and can be calculated using the Christoffel symbols and the partial derivatives of the metric tensor.

The metric tensor in general relativity describes the geometry of spacetime. The first covariant derivative of the metric tensor, denoted as ∇μgνρ, represents the change of the metric tensor components along a particular direction specified by the index μ. It is used in various calculations involving curvature and geodesic equations.

To calculate the first covariant derivative, we can use the Christoffel symbols, which are related to the metric tensor and its partial derivatives. The Christoffel symbols can be expressed as:

Γλμν = (1/2) gλσ (∂μgσν + ∂νgμσ - ∂σgμν)

Then, the first covariant derivative of the metric tensor is given by:

∇μgνρ = ∂μgνρ - Γλμν gλρ - Γλμρ gνλ

By substituting the appropriate Christoffel symbols and metric tensor components into the equation, we can calculate the first covariant derivative. This operation is essential in understanding the curvature of spacetime and solving field equations in general relativity.

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1. The temperature on a digital thermometer reads 150 C what is the uncertainty (error) in the measurement? 2. The accepted value for the speed of light in vacuum is 2.998x10^8 m/s. Assume that you have performed an experiment to determine the speed of light and obtained an average value of 2.977x10^8 m/s. Calculate the percent difference between the experimental and accepted value for the speed of light.

Answers

1. The uncertainty (error) in the temperature measurement of 150°C is ±0.1°C.

2. The percent difference between the experimental and accepted value for the speed of light is approximately 0.700%.

1. The uncertainty in the measurement can be determined by considering the least count or precision of the digital thermometer. If we assume that the least count is ±0.1°C, then the uncertainty (error) in the measurement is ±0.1°C.

2. To calculate the percent difference between the experimental and accepted value for the speed of light, we can use the formula:

  Percent Difference = |(Experimental Value - Accepted Value) / Accepted Value| * 100

  Substituting the given values, we have:

  Percent Difference = |(2.977x10⁸ m/s - 2.998x10⁸ m/s) / 2.998x10⁸ m/s| * 100

  = |(-0.021x10⁸ m/s) / 2.998x10⁸ m/s| * 100

  = |(-0.021/2.998) * 100|

  = |-0.0070033356| * 100

  = 0.70033356%

Therefore, the percent difference between the experimental and accepted value for the speed of light is approximately 0.700%.

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Three deer, A, B, and C, are grazing in a field. Deer B is located 62.4 m from deer A at an angle of 51.9" north of west. Deer C is located 76,4° north of east relative to deer A. The distance between deer B and is 94.2 m. What is the distance between deer A and C (Hint: Consider the laws of sines and cosines given in Appendix E.)

Answers

Answer:

The distance between deer A and C is approximately 122.6 meters.

To find the distance between deer A and C, we can use the law of cosines. According to the given information, we have a triangle formed by deer A, deer B, and deer C.

Let's denote the distance between deer A and C as dAC. Using the law of cosines, we have:

dAC² = dAB² + dBC² - 2(dAB)(dBC)cosθ

where:

dAB is the distance between deer A and B (62.4 m),

dBC is the distance between deer B and C (94.2 m),

θ is the angle between dAB and dBC.

Now, we need to find θ. Since deer B is located north of west, and deer C is located north of east relative to deer A,

we can infer that the angle θ is 180° - 51.9° - 76.4° = 52.7°.

Substituting the values into the equation, we have:

dAC² = (62.4 m)² + (94.2 m)² - 2(62.4 m)(94.2 m)cos(52.7°)

Calculating:

dAC ≈ 122.6 m

Therefore, the distance between deer A and C is approximately 122.6 meters.

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2. Two closeby speakers produce sound waves. One of the speakers vibrates at 400 Hz. What would be the frequency of the other speaker, which produces 10 Hz of beats? A. 10 Hz B. 390 Hz C. 410 Hz

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Summary:

The frequency of the other speaker would be 390 Hz. When two closeby speakers produce sound waves, a phenomenon known as beats can occur. Beats are the periodic variations in the intensity or loudness of sound that result from the interference of two waves with slightly different frequencies.

Explanation:

In this case, if one speaker vibrates at 400 Hz and the beats have a frequency of 10 Hz, it means that the frequency of the other speaker is slightly different. The beat frequency is the difference between the frequencies of the two speakers. So, by subtracting the beat frequency of 10 Hz from the frequency of one speaker (400 Hz), we find that the frequency of the other speaker is 390 Hz.

To understand this concept further, let's delve into the explanation. When two sound waves with slightly different frequencies interact, they undergo constructive and destructive interference, resulting in a periodic variation in the amplitude of the resulting wave. This variation is what we perceive as beats. The beat frequency is equal to the absolute difference between the frequencies of the two sound waves. In this case, the given speaker has a frequency of 400 Hz, and the beat frequency is 10 Hz. By subtracting the beat frequency from the frequency of the given speaker (400 Hz - 10 Hz), we find that the frequency of the other speaker is 390 Hz. This frequency creates the interference pattern that produces the 10 Hz beat frequency when combined with the 400 Hz wave. Therefore, the correct answer is B. 390 Hz.

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(a) A projectile is shot from the ground level with an initial speed of 22 m/s at an angle of 40 ∘ above the horizontal. Finally, the projectile lands at the same ground level. (i) Calculate the maximum height reached by the projectile with respect to the ground level. (3 marks) (ii) Determine the range of the projectile as measured from the launching point. (3 marks) (b) The actual weight of an iron anchor is 6020 N in air and its apparent weight is 5250 N in water. Given that the density of water is rho water ​ =1×10 3 kg/m 3 . (i) Calculate the volume of the iron anchor. (3 marks ) (ii) Calculate the density of the iron anchor (3 marks) (c) Two vectors are given as: P =2 i ^ −4  ^ ​ +5 k ^ and Q ​ =7  ^ −3  ^ ​ −6 k ^ . Determine (i) P ⋅ Q ​ (3 marks) (ii) angle between P and Q ​ , (4 marks) (iii) P × Q ​ , and (3 marks) (iv) 3 P − Q ​ . (3 marks)

Answers

a)

i) The maximum height reached by the projectile with respect to the ground level can be calculated as follows:

Given, the initial speed of the projectile = u = 22 m/s

Angle of projection = θ = 40°

The horizontal component of velocity, v_{x} = u cosθ = 22 cos40° = 16.8 m/s

The vertical component of velocity, v_{y} = u sinθ = 22 sin40° = 14.2 m/s

Acceleration due to gravity, g = 9.8 m/s²

At the maximum height, the vertical component of velocity becomes zero.

Using the following kinematic equation: v^{2} = u^{2} + 2as

At maximum height, v = 0, u = v_{y}, and a = -g

Substituting the values, we get: 0 = (14.2)² - 2 × 9.8 × s⇒ s = 10.89 m

Therefore, the maximum height reached by the projectile is 10.89 m.

ii) The range of the projectile can be calculated as follows:

Using the following kinematic equations:

v_{x} = u cosθ (horizontal motion)S_{x} = (u cosθ)t (horizontal motion)t = 2u sinθ/g (time of flight)S_{y} = u sinθt - 0.5gt² (vertical motion)

Substituting the values, we get: S_{x} = 16.8 × (2 × 22 sin40°)/9.8 = 44.1 m

Therefore, the range of the projectile is 44.1 m.

b)

i) The volume of the iron anchor can be calculated using the following formula:

Volume of the object = mass of the object/density of the object

Given, the actual weight of the iron anchor in air = 6020 N

Apparent weight of the iron anchor in water = 5250 N

Density of water, ρ_{water} = 1000 kg/m³

The buoyant force acting on the iron anchor can be calculated as follows:

Buoyant force = Weight of the object in air - Apparent weight of the object in water

Buoyant force = 6020 - 5250 = 770 N

The buoyant force is equal to the weight of the water displaced by the iron anchor.

Therefore, the volume of the iron anchor can be calculated as follows:

Volume of the iron anchor = Buoyant force/density of water

Volume of the iron anchor = 770/1000 = 0.77 m³

Therefore, the volume of the iron anchor is 0.77 m³.

ii) The density of the iron anchor can be calculated using the following formula:

Density of the object = Mass of the object/Volume of the object

Given, the actual weight of the iron anchor in air = 6020 N

Density of water, ρ_{water} = 1000 kg/m³

Volume of the iron anchor = 0.77 m³

Using the following formula to calculate the mass of the iron anchor:

Weight of the iron anchor = Mass of the iron anchor × g6020 N = Mass of the iron anchor × 9.8 m/s²

Mass of the iron anchor = 614.29 kg

Therefore, the density of the iron anchor can be calculated as follows:

Density of the iron anchor = 614.29 kg/0.77 m³

Density of the iron anchor = 798.7 kg/m³

Therefore, the density of the iron anchor is 798.7 kg/m³.

c)

i) The dot product of the two vectors P and Q can be calculated using the following formula:

P · Q = P_{x}Q_{x} + P_{y}Q_{y} + P_{z}Q_{z}

Given, P = 2i - 4j + 5k and Q = 7i - 3j - 6k

Substituting the values, we get:

P · Q = (2 × 7) + (-4 × -3) + (5 × -6)P · Q = 14 + 12 - 30P · Q = -4

Therefore, P · Q = -4.

ii) The angle between two vectors P and Q can be calculated using the following formula:

cosθ = (P · Q)/(|P||Q|)

Given, P = 2i - 4j + 5k and Q = 7i - 3j - 6k

Substituting the values, we get:|P| = √(2² + (-4)² + 5²) = √45 = 6.71|Q| = √(7² + (-3)² + (-6)²) = √94 = 9.7cosθ = (-4)/(6.71 × 9.7)cosθ = -0.044θ = cos⁻¹(-0.044)θ = 91.13°

Therefore, the angle between vectors P and Q is 91.13°.

iii) The cross product of the two vectors P and Q can be calculated using the following formula:

P × Q = |P||Q| sinθ n

Given, P = 2i - 4j + 5k and Q = 7i - 3j - 6kθ = 91.13° (from part ii)

Substituting the values, we get:

P × Q = 6.71 × 9.7 × sin91.13° n

P × Q = -64.9n

Therefore, the cross product of vectors P and Q is -64.9n. (n represents the unit vector in the direction perpendicular to the plane containing the two vectors).

iv) The vector 3P - Q can be calculated as follows:

3P - Q = 3(2i - 4j + 5k) - (7i - 3j - 6k)3P - Q = 6i - 12j + 15k - 7i + 3j + 6k3P - Q = -i - 9j + 21k

Therefore, the vector 3P - Q is -i - 9j + 21k.

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Transcribed image text: Buttercup is on a frictionless sled that is attached to a spring on horiontal ground. You pull the sled out 1.6 m to the right and release the sled from rest. The spring has a spring constant of 521 N/m and Buttercup and the sled have a combined mass of 53 kg. Assume the positive x-direction is to the right, that Buttercup and the sled were at x=0m before you pulled them to the right. Help on how to format answers: units a. What is Buttercup's position after oscillating for 8.1 s? Buttercup's position is i. b. What is Buttercup's velocity after oscillating for 8.1 s?

Answers

(a) Buttercup's position after oscillating for 8.1 s is approximately -1.576 m.

(b) Buttercup's velocity after oscillating for 8.1 s is approximately 0.567 m/s.

To determine Buttercup's position and velocity after oscillating for 8.1 s, we need to consider the principles of harmonic motion.

Amplitude (A) = 1.6 m (maximum displacement from equilibrium position)

Spring constant (k) = 521 N/m

Mass (m) = 53 kg

Time (t) = 8.1 s

a) Position:

The equation for the position of an object undergoing simple harmonic motion is given by:

x(t) = A * cos(ωt + φ)

Where:

x(t) is the position at time t,

A is the amplitude,

ω is the angular frequency, and

φ is the phase constant.

To find the position at t = 8.1 s, we need to determine the angular frequency and phase constant.

The angular frequency is given by:

ω = sqrt(k/m)

Substituting the values, we have:

ω = sqrt(521 N/m / 53 kg)

ω ≈ 2.039 rad/s

Since Buttercup is released from rest, the phase constant φ is 0.

Now we can calculate the position:

x(8.1) = 1.6 m * cos(2.039 rad/s * 8.1 s)

x(8.1) ≈ 1.6 m * cos(16.479 rad)

x(8.1) ≈ 1.6 m * (-0.985)

x(8.1) ≈ -1.576 m

Therefore, Buttercup's position after oscillating for 8.1 s is approximately -1.576 m.

b) Velocity:

The velocity of an object undergoing simple harmonic motion is given by:

v(t) = -A * ω * sin(ωt + φ)

To find the velocity at t = 8.1 s, we can use the same values of ω and φ.

v(8.1) = -1.6 m * 2.039 rad/s * sin(2.039 rad/s * 8.1 s)

v(8.1) ≈ -1.6 m * 2.039 rad/s * sin(16.479 rad)

v(8.1) ≈ -1.6 m * 2.039 rad/s * (-0.173)

v(8.1) ≈ 0.567 m/s

Therefore, Buttercup's velocity after oscillating for 8.1 s is approximately 0.567 m/s.

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We have a rare sample of Unobtainium which has a half life of 54
hours and is currently measuring 1440 uCi. How radioactive will it
be in 18 days?

Answers

The given sample of Unobtainium has a half-life of 54 hours and is currently measuring 1440 uCi. The problem is asking us to determine how radioactive the sample will be in 18 days.

To solve the given problem, we will first find the decay constant using the half-life formula, which is given as follows:Half-life (t1/2) = 0.693/λWhere λ is the decay constant.To find λ, we will rearrange the above formula as follows:

λ = 0.693/t1/2λ = 0.693/54λ

= 0.01283 per hourThe decay constant of the given Unobtainium sample is 0.01283 per hour.

Now, we will use the exponential decay formula to find the radioactive decay of the sample in 18 days. The formula is given as:A = A0 e-λtWhere A is the current activity of the sample, A0 is the initial activity of the sample, e is the mathematical constant, t is the time elapsed, and λ is the decay constant.We know that the current activity of the sample (A) is 1440 uCi and that we need to find its activity after 18 days. We can convert 18 days into hours by multiplying it by 24 as follows:

18 days × 24 hours/day =

432 hours

Now, we will substitute the given values into the exponential decay formula and solve for A

:A = A0 e-λtA =

1440 e-0.01283(432)A ≈

43.85 uCi

Therefore, the sample of Unobtainium will be radioactive at a rate of approximately 43.85 uCi after 18 days.

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The heating coil in an electric bea pot is made of nichrome wire with a radius of 0.400 mm. If the coil draws a current or 5.60 A when there is a 120 V potential oference across ta ende, find the following. (Take the resistivity of nicome to be 1.50 X 100m) (a) resistance of the col (in) (1) length or wire used to win the col tinm) m

Answers

The resistance of the coil is approximately 21.43 Ω, and the length of wire used to wind the coil is approximately 0.071 m.

To find the resistance of the coil, we can use the formula:

Resistance (R) = Resistivity (ρ) * Length (L) / Cross-sectional area (A)

Given the resistivity of nichrome wire as 1.50 × 10^−6 Ω·m and the radius of the wire as 0.400 mm, we can calculate the cross-sectional area (A) using the formula:

[tex]A = π * r^2[/tex]

where r is the radius of the wire.

Let's calculate the cross-sectional area first:

[tex]A = π * (0.400 mm)^2[/tex]

[tex]= π * (0.400 × 10^−3 m)^2[/tex]

[tex]≈ 5.03 × 10^−7 m^2[/tex]

Now, we can calculate the resistance (R) of the coil using the given formula:

[tex]R = ρ * L / A[/tex]

To find the length of the wire used in the coil (L), we rearrange the formula:

[tex]L = R * A / ρ[/tex]

Given that the current drawn by the coil is 5.60 A and the potential difference across the coil is 120 V, we can use Ohm's Law to find the resistance:

[tex]R = V / I[/tex]

Now, we can substitute the values into the formula for the length (L):

[tex]L = (21.43 Ω) * (5.03 × 10^−7 m^2) / (1.50 × 10^−6 Ω·m)[/tex]

Simplifying:

L ≈ 0.071 m

Therefore, the resistance of the coil is approximately 21.43 Ω, and the length of wire used to wind the coil is approximately 0.071 m.

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Suppose you want to operate an ideal refrigerator with a cold temperature of -12.3°C, and you would like it to have a coefficient of performance of 7.50. What is the hot reservoir temperature for such a refrigerator?

Answers

An ideal refrigerator operating with a cold temperature of -12.3°C and a coefficient of performance of 7.50 can be analyzed with the help of

Carnot's refrigeration cycle

.


The coefficient of performance is a measure of the efficiency of a refrigerator.

It represents the ratio of the heat extracted from the cold reservoir to the work required to operate the refrigerator.

Coefficient of performance

(COP) = Heat extracted from cold reservoir / Work inputSince the refrigerator is ideal, it can be assumed that it operates on a Carnot cycle, which consists of four stages: compression, rejection, expansion, and absorption.

The Carnot cycle is a reversible cycle, which means that it can be

operated

in reverse to act as a heat engine.Carnot's refrigeration cycle is represented in the PV diagram as follows:PV diagram of Carnot's Refrigeration CycleThe hot reservoir temperature (Th) of the refrigerator can be determined by using the following formula:COP = Th / (Th - Tc)Where Th is the temperature of the hot reservoir and Tc is the temperature of the cold reservoir.

Substituting

the values of COP and Tc in the above equation:7.50 = Th / (Th - (-12.3))7.50 = Th / (Th + 12.3)Th + 12.3 = 7.50Th60.30 = 6.50ThTh = 60.30 / 6.50 = 9.28°CTherefore, the hot reservoir temperature required to operate the ideal refrigerator with a cold temperature of -12.3°C and a coefficient of performance of 7.50 is 9.28°C.

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Problem# 12 (Please Show Work 20 points) (a) What is the angle between a wire carrying an 9.00-A current and the 1.20-T field it is in if 50.0 cm of the wire experiences a magnetic force of 3.40 N? (b) What is the force on the wire if it is rotated to make an angle of with the field?

Answers

a) Angle: 0.377 radians or 21.63 degrees. b) Force: I * L * B * sin().

a) To find the angle between the wire carrying a current and the magnetic field, we can use the formula for the magnetic force on a current-carrying wire:

F = I * L * B * sin(theta)

Where:

- F is the magnetic force on the wire,

- I is the current in the wire,

- L is the length of the wire segment experiencing the force,

- B is the magnetic field strength,

- theta is the angle between the wire and the magnetic field.

Given:

- Current (I) = 9.00 A

- Length (L) = 50.0 cm = 0.50 m

- Magnetic force (F) = 3.40 N

- Magnetic field strength (B) = 1.20 T

Rearranging the formula, we can solve for the angle theta:

theta = arcsin(F / (I * L * B))

Substituting the given values into the equation, we find:

theta = arcsin(3.40 N / (9.00 A * 0.50 m * 1.20 T))

Calculating this expression, we get:

theta ≈ 0.377 radians or 21.63 degrees

Therefore, the angle between the wire carrying the current and the magnetic field is approximately 0.377 radians or 21.63 degrees.

b) To find the force on the wire when it is rotated to make an angle with the magnetic field, we can use the same formula as in part (a), but with the new angle:

F' = I * L * B * sin()

Given:

- Angle (theta) = (angle with the field)

Substituting these values into the formula, we can calculate the force on the wire when it is rotated:

F' = 9.00 A * 0.50 m * 1.20 T * sin()

(b) To determine the force on the wire when it is rotated to make an angle (θ) with the magnetic field, we can use the same formula for the magnetic force:

F = BILsinθ

Given that the magnetic field strength (B) is 1.20 T, the current (I) is 9.00 A, and the angle (θ) is provided, we can substitute these values into the formula:

F = (1.20 T) * (9.00 A) * L * sinθ

The force on the wire depends on the length of the wire (L), which is not provided in the given information. If the length of the wire is known, you can substitute that value into the formula to calculate the force on the wire when it is rotated to an angle θ with the field.

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Find the total surface area of the washer, rounded to one
decimal place, for x = 14 mm and y = 24 mm. Hint: Think of the
washer as a cylinder through which a hole has been drilled.

Answers

The total surface area of the washer, considering the outer and inner cylinders, is approximately 1051.4 mm². The outer cylinder contributes to the surface area while the inner cylinder, representing the hole, does not affect it.

To find the total surface area of the washer, we need to calculate the surface area of the outer cylinder and subtract the surface area of the inner cylinder.

The surface area of a cylinder is given by the formula:

[tex]A_{cylinder[/tex]= 2πrh

where r is the radius of the cylinder's base and h is the height of the cylinder.

In this case, the washer can be seen as a cylinder with a hole drilled through it, so we need to calculate the surface areas of both the outer and inner cylinders.

Let's calculate the total surface area of the washer:

Calculate the surface area of the outer cylinder:

Given x = 14 mm, the radius of the outer cylinder ( [tex]r_{outer[/tex] ) is half of x, so  [tex]r_{outer[/tex] = x/2 = 14/2 = 7 mm.

The height of the outer cylinder ([tex]h_{outer[/tex]) is y = 24 mm.

[tex]A_{outer_{cylinder[/tex]  = 2π  [tex]r_{outer[/tex][tex]h_{outer[/tex] = 2π(7)(24) ≈ 1051.4 mm² (rounded to one decimal place).

Calculate the surface area of the inner cylinder:

Given the inner radius (r_inner) is 7 mm less than the outer radius, so r_inner = r_outer - 7 = 7 - 7 = 0 mm (since the inner hole has no radius).

The height of the inner cylinder ([tex]h_{inner[/tex]) is the same as the outer cylinder, y = 24 mm.

[tex]A_{inner_{cylinder[/tex] = 2π [tex]r_{inner[/tex] [tex]h_{inner[/tex] = 2π(0)(24) = 0 mm².

Subtract the surface area of the inner cylinder from the surface area of the outer cylinder to get the total surface area of the washer:

Total surface area = [tex]A_{outer_{cylinder[/tex] -  [tex]A_{inner_{cylinder[/tex]  = 1051.4 - 0 = 1051.4 mm².

Therefore, the total surface area of the washer, rounded to one decimal place, is approximately 1051.4 mm².

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The
speed of a car is found by dividing the distance traveled by the
time required to travel that distance. Consider a car that traveled
18.0 miles in 0.969 hours. What's the speed of car in km / h
(k

Answers

The speed of the car is approximately 29.02 km/h, given that it traveled 18.0 miles in 0.969 hours.

To convert the speed of the car from miles per hour to kilometers per hour, we need to use the conversion factor that 1 mile is equal to 1.60934 kilometers.

Given:

Distance traveled = 18.0 milesTime taken = 0.969 hours

To calculate the speed of the car, we divide the distance traveled by the time taken:

Speed (in miles per hour) = Distance / Time

Speed (in miles per hour) = 18.0 miles / 0.969 hours

Now, we can convert the speed from miles per hour to kilometers per hour by multiplying it by the conversion factor:

Speed (in kilometers per hour) = Speed (in miles per hour) × 1.60934

Let's calculate the speed in kilometers per hour:

Speed (in kilometers per hour) = (18.0 miles / 0.969 hours) × 1.60934

Speed (in kilometers per hour) = 29.02 km/h

Therefore, the speed of the car is approximately 29.02 km/h.

The complete question should be:

The speed of a car is found by dividing the distance traveled by the time required to travel that distance. Consider a car that traveled 18.0 miles in 0.969 hours. What's the speed of car in km / h (kilometer per hour)?

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The angular frequency (w') of a damped oscillator is half of the angular frequency of the undamped oscillator (w) of the same system. The mass of the oscillator is 2 kg and force constant K = 200 N/m. (i) What is the damping coefficient (p)? (ii) Calculate the time when the energy of the oscillator drops to one half of its initial undamped value. (iii) Calculate the amplitude drop with respect to initial amplitude during the above time found in (ii).

Answers

(i) The damping coefficient (p) of the oscillator is 10 kg/s.  (ii) The time when the energy of the oscillator drops to one half of its initial undamped value is approximately 1.04 seconds. (iii) The amplitude of the oscillator drops to approximately 0.293 times its initial value.

(i) In a damped oscillator, the relationship between the angular frequency (w) and the damping coefficient (p) is given by p = 2m(w - w'), where m is the mass of the oscillator. Substituting the given values, we have p = 2(2 kg)((200 N/m) - (0.5w)) = 10 kg/s.

(ii) The energy of an undamped oscillator is given by E = 0.5mw^2A^2, where A is the initial amplitude. In a damped oscillator, the energy decreases exponentially with time. The time taken for the energy to drop to one half of its initial undamped value is given by t = (1/p)ln(2). Substituting the value of p, we find t ≈ (1/10 kg/s)ln(2) ≈ 1.04 seconds.

(iii) The amplitude of the oscillator in a damped system decreases exponentially with time and can be expressed as A = A₀e^(-pt/2m), where A₀ is the initial amplitude. Substituting the values of p, t, and m, we have A = A₀e^(-1.04s/4kg) ≈ 0.293A₀. Therefore, the amplitude drops to approximately 0.293 times its initial value during the time found in (ii).

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A circular coil of diameter 14.0 cm, with 43 turns is in a magnetic field of 0.600 Tesla. Initially the field is perpendicular to the plane of the coil. If the coil is rotated in 17.0 ms so its plane is parallel to the field, find the magnitude of the average induced emf in volts.
A 110-turn coil of resistance 3.60 ohms and cross sectional area 17.5 cm² lies in the plane of the page. An external magnetic field of 0.900 T is directed out of the plane of the page. The external I decreases to 0.300 T in 11.7 milliseconds. What is the magnitude of the induced current (in Amperes) in the coil?

Answers

The magnitude of the average induced emf in volts is 0.54V and the magnitude of the induced current (in Amperes) in the coil is 2.49 A

Diameter (d) = 14.0 cm, No of turns (N) = 43, Magnetic field (B) = 0.600 TeslaTime (t) = 17.0 ms

Firstly, calculate the area of the circular coil using the given diameter.

Area of the coil (A) = πr²where r = d/2= 7 cm

Therefore, A = π(7 cm)²= 153.94 cm², Number of turns per unit area isN/A = 43/153.94 = 0.279 turns/cm²

When the coil is perpendicular to the magnetic field, the flux linked with the coil is zero. When it is parallel, the flux is maximum.

The magnetic flux linkage change is given byΔΦ = BAN ΔΦ = B(43/A)

ΔΦ = (0.6 Tesla)(43/153.94 cm²)

ΔΦ = 0.0945 Wb

Therefore, the average induced emf (ε) is ε = ΔΦ/Δt

ε = 0.0945 Wb/ (17.0 × 10-3 s)

ε = 5.56 V

Therefore, the magnitude of the average induced emf in volts is 0.54V.

The solution to the second part of the question is as follows:

Given:

Number of turns (N) = 110, Resistance (R) = 3.60 ohms, Cross-sectional area (A) = 17.5 cm,

²Initial magnetic field (B1) = 0.900 T

Final magnetic field (B2) = 0.300 T

Time (t) = 11.7 ms

The induced emf (ε) can be given by

ε = -N dΦ/dt, where dΦ/dt is the rate of change of flux linkage Φ = BA

Φ = (0.9 T)(17.5 × 10-4 m²)

Φ = 1.575 × 10-4 Wb

For the final magnetic field, Φ = BA

Φ = (0.3 T)(17.5 × 10-4 m²)

Φ = 5.25 × 10-5 Wb

Therefore, ΔΦ = 1.05 × 10-4 Wb

Δt = 11.7 × 10-3 s

ε = ΔΦ/Δt

ε = (1.05 × 10-4 Wb)/(11.7 × 10-3 s)

ε = - 8.97 V

Therefore, the magnitude of the induced current (in Amperes) in the coil is 2.49 A (approx).

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DEPARTMENT OF PHYSICS NO. 3: R. (12 POINTS) A projectile is launched from the origin with an initial velocity 3 = 207 + 20. m/s. Find the (a) (2 points) initial projection angle, (b) (2 points) velocity vector of the projectile after 3 seconds of launching (c) (3 points) position vector of the projectile after 3 seconds of launching, (d) (2 points) time to reach the maximum height, (e) (1 point) time of flight (1) (2 points) maximum horizontal range reached.

Answers

A projectile is launched from the origin with an initial velocity 3 = 207 + 20. m/s. Therefore :

(a) The initial projection angle is 53.13°.

(b) The velocity vector of the projectile after 3 seconds of launching is (20cos(53.13), 20sin(53.13)) = (14.24, 14.14) m/s.

(c) The position vector of the projectile after 3 seconds of launching is (14.243, 14.143) = (42.72, 42.42) m.

(d) The time to reach the maximum height is 1.5 seconds.

(e) The time of flight is 3 seconds.

(f) The maximum horizontal range reached is 76.6 meters.

Here are the steps involved in solving for each of these values:

(a) The initial projection angle can be found using the following equation:

tan(Ф) = [tex]v_y/v_x[/tex]

where [tex]v_y[/tex] is the initial vertical velocity and [tex]v_x[/tex] is the initial horizontal velocity.

In this case, [tex]v_y[/tex] = 20 m/s and [tex]v_x[/tex] = 20 m/s. Therefore, Ф = [tex]\tan^{-1}\left(\frac{20}{20}\right)[/tex] = 53.13°.

(b) The velocity vector of the projectile after 3 seconds of launching can be found using the following equation:

v(t) = v₀ + at

where v(t) is the velocity vector at time t, v₀ is the initial velocity vector, and a is the acceleration vector.

In this case, v₀ = (20cos(53.13), 20sin(53.13)) and a = (0, -9.8) m/s². Therefore, v(3) = (14.24, 14.14) m/s.

(c) The position vector of the projectile after 3 seconds of launching can be found using the following equation:

r(t) = r₀ + v₀t + 0.5at²

where r(t) is the position vector at time t, r₀ is the initial position vector, v0 is the initial velocity vector, and a is the acceleration vector.

In this case, r₀ = (0, 0) and v₀ = (14.24, 14.14) m/s. Therefore, r(3) = (42.72, 42.42) m.

(d) The time to reach the maximum height can be found using the following equation:

v(t) = 0

where v(t) is the velocity vector at time t.

In this case, v(t) = (0, -9.8) m/s. Therefore, t = 1.5 seconds.

(e) The time of flight can be found using the following equation:

t = 2v₀ / g

where v₀ is the initial velocity and g is the acceleration due to gravity.

In this case, v₀ = 20 m/s and g = 9.8 m/s². Therefore, t = 3 seconds.

(f) The maximum horizontal range reached can be found using the following equation:

R = v² / g

where R is the maximum horizontal range, v is the initial velocity, and g is the acceleration due to gravity.

In this case, v = 20 m/s and g = 9.8 m/s². Therefore, R = 76.6 meters.

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A platinum cube of mass 4.4 kg attached to a spring with spring constant 7.2 N/m is oscillating back and forth and reaches a maximum speed of 3.3 m/s. What is the amplitude of the oscillation of the cube in meters? Ignore friction between the cube and the level surface on which it is oscillating.

Answers

The amplitude of the oscillation of the platinum cube is approximately 2.578 meters.

To find the amplitude of the oscillation, we can use the equation for the maximum velocity of an object undergoing simple harmonic motion:

v_max = Aω,

where:

v_max is the maximum velocity,A is the amplitude of the oscillation, andω is the angular frequency.

The angular frequency can be calculated using the equation:

ω = √(k/m),

where:

k is the spring constant, andm is the mass of the cube.

Given:

v_max = 3.3 m/s,k = 7.2 N/m, andm = 4.4 kg.

Let's substitute these values into the equations to find the amplitude:

ω = √(k/m) = √(7.2 N/m / 4.4 kg) ≈ √1.6364 ≈ 1.28 rad/s.

Now we can find the amplitude:

v_max = Aω,

3.3 m/s = A * 1.28 rad/s.

Solving for A:

A = 3.3 m/s / 1.28 rad/s ≈ 2.578 m.

Therefore, the amplitude of the oscillation is approximately 2.578 meters.

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A woman is standing on a bathroom scale in an elevator that is not moving. The balance reads 500 N. The elevator then moves downward at a constant speed of 5 m/s. What is the reading on the scale while the elevator is descending at constant speed?
d. 500N
e. 750N
b. 250N
c. 450N
a. 100N
Two point-shaped masses m and M are separated by a distance d. If the separation d remains fixed and the masses are increased to the values ​​3m and 3M respectively, how will the gravitational force between them change?
d. The force will be nine times greater.
b. The force will be reduced to one ninth.
e. It is impossible to determine without knowing the numerical values ​​of m, M, and d.
c. The force will be three times greater.
a. The force will be reduced to one third.

Answers

The reading on the scale while the elevator is descending at a constant speed is 500N (d). The gravitational force between the masses will be nine times greater when the masses are increased to 3m and 3M (d).

When the elevator is not moving, the reading on the scale is 500N, which represents the normal force exerted by the floor of the elevator on the woman. This normal force is equal in magnitude and opposite in direction to the gravitational force acting on the woman due to her weight.

When the elevator moves downward at a constant speed of 5 m/s, it means that the elevator and everything inside it, including the woman, are experiencing the same downward acceleration. In this case, the woman and the scale are still at rest relative to each other because the downward acceleration cancels out the gravitational force.

As a result, the reading on the scale remains the same at 500N. This is because the normal force provided by the scale continues to balance the woman's weight, preventing any change in the scale reading.

Therefore, the reading on the scale while the elevator is descending at a constant speed remains 500N, which corresponds to option d. 500N.

Regarding the gravitational force between the point-shaped masses, according to Newton's law of universal gravitation, the force between two masses is given by:

F = G × (m1 × m2) / r²,

where

F is the gravitational forceG is the gravitational constantm1 and m2 are the massesr is the separation distance between the masses

In this case, the separation distance d remains fixed, but the masses are increased to 3m and 3M. Plugging these values into the equation, we get:

New force (F') = G × (3m × 3M) / d² = 9 × (G × m × M) / d² = 9F,

where F is the original force between the masses.

Therefore, the gravitational force between the masses will be nine times greater when the masses are increased to 3m and 3M, which corresponds to option d. The force will be nine times greater.

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Determine the maximum vertical height h which the rollercoaster will reach on the second slope. Include an FBD for the rollercoaster while it is ascending (going up) the slope on the right. Use conservation of energy.

Answers

To determine the maximum vertical height the rollercoaster will reach on the second slope, we can use the principle of conservation of energy.  The rollercoaster will not reach any additional height on the second slope.

Using the principle of conservation of energy, we equate the initial kinetic energy of the rollercoaster to the final potential energy at the maximum height. We assume negligible energy losses due to friction or air resistance.

1. Initial kinetic energy:

The rollercoaster's initial kinetic energy is given by

K = 1/2 * m * v^2, where

m is the mass of the rollercoaster  

v is its initial velocity.

2. Final potential energy:

At the maximum height, the rollercoaster's potential energy is given by

P = m * g * h, where

m is the mass

g is the acceleration due to gravity

h is the height.

Since the rollercoaster starts at the top of the first slope, we can consider its initial kinetic energy to be zero since it comes to rest momentarily before ascending the second slope. Therefore, we have:

0 = m * g * h

Solving for h, we find that the maximum vertical height the rollercoaster will reach on the second slope is h = 0.

In other words, the rollercoaster will not reach any additional height on the second slope.

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The density of glycerin is 20 g/cm³ at 20 °C. Find the density of glycerin at 60 °C. The volume coefficient of glycerin is 5.1 x 10-4 °C-¹. A) 19.6 g/cm³ B 21.2 g/cm³ C 20.12 g/cm³ D 20 g/cm³

Answers

The correct option is D) 20 g/cm³.

The volume coefficient of glycerin is 5.1 x 10-4 °C-¹.

The temperature difference is 40°C (60°C - 20°C).

We can use the formula for calculating thermal expansion to calculate the new volume of glycerin.ΔV = V₀αΔT

Where, ΔV is the change in volume V₀ is the initial volume α is the volume coefficient ΔT is the temperature difference

V₀ = m/ρ₀

where m is the mass of the glycerin and ρ₀ is the density of glycerin at 20°C.

Now, we can substitute the values into the formula for calculating ΔV.ΔV = (m/ρ₀) α ΔT

Now, we can calculate the new volume of glycerin at 60°C.V₁ = V₀ + ΔV

Where V₁ is the new volume at 60°C, and V₀ is the initial volume at 20°C.ρ = m/V₁

Now, we can calculate the density of glycerin at 60°C.

ρ = m/V₁ρ = m/(V₀ + ΔV)

ρ = m/[m/ρ₀ + (m/ρ₀) α ΔT]ρ = 1/[1/ρ₀ + α ΔT]

ρ = 1/[1/20 + (5.1 x 10-4)(40)]

ρ = 1/[1/20 + 0.0204]

ρ = 1/[0.0504]

ρ = 19.84 g/cm³

Therefore, the density of glycerin at 60°C is 19.84 g/cm³, which rounds off to 19.8 g/cm³ (approximately).

Hence, the correct option is D) 20 g/cm³.

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Two identical diverging lenses are separated by 15.1cm. The focal length of each lens is -7.81cm. An object is located 3.99cm to the left of the lens that is on the left. Determine the final image distance relative to the lens on the right.

Answers

Given the following conditionsTwo identical diverging lenses separated by 15.1cm.

The focal length of each lens is -7.81cm.

An object is located 3.99cm to the left of the lens that is on the left.

The image formed is virtual and erect as both the lenses are diverging lenses.

As the final image distance relative to the lens on the right is to be determined, it is easier to calculate it if the image distance relative to the left lens is found first.

Using the lens formula,

1/f = 1/v - 1/u

where,f is the focal length of the lens

u is the distance of the object from the lens

v is the distance of the image from the lens.

The object distance from the lens,

u = -3.99 cm (since it is on the left of the lens, it is taken as negative).

The focal length of the lens,

f = -7.81cm.

The image distance,

v = 1/f + 1/u

= 1/-7.81 - 1/-3.99

= -0.413 cm

As the image is virtual and erect, its distance from the lens is taken as positive.

Hence, the image is at a distance of 0.413cm from the left lens.

Now, using the formula for the combination of thin lenses,

1/f = 1/f₁ + 1/f₂ - d/f₁f₂

where,d is the distance between the two lenses

f₁ is the focal length of the first lens

f₂ is the focal length of the second lens.

Both lenses are identical and have the same focal length,

f₁ = f₂

= -7.81 cm.

The distance between the lenses,

d = 15.1 cm.

Substituting the values,

1/f = 1/-7.81 + 1/-7.81 - 15.1/-7.81×-7.81

= -0.258 cm⁻¹

The image distance relative to the lens on the right,

v₂ = f / (1/f - 2/f - d)

= -7.81 / (1/-0.258 - 2/-7.81 - 15.1/-7.81×-7.81)

= -3.33cm

Therefore, the final image distance relative to the lens on the right is -3.33cm.

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Q11 A square with a mass and length L has a moment of inertia of lo when rotating about an axis perpendicular to its surface as show (left image). A mass M is attached to one corner of the square. What is the new moment of inertia about the same axis? M M22 A. lot بت 4 M22 L

Answers

The moment of inertia of a square with a mass and length L about an axis perpendicular to its surface is given by lo. When a mass M is attached to one corner of the square, the new moment of inertia about the same axis is different.

The correct answer to the question is not provided in the given options, as the new moment of inertia depends on the position and distribution of the added mass.

To determine the new moment of inertia when a mass M is attached to one corner of the square, we need to consider the distribution of mass and the axis of rotation. The added mass will affect the overall distribution of mass and thus change the moment of inertia.

However, the specific details regarding the location and distribution of the added mass are not provided in the question. Therefore, it is not possible to determine the new moment of inertia without this information. None of the options A, B, or any other option provided in the question can be considered the correct answer.

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An 80 kg crate is being pushed across a floor with a force of 254.8 N. If μkμk= 0.2, find the acceleration of the crate.

Answers

With a force of 254.8 N and a coefficient of kinetic friction of 0.2, the crate's acceleration is found to be approximately 1.24 m/s².

To find the acceleration of the crate, we can apply Newton's second law of motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration (F = ma). In this case, the force pushing the crate is given as 254.8 N.

The force of friction opposing the motion of the crate is the product of the coefficient of kinetic friction (μk) and the normal force (N). The normal force is equal to the weight of the crate, which can be calculated as the mass (80 kg) multiplied by the acceleration due to gravity (9.8 m/s²).

The formula for the force of friction is given by f = μkN. Substituting the values, we get f = 0.2 × (80 kg × 9.8 m/s²).

The net force acting on the crate is the difference between the applied force and the force of friction: Fnet = 254.8 N - f.

Finally, we can calculate the acceleration using Newton's second law: Fnet = ma. Rearranging the equation, we have a = Fnet / m. Substituting the values, we get a = (254.8 N - f) / 80 kg.

By evaluating the expression, we find that the acceleration of the crate is approximately 1.24 m/s². This means that for every second the crate is pushed, its velocity will increase by 1.24 meters per second.

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A skater spins at an initial angular velocity of 11 rads/s with his arms outstretched. The skater then lowers his arms, thereby decreasing his moment of inertia by a factor 5. What is the skater's final angular velocity? Assume that any friction between the skater's skates and the ice is negligible.

Answers

The skater's final angular velocity is 55 rad/s.

We can apply the principle of conservation of angular momentum to solve this problem. According to this principle, the initial and final angular momentum of the skater will be equal.

The formula for angular momentum is given by:

L = I * ω

where

L is the angular momentum,

I is the moment of inertia, and

ω is the angular velocity.

The skater starts with an angular velocity of 11 rad/s and his arms are outstretched. [tex]I_i_n_i_t_i_a_l[/tex] will be used to represent the initial moment of inertia.

The skater's moment of inertia now drops by a factor of 5 as he lowers his arms. Therefore, [tex]I_f_i_n_a_l[/tex]= [tex]I_i_n_i_t_i_a_l[/tex] / 5 can be used to express the final moment of inertia.

According to the conservation of angular momentum:

[tex]L_i=L_f[/tex]     (where i= initial, f= final)

[tex]I_i *[/tex]ω[tex]_i[/tex] = I[tex]_f[/tex] *ω[tex]_f[/tex]

Substituting the given values:

[tex]I_i[/tex]* 11 = ([tex]I_i[/tex] / 5) * ω_f

11 = ω[tex]_f[/tex] / 5

We multiply both the sides by 5.

55 = ω[tex]_f[/tex]

Therefore, the skater's final angular velocity is 55 rad/s.

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Example 8 A planet orbits a star in a year of length 4.37 x 10's, in a nearly circular orbit of radius 2.94 x 1011 m. With respect to the star, determine (a) the angular speed of the planet, (b) the tangential speed of the planet, and (c) the magnitude of the planet's centripetal acceleration. (a) Number Units m m (b) Number Units m/s (c) Number Units m/ s2

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(a) The angular speed of the planet is approximately 0.144 rad/s.

(b) The tangential speed of the planet is approximately 1.27 x 10⁴ m/s.

(c) The magnitude of the planet's centripetal acceleration is approximately 5.50 x 10⁻³ m/s².

(a) The angular speed of an object moving in a circular path is given by the equation ω = 2π/T, where ω represents the angular speed and T is the time period. In this case, the time period is given as 4.37 x 10⁶ s, so substituting the values, we have ω = 2π/(4.37 x 10⁶) ≈ 0.144 rad/s.

(b) The tangential speed of the planet can be calculated using the formula v = ωr, where v represents the tangential speed and r is the radius of the orbit. Substituting the given values, we get v = (0.144 rad/s) × (2.94 x 10¹¹ m) ≈ 1.27 x 10⁴ m/s.

(c) The centripetal acceleration of an object moving in a circular path is given by the equation a = ω²r. Substituting the values, we get a = (0.144 rad/s)² × (2.94 x 10¹¹ m) ≈ 5.50 x 10⁻³ m/s².

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What is the voltage of a battery that will charge a 2.0 μF capacitor to ± 54 μC?

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The voltage of a battery that will charge a 2.0 μF capacitor to ± 54 μC is 54 V. The capacitance formula is Q = CV where Q is the charge stored in the capacitor, C is the capacitance of the capacitor and V is the voltage across the capacitor.

The charge of a capacitor is given as Q = ±54 μC, and the capacitance of the capacitor is given as C = 2.0 μF. Therefore, the formula can be rearranged to solve for voltage as follows:Q = CV ⇒ V = Q/C

Since the charge is ±54 μC and the capacitance is 2.0 μF, thenV = ±54 μC/2.0 μFV = ±27 VThe voltage across the capacitor is either 27 V or -27 V.

Thus, the voltage of a battery that will charge a 2.0 μF capacitor to ± 54 μC is 54 V.

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Final answer:

The battery required to charge a 2.0 μF capacitor to ± 54 μC will need to provide a voltage of 27 volts. This calculation is based on the formula Q=CV.

Explanation:

The voltage of a battery used to charge a capacitor can be determined using the formula Q=CV where:

Q is the charge in Coulombs (C), C is the capacitance in farads (F), and V is the voltage in Volts (V).

Given that C = 2.0 μF and the absolute Q = 54 μC, we can rearrange the formula to solve for V:

V = Q/C

This gives us V = 54 μC/2.0 μF = 27 volts.

Therefore, a battery providing 27 volts will charge a 2.0 μF capacitor to ± 54 μC.

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For the following questions, you may use any resources you wish to answer them. You must write your solutions by hand, cite all your references, and show all your calculations [a] Write a calculation-based question appropriate for this study guide about the deformation in tension of a biological substance whose Young's modulus is given in the OpenStax College Physics textbook, if its length changes by X percent. Then answer it. Your solution should be significant to three figures. Y = 3.301 W=1301 [b] In Example 5.5 (Calculating Force Required to Deform) of Chapter 5.3 (Elasticity: Stress and Strain) of the OpenStax College Physics textbook, replace the amount the nail bends with Y micrometers. Then solve the example, showing your work [c] In Example 5.6 (Calculating Change in Volume) of that same chapter, replace the depth with w meters. Find out the force per unit area at that depth, and then solve the example. Cite any sources you use and show your work. Your answer should be significant to three figures.

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Answer:

a.) A biological substance with Young's modulus of 3.301 GPa has a tensile strain of 1.301 if its length is increased by 1301%.

b.) The force required to bend a nail by 100 micrometers is 20 N.

c.) The stress at a depth of 1000 meters is 10^8 Pa, which is equivalent to a pressure of 100 MPa.

Explanation:

a.) The tensile strain in the substance is given by the equation:

strain = (change in length)/(original length)

In this case, the change in length is X = 1301% of the original length.

Therefore, the strain is:

strain = (1301/100) = 1.301

The Young's modulus is a measure of how much stress a material can withstand before it deforms. In this case, the Young's modulus is Y = 3.301 GPa. Therefore, the stress in the substance is:

stress = (strain)(Young's modulus) = (1.301)(3.301 GPa) = 4.294 GPa

The stress is the force per unit area. Therefore, the force required to deform the substance is:

force = (stress)(area) = (4.294 GPa)(area)

The area is not given in the problem, so the force cannot be calculated. However, the strain and stress can be calculated, which can be used to determine the amount of deformation that has occurred.

b.) The force required to bend the nail is given by the equation:

force = (Young's modulus)(length)(strain)

In this case, the Young's modulus is Y = 200 GPa, the length of the nail is L = 10 cm, and the strain is ε = 0.001.

Therefore, the force is:

force = (200 GPa)(10 cm)(0.001) = 20 N

The force of 20 N is required to bend the nail by 100 micrometers.

c.) The force per unit area at a depth of w = 1000 meters is given by the equation:

stress = (weight density)(depth)

In this case, the weight density of water is ρ = 1000 kg/m^3, and the depth is w = 1000 meters.

Therefore, the stress is:

stress = (1000 kg/m^3)(1000 m) = 10^8 Pa

The stress of 10^8 Pa is equivalent to a pressure of 100 MPa.

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Is it possible for two objects to be in thermal equilibrium if they are not in contact with each other? Explain.

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It is not possible for two objects to be in thermal equilibrium if they are not in contact with each other. Thermal equilibrium occurs when two objects reach the same temperature and there is no net flow of heat between them. Heat is the transfer of thermal energy from a hotter object to a colder object.

When two objects are in contact with each other, heat can be transferred between them through conduction, convection, or radiation. Conduction is the transfer of heat through direct contact, convection is the transfer of heat through the movement of fluids, and radiation is the transfer of heat through electromagnetic waves.

If two objects are not in contact with each other, there is no medium for heat to transfer between them.

Therefore, they cannot reach the same temperature and be in thermal equilibrium. Even if the objects are at the same temperature initially, without any means of heat transfer, their temperatures will not change and they will not be in thermal equilibrium.

For example, let's consider two metal blocks, each initially at a temperature of 150 degrees Celsius. If the blocks are not in contact with each other and there is no medium for heat transfer, they will remain at 150 degrees Celsius and not reach thermal equilibrium.

In conclusion, for two objects to be in thermal equilibrium, they must be in contact with each other or have a medium through which heat can be transferred.

Without contact or a medium for heat transfer, the objects cannot reach the same temperature and therefore cannot be in thermal equilibrium.

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