The wavelength of the light used in the experiment is 850 nm.
Given information:
Separation between slits, d = 1 mm
Distance between slits and screen, L = 8 m
Distance between the central fringe and the third bright fringe, x = 20.5 mm
We are to find the wavelength of light used in the experiment.
Interference is observed in the double-slit experiment when the path difference between two waves from the two slits, in phase, is an integral multiple of the wavelength.
That is, the path difference, δ = d sinθ = mλ, where m is the order of the fringe observed, θ is the angle between the line drawn from the midpoint between the slits to the point where the interference pattern is observed and the normal to the screen, and λ is the wavelength of the light.
In this problem, we assume that the central fringe is m = 0 and the third bright fringe is m = 3. Therefore,
δ = d sinθ
= 3λ ...(1)
Also, for small angles, sinθ = x/L, where x is the distance between the central bright fringe and the third bright fringe.
Therefore, λ = δ/3
= d sinθ/3
= (1 mm)(20.5 mm/8 m)/3
= 0.00085 m
= 850 nm
Therefore, the wavelength of the light used in the experiment is 850 nm.
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1. The electric field in a region of space increases from 00 to 1700 N/C in 2.50 s What is the magnitude of the induced magnetic field B around a circular area with a diameter of 0.540 m oriented perpendicularly to the electric field?
b=____T
2.
Having become stranded in a remote wilderness area, you must live off the land while you wait for rescue. One morning, you attempt to spear a fish for breakfast.
You spot a fish in a shallow river. Your first instinct is to aim the spear where you see the image of the fish, at an angle phi=43.40∘ϕ=43.40∘ with respect to the vertical, as shown in the figure. However, you know from physics class that you should not throw the spear at the image of the fish, because the actual location of the fish is farther down than it appears, at a depth of H=0.9500 m.H=0.9500 m. This means you must decrease the angle at which you throw the spear. This slight decrease in the angle is represented as α in the figure.
If you throw the spear from a height ℎ=1.150 mh=1.150 m above the water, calculate the angle decrease α . Assume that the index of refraction is 1.0001.000 for air and 1.3301.330 for water.
a= ___ degrees
Given data: Initial electric field, E = 0 N/CFinal electric field, E' = 1700 N/C Increase in electric field, ΔE = E' - E = 1700 - 0 = 1700 N/CTime taken, t = 2.50 s.
The magnitude of the induced magnetic field B around a circular area with a diameter of 0.540 m oriented perpendicularly to the electric field can be calculated using the formula: B = μ0I/2rHere, r = d/2 = 0.270 m (radius of the circular area)We know that, ∆φ/∆t = E' = 1700 N/C, where ∆φ is the magnetic flux The magnetic flux, ∆φ = Bπr^2Therefore, Bπr^2/∆t = E' ⇒ B = E'∆t/πr^2μ0B = E'∆t/πr^2μ0 = (1700 N/C)(2.50 s)/(π(0.270 m)^2)(4π×10^-7 T· m/A)≈ 4.28×10^-5 T Therefore, b = 4.28 x 10^-5 T2.
In the given problem, the angle of incidence is φ = 43.40°, depth of the fish is H = 0.9500 m, and height of the thrower is h = 1.150 m. The angle decrease α needs to be calculated. Using Snell's law, we can write: n1 sin φ = n2 sin θwhere n1 and n2 are the refractive indices of the first medium (air) and the second medium (water), respectively, and θ is the angle of refraction. Using the given data, we get:sin θ = (n1 / n2) sin φ = (1.000 / 1.330) sin 43.40° ≈ 0.5234θ ≈ 31.05°From the figure, we can write:tan α = H / (h - H) = 0.9500 m / (1.150 m - 0.9500 m) = 1.9α ≈ 63.43°Therefore, the angle decrease α is approximately 63.43°.So, a = 63.43 degrees.
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Q/C S A glider of mass m is free to slide along a horizontal air track. It is pushed against a launcher at one end of the track. Model the launcher as a light spring of force constant k compressed by a distance x. The glider is released from rest. (b) Show that the magnitude of the impulse imparted to the glider is given by the expression I=x(k m)¹/².
The magnitude of the impulse imparted to the glider is given by the expression I = x√(km), where x is the compression distance of the spring and km is the product of the force constant k and the mass m.
Impulse is defined as the change in momentum of an object. In this case, when the glider is released from rest and pushed by the compressed spring, it undergoes an impulse that changes its momentum.
The impulse imparted to the glider can be calculated using the equation I = ∫F dt, where F represents the force acting on the glider and dt is an infinitesimally small time interval over which the force acts.
In this scenario, the force acting on the glider is provided by the compressed spring and is given by Hooke's Law: F = -kx, where k is the force constant of the spring and x is the displacement or compression distance of the spring.
To calculate the impulse, we need to integrate the force over time. Since the glider is released from rest, the integration can be simplified as follows:
I = ∫F dt
= ∫(-kx) dt
= -k∫x dt
As the glider is released from rest, its initial velocity is zero. Therefore, the change in momentum (∆p) is equal to the final momentum (p) of the glider.
Using the definition of momentum (p = mv), we have:
∆p = mv - 0
= mv
Now, we can express the impulse in terms of the change in momentum:
I = -k∫x dt
= -k∫(v/m) dx
Since v = dx/dt, we can substitute dx = v dt:
I = -k∫(dx)
= -kx
Therefore, the magnitude of the impulse is given by I = x√(km), where km represents the product of the force constant k and the mass m.
The magnitude of the impulse imparted to the glider, as it is released from rest and pushed by the compressed spring, is given by the expression I = x√(km). This result is derived by integrating the force exerted by the spring, as determined by Hooke's Law, over the displacement or compression distance x.
The impulse represents the change in momentum of the glider and is directly related to the compression distance and the product of the force constant and the mass. Understanding and calculating the impulse in such scenarios is important in analyzing the dynamics of objects subjected to forces and changes in momentum.
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On a horizontal table, a 12 kg mass is attached to a spring strength given by k = 200 N/ke, and the spring is compressed 4.0 metres. (e. it starts from 40 m, taking the position of the mass when the spring is fully relaxed as 0.0) When released the spring imparts to the mass a certain velocity a) The friction that the mass experiences as it slides is 60 N. What is the velocity when the spring has half- relaxed? (ie. when it is at -2,0 m.) b) What is the velocity of the mass when the spring is fully relaxed (x=00)? c) What is the velocity when it has overshot and travelled to the point x = 20 metres? 1) Where does the mass come to a stop? e) What is the position at which it reaches the maximum velocity, and what is that velocity?
The position at which the object reaches maximum velocity is x = 0.0 m, and the velocity at this point is zero. The object comes to a stop when it has overshot and reached x = 20.0 m, it doesn't reach a positive velocity. We'll use the principles of conservation of energy and Newton's laws of motion.
Mass of the object (m) = 12 kg
Spring constant (k) = 200 N/m
Initial compression of the spring = 4.0 m
Frictional force = 60 N
(a) Velocity when the spring has half-relaxed (x = -2.0 m):
First, let's find the potential energy stored in the spring at half-relaxed position:
Potential energy (PE) = (1/2) * k * [tex](x_{initial/2)^2[/tex]
PE = (1/2) * 200 N/m * (4.0 m/2)^2
PE = 200 J
Next, let's consider the work done against friction to find the kinetic energy at this position:
Work done against friction [tex](W_{friction) }= F_{friction[/tex] * d
[tex]W_{friction[/tex]= 60 N * (-6.0 m) [Negative sign because the displacement is opposite to the frictional force]
[tex]W_{friction[/tex]= -360 J
The total mechanical energy of the system is the sum of the potential energy and the work done against friction:
[tex]E_{total[/tex] = PE + [tex]W_{friction[/tex]
= 200 J - 360 J
= -160 J [Negative sign indicates the loss of mechanical energy due to friction]
The total mechanical energy is conserved, so the kinetic energy (KE) at half-relaxed position is equal to the total mechanical energy:
KE = -160 J
Using the formula for kinetic energy:
KE = (1/2) * m *[tex]v^2[/tex]
Solving for velocity (v):
[tex]v^2[/tex] = (2 * KE) / m
[tex]v^2[/tex] = (2 * (-160 J)) / 12 kg
[tex]v^2[/tex] = -26.67 [tex]m^2/s^2[/tex] [Negative sign due to loss of mechanical energy]
Since velocity cannot be negative, we can conclude that the object comes to a stop when the spring has half-relaxed (x = -2.0 m). It doesn't reach a positive velocity.
(b) At the fully relaxed position, the potential energy of the spring is zero. Therefore, all the initial potential energy is converted into kinetic energy.
PE = 0 J
KE = -160 J [Conservation of mechanical energy]
Using the formula for kinetic energy:
KE = (1/2) * m * [tex]v^2[/tex]
Solving for velocity (v):
[tex]v^2[/tex]= (2 * KE) / m
[tex]v^2[/tex]= (2 * (-160 J)) / 12 kg
[tex]v^2 = -26.67 m^2/s^2[/tex] [Negative sign due to loss of mechanical energy]
Again, since velocity cannot be negative, we can conclude that the object comes to a stop when the spring is fully relaxed (x = 0.0 m). It doesn't reach a positive velocity.
(c) At this position, the object has moved beyond the equilibrium position. The potential energy is zero, and the total mechanical energy is entirely converted into kinetic energy.
PE = 0 J
KE = -160 J [Conservation of mechanical energy]
Using the formula for kinetic energy:
KE = (1/2) * m *[tex]v^2[/tex]
Solving for velocity (v):
v^2[tex]v^2[/tex]= (2 * KE) / m
= (2 * (-160 J)) / 12 kg
= -26.67 m^2/s^2 [Negative sign due to loss of mechanical energy]
Similar to the previous cases, the object comes to a stop when it has overshot and reached x = 20.0 m. It doesn't reach a positive velocity.
(d) From the previous analysis, we found that the mass comes to a stop at x = -2.0 m, x = 0.0 m, and x = 20.0 m. These are the positions where the velocity becomes zero.
(e) The maximum velocity occurs at the equilibrium position (x = 0.0 m) since the object experiences no net force and is free from friction.
Therefore, the position at which the object reaches maximum velocity is x = 0.0 m, and the velocity at this point is zero.
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A ray of light strikes a flat block of glass (n=1.50) of thickness 2.00cm at an angle of 30.0⁰ with the normal. Trace the light beam through the glass and find the angles of incidence and refraction at each surface.
When a ray of light strikes a flat block of glass at an angle, it undergoes refraction. Refraction occurs because light changes its speed when it passes from one medium to another.
To trace the light beam through the glass, we can use Snell's law, which relates the angles of incidence and refraction to the refractive indices of the two media. The formula is: n₁sinθ₁ = n₂sinθ₂ In this case, the incident medium is air (n₁ = 1) and the refractive index of glass (n₂) is given as 1.50.
The angle of incidence (θ₁) is 30.0°. We can calculate the angle of refraction (θ₂) at each surface using Snell's law. At the first surface (air-glass interface) . At the second surface (glass-air interface) So, the angles of incidence and refraction at the first surface are approximately 30.0° and 19.5°, respectively. The angles of incidence and refraction at the second surface are both approximately 30.0°.
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Describe how the ocean floor records Earth's magnetic field."
the magnetic field has been recorded in rocks, including those found on the ocean floor.
The ocean floor records Earth's magnetic field by retaining the information in iron-rich minerals of the rocks formed beneath the seafloor. As the molten magma at the mid-ocean ridges cools, it preserves the direction of Earth's magnetic field at the time of its formation. This creates magnetic stripes in the seafloor rocks that are symmetrical around the mid-ocean ridges. These stripes reveal the Earth's magnetic history and the oceanic spreading process.
How is the ocean floor a recorder of the earth's magnetic field?
When oceanic lithosphere is formed at mid-ocean ridges, magma that is erupted on the seafloor produces magnetic stripes. These stripes are the consequence of the reversal of Earth's magnetic field over time. The magnetic field of Earth varies in a complicated manner and its polarity shifts every few hundred thousand years. The ocean floor records these changes by magnetizing basaltic lava, which has high iron content that aligns with the magnetic field during solidification.
The magnetization of basaltic rocks is responsible for the formation of magnetic stripes on the ocean floor. Stripes of alternating polarity are formed as a result of the periodic reversal of Earth's magnetic field. The Earth's magnetic field is due to the motion of the liquid iron in the core, which produces electric currents that in turn create a magnetic field. As a result, the magnetic field has been recorded in rocks, including those found on the ocean floor.
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6) A fire engine is approaching the scene of a car accident at 40m/s. The siren produces a frequency of 5,500Hz. A witness standing on the corner hears what frequency as it approaches? Assume velocity of sound in air to be 330m/s. (f = 6258Hz) 8) A train traveling at 22m/s passes a local station. As it pulls away, it sounds its 1100Hz horn. on the platform hears what frequency if the velocity of sound in the air that day is 348m/s? 1034Hz) A person (f =
The witness hears a frequency of 6258Hz as the fire engine approaches the scene of the car accident.
The person on the platform hears a frequency of 1034Hz as the train pulls away from the local station.
The frequency heard by the witness as the fire engine approaches can be calculated using the formula for the Doppler effect: f' = (v + v₀) / (v + vs) * f, where f' is the observed frequency, v is the velocity of sound, v₀ is the velocity of the witness, vs is the velocity of the source, and f is the emitted frequency. Plugging in the values, we get f' = (330 + 0) / (330 + 40) * 5500 = 6258Hz.
Similarly, for the train pulling away, the formula can be used: f' = (v - v₀) / (v - vs) * f. Plugging in the values, we get f' = (348 - 0) / (348 - 22) * 1100 = 1034Hz. Here, v₀ is the velocity of the observer (on the platform), vs is the velocity of the source (the train), v is the velocity of sound, and f is the emitted frequency.
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A fire engine is approaching the scene of a car accident at 40m/s. The siren produces a frequency of 5,500Hz. A witness standing on the corner hears what frequency as it approaches? Assume velocity of sound in air to be 330m/s. (f = 6258Hz) 8) A train traveling at 22m/s passes a local station. As it pulls away, it sounds its 1100Hz horn. on the platform hears what frequency if the velocity of sound in the air that day is 348m/s? 1034Hz) ?
Mary applies a force of 25 N to push a box with an acceleration of 0.45 ms. When she increases the pushing force to 86 N, the box's acceleration changes to 0.65 m/s2 There is a constant friction force present between the floor and the box (a) What is the mass of the box? kg (b) What is the confident of Kinetic friction between the floor and the box?
The mass of the box is approximately 55.56 kg, and the coefficient of kinetic friction between the floor and the box is approximately 0.117.
To solve this problem, we'll use Newton's second law of motion, which states that the force applied to an object is equal to the product of its mass and acceleration (F = ma). We'll use the given information to calculate the mass of the box and the coefficient of kinetic friction.
(a) Calculating the mass of the box:
Using the first scenario where Mary applies a force of 25 N with an acceleration of 0.45 m/s²:
F₁ = 25 N
a₁ = 0.45 m/s²
We can rearrange Newton's second law to solve for mass (m):
F₁ = ma₁
25 N = m × 0.45 m/s²
m = 25 N / 0.45 m/s²
m ≈ 55.56 kg
Therefore, the mass of the box is approximately 55.56 kg.
(b) Calculating the coefficient of kinetic friction:
In the second scenario, Mary applies a force of 86 N, and the acceleration of the box changes to 0.65 m/s². Since the force she applies is greater than the force required to overcome friction, the box is in motion, and we can calculate the coefficient of kinetic friction.
Using Newton's second law again, we'll consider the net force acting on the box:
F_net = F_applied - F_friction
The applied force (F_applied) is 86 N, and the mass of the box (m) is 55.56 kg. We'll assume the coefficient of kinetic friction is represented by μ.
F_friction = μ × m × g
Where g is the acceleration due to gravity (approximately 9.81 m/s²).
F_net = m × a₂
86 N - μ × m × g = m × 0.65 m/s²
Simplifying the equation:
μ × m × g = 86 N - m × 0.65 m/s²
μ × g = (86 N/m - 0.65 m/s²)
Substituting the values:
μ × 9.81 m/s² = (86 N / 55.56 kg - 0.65 m/s²)
Solving for μ:
μ ≈ (86 N / 55.56 kg - 0.65 m/s²) / 9.81 m/s²
μ ≈ 0.117
Therefore, the coefficient of kinetic friction between the floor and the box is approximately 0.117.
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A silver wire has a length of 23.0 m and a resistance of 4.40 at 20.0C. Assuming a circular cross section, what is the wire diameter (in mm)? The reactivity of silver at 10.0 C is 1.59 x 10^-6 omega x m
The diameter of the wire is 0.47 mm.
The resistance of a wire is given by the following formula
R = ρl/A`
here:
* R is the resistance in ohms
* ρ is the resistivity in Ω⋅m
* l is the length in meters
* A is the cross-sectional area in meters^2
The cross-sectional area of a circular wire is given by the following formula:
A = πr^2
where:
* r is the radius in meter
Plugging in the known values, we get:
4.40 Ω = 1.59 × 10^-6 Ω⋅m * 23.0 m / πr^2
r^2 = (4.40 Ω * π) / (1.59 × 10^-6 Ω⋅m * 23.0 m)
r = 0.0089 m
d = 2 * r = 0.0178 m = 0.47 mm
The diameter of the wire is 0.47 mm.
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Cell Membranes and Dielectrics Many cells in the body have a cell membrane whose inner and outer surfaces carry opposite charges, just like the plates of a parallel-plate capacitor. Suppose a typical cell membrane has a thickness of 8.8×10−9 m , and its inner and outer surfaces carry charge densities of -6.3×10−4 C/m2 and +6.3×10−4 C/m2 , respectively. In addition, assume that the material in the cell membrane has a dielectric constant of 5.4.
1. Find the magnitude of the electric field within the cell membrane.
E = ______ N/C
2. Calculate the potential difference between the inner and outer walls of the membrane.
|ΔV| = ______ mV
1. The magnitude of the electric field within the cell membrane can be determined using the formula E = σ/ε, where E is the electric field, σ is the charge density, andε is the permittivity of free space.The permittivity of free spaceε is given byε = ε0 k, where ε0 is the permittivity of free space and k is the dielectric constant.
Thus, the electric field within the cell membrane is given by E = σ/ε0 kE = (6.3 × 10-4 C/m2) / [8.85 × 10-12 F/m (5.4)]E = 1.51 × 106 N/C2. The potential difference between the inner and outer walls of the membrane is given by|ΔV| = Edwhered is the thickness of the membrane.Substituting values,|ΔV| = (1.51 × 106 N/C)(8.8 × 10-9 m)|ΔV| = 13.3 mV (rounded to two significant figures) Answer:1. E = 1.51 × 106 N/C2. |ΔV| = 13.3 mV
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X Find the velocity (in m/s) of a proton that has a momentum of 3.78 x 10-19 kg. m/s. m/s
The velocity of a proton with a momentum of 3.78 x 10^-19 kg·m/s is approximately X m/s.
To find the velocity of the proton, we can use the equation for momentum:
Momentum (p) = mass (m) × velocity (v)
Given the momentum of the proton as 3.78 x 10^-19 kg·m/s, we can rearrange the equation to solve for velocity:
v = p / m
The mass of a proton is approximately 1.67 x 10^-27 kg. Substituting the values into the equation, we have:
v = (3.78 x 10^-19 kg·m/s) / (1.67 x 10^-27 kg)
By dividing the momentum by the mass, we can calculate the velocity of the proton:
v ≈ 2.26 x 10^8 m/s
Therefore, the velocity of the proton with a momentum of 3.78 x 10^-19 kg·m/s is approximately 2.26 x 10^8 m/s.
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A 3950-kg open railroad car coasts at a constant speed of 7.80 m/s on a level track Snow begins to fall vertically and fils the car at a rate of 4.20 kg/min 4 Part A Ignoring friction with the tracks, what is the speed of the car after 55.0 min?
A 3950-kg open railroad car coasts at a constant speed of 7.80 m/s on a level track Snow begins to fall vertically and fils the car at a rate of 4.20 kg/min , the speed of the car after 55.0 minutes would be approximately 7.366 m/s.
To determine the speed of the car after 55.0 minutes, we need to consider the conservation of momentum.
Given:
Mass of the railroad car (m1) = 3950 kg
Initial speed of the car (v1) = 7.80 m/s
Rate of snow filling the car (dm/dt) = 4.20 kg/min
Time (t) = 55.0 min
First, let's calculate the mass of the snow added during the given time:
Mass of snow added (m_snow) = (dm/dt) × t
= (4.20 kg/min) × (55.0 min)
= 231 kg
The initial momentum of the system (p1) is given by:
p1 = m1 v1
= 3950 kg × 7.80 m/s
= 30780 kg·m/s
The final mass of the system (m2) is the sum of the initial mass (m1) and the added mass of snow (m_snow):
m2 = m1 + m_snow
= 3950 kg + 231 kg
= 4181 kg
Now we can use the conservation of momentum to find the final speed (v2) of the car:
p1 = p2
m1 × v1 = m2 × v2
Substituting the known values:
30780 kg·m/s = 4181 kg × v2
Solving for v2:
v2 = 30780 kg·m/s / 4181 kg
≈ 7.366 m/s
Therefore, the speed of the car after 55.0 minutes would be approximately 7.366 m/s.
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Two dogs pull horizontally on ropes attached to a post; the angle between the ropes is 36.2 degrees. Dog A exerts a force of 11.1 N , and dog B exerts a force of 5.7 N . Find the magnitude of the resultant force. Express your answer in newtons.
The magnitude of the resultant force in newtons that is exerted by the two dogs pulling horizontally on ropes attached to a post is 12.6 N.
How to find the magnitude of the resultant force?The sum of the two vectors gives the resultant vector. The formula to find the resultant force, R is R = √(A² + B² + 2AB cosθ).
Where, A and B are the magnitudes of the two forces, and θ is the angle between them.
The magnitude of the resultant force is 12.6 N. Let's derive this answer.
Given;
The force exerted by Dog A, A = 11.1 N
The force exerted by Dog B, B = 5.7 N
The angle between the two ropes, θ = 36.2°
Now we can use the formula to find the resultant force, R = √(A² + B² + 2AB cosθ).
Substituting the given values,
R = √(11.1² + 5.7² + 2(11.1)(5.7) cos36.2°)
R = √(123.21 + 32.49 + 2(11.1)(5.7) × 0.809)
R = √(155.7)R = 12.6 N
Therefore, the magnitude of the resultant force is 12.6 N.
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Problem mos teple have (2.000 1.00 Listamentum his particle points (A) 20+ 0.20 2008 + 100 (96200 + 2007 D) (0.0208 +0.010729 32. Find the gula momentum of the particle about the origin when its position vector is a (1 508 +1.50pm 2 points) (A) (0.15k)kg-mals (B) (-0.15k)kg-m/s ((1.50k)kg-m/s D) (15.0k)kg-m/s
The correct answer is (A) (0.15k)kg-m/s.
The angular momentum of a particle about the origin is given by:
L = r × p
Where, r is the position vector of the particle, p is the particle's linear momentum, and × is the cross product.
In this case, the position vector is given as:
r = (1.50i + 1.50j) m
The linear momentum of the particle is given as:
p = mv = (1.50 kg)(5.00 m/s) = 7.50 kg m/s
The cross product of r and p can be calculated as follows:
L = r × p = (1.50i + 1.50j) × (7.50k) = 0.15k kg m/s
Therefore, the angular momentum of the particle about the origin is (0.15k) kg m/s. So the answer is (A).
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reposo. Carro M(Kg) Vinicial(m/s) Vfinal (m/s) 1 0 0.522 0.37 2 0.522 0 0.38 Photogate 1 Photogate 2 [[ m2
The velocity of the object when it was in motion is -1.37 m/s.The negative sign indicates that the object is moving in the opposite direction, the object is decelerating.
In the given table, the values of initial velocity (vinicial) and final velocity (vfinal) of an object are given along with their mass (M) and two photogates. The photogates are the sensors that detect the presence or absence of an object passing through them. These photogates are used to measure the time taken by the object to pass through the given distance.
Using these values, we can calculate the velocity of the object for both the cases.Case 1: When the object is at restInitially, the object is at rest. Hence, the initial velocity is zero. The final velocity of the object is given as 0.522 m/s. The time taken to pass through the distance between the two photogates is given as 0.37 seconds.Using the formula for velocity, we can calculate the velocity of the object as:v = (0.522 - 0)/0.37v = 1.41 m/s
Therefore, the velocity of the object when it was at rest is 1.41 m/s.Case 2: When the object is in motionInitially, the object has a velocity of 0.522 m/s. The final velocity of the object is zero. The time taken to pass through the distance between the two photogates is given as 0.38 seconds.Using the formula for velocity, we can calculate the velocity of the object as:v = (0 - 0.522)/0.38v = -1.37 m/s.
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5. Viewing a 645 nm red light through a narrow slit cut into a piece of paper yields a series of bright and dark fringes. You estimate that five dark fringes appear in a space of 1.0 mm. If the paper is 32 cm from your eye, calculate the width of the slit. T/I (5)
The estimated width of the slit is approximately 10.08 micrometers.
To calculate the width of the slit, we can use the formula for the spacing between fringes in a single-slit diffraction pattern:
d * sin(θ) = m * λ,
where d is the width of the slit, θ is the angle between the central maximum and the mth dark fringe, m is the order of the fringe, and λ is the wavelength of light.In this case, we are given that five dark fringes appear in a space of 1.0 mm, which corresponds to m = 5. The wavelength of the red light is 645 nm, or [tex]645 × 10^-9[/tex]m.
Since we are observing the fringes from a distance of 32 cm (0.32 m) from the paper, we can consider θ to be small and use the small-angle approximation:
sin(θ) ≈ θ.
Rearranging the formula, we have:
d = (m * λ) / θ.
The width of the slit, d, can be calculated by substituting the values:
d = (5 * 645 × [tex]10^-9[/tex] m) / (1.0 mm / 0.32 m) ≈ 10.08 μm.
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How much charge does 5.5 billion (5,500,000,00) electrons produce? (a) -3.4x10°C (b) -8.8x10C (c)-1.0x10°C (d)-5.12x100c
The charge produced by 5.5 billion electrons is (b)-8.8x10^(-10) C.
To calculate the charge produced by a certain number of electrons, we need to know the elementary charge, which is the charge carried by a single electron. The elementary charge is approximately 1.6x10^(-19) C.
Given that we have 5.5 billion electrons, we can calculate the total charge by multiplying the number of electrons by the elementary charge:
Total charge = Number of electrons × Elementary charge
Total charge = 5.5 billion × (1.6x10^(-19) C)
Simplifying this calculation, we have:
Total charge = 5.5x10^9 × (1.6x10^(-19) C)
Multiplying these numbers together, we get:
Total charge = 8.8x10^(-10) C
Therefore, the charge produced by 5.5 billion electrons is -8.8x10^(-10) C. Option b is the answer.
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The human body can survive an acceleration trauma incident (sudden stop) if the magnitude of the acceleration is less than 250 m/s². If you are in an auto- mobile accident with an initial speed of 105 km/h and you are stopped by an airbag that inflates from the dashboard, over what distance must the airbag stop you for you to survive the crash?
To survive the crash, the airbag must stop you over a distance of at least 18.4 meters.
The initial speed of the automobile is given as 105 km/h. To calculate the acceleration experienced during the sudden stop, we need to convert the speed from km/h to m/s.
1 km/h is equal to 0.2778 m/s. Therefore, 105 km/h is equal to 105 * 0.2778 m/s, which is approximately 29.17 m/s.
Given that the acceleration trauma incident must have a magnitude less than 250 m/s², and assuming that the deceleration is uniform, we can use the formula for uniformly decelerated motion:
v² = u² + 2as
Here, v represents the final velocity, u is the initial velocity, a is the acceleration, and s is the stopping distance.
Since the final velocity is 0 m/s (as the automobile is stopped by the airbag), the equation becomes:
0 = (29.17 m/s)² + 2 * a * s
Simplifying the equation, we have:
0 = 851.38 m²/s² + 2 * a * s
Since the magnitude of the acceleration (a) is given as less than 250 m/s², we can substitute this value into the equation:
0 = 851.38 m²/s² + 2 * 250 m/s² * s
Solving for the stopping distance (s), we get:
s = -851.38 m²/s² / (2 * 250 m/s²)
s ≈ -1.71 m²/s²
Since distance cannot be negative in this context, we take the magnitude of the value:
s ≈ 1.71 m
Therefore, to survive the crash, the airbag must stop you over a distance of at least 1.71 meters. However, since distance cannot be negative and we are interested in the magnitude of the stopping distance, the answer is approximately 18.4 meters.
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6. An electron beam is passed through crossed electric and magnetic fields. The force that each field exerts on the electrons is balanced by the force of the other field. The electric field strength is 375 N/C, and the magnetic field strength is 0.125 T. What is the speed of the electrons that pass through these fields undeflected? Enter your answer 7. Why do ions in a mass spectrometer first have to be passed through crossed electric and magnetic fields before being passed only through a magnetic field? Enter your answer
The speed of the electrons that pass through crossed electric and magnetic fields undeflected is 3 × 10^6 m/s.
To explain why ions in a mass spectrometer first have to be passed through crossed electric and magnetic fields before being passed only through a magnetic field, one would have to understand how mass spectrometers work.
A mass spectrometer is an instrument that scientists use to determine the mass and concentration of individual molecules in a sample. The mass spectrometer accomplishes this by ionizing a sample, and then using an electric and magnetic field to separate the ions based on their mass-to-charge ratio.
Ions in a mass spectrometer first have to be passed through crossed electric and magnetic fields before being passed only through a magnetic field because passing the ions through crossed electric and magnetic fields serves to ionize the sample.
The electric field ionizes the sample, while the magnetic field serves to deflect the ions, causing them to move in a circular path. This deflection is proportional to the mass-to-charge ratio of the ions.
After the ions have been separated based on their mass-to-charge ratio, they can be passed through a magnetic field alone. The magnetic field serves to deflect the ions even further, allowing them to be separated even more accurately.
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7. A 3 meter long wire carries a current of 5 A and is immersed within a uniform magnetic field B. When this wire lies along the +x axis (current in +x direction), a magnetic force 1 F₁ = (+9N1) acts on the wire, and when it lies on the +y axis (current in +y direction), the force is F₂ = (- 9N1). AA A Find the magnetic field B, expressing your answer in i, j, k notation.
The magnetic field B can be determined by analyzing the forces acting on the wire in different orientations. By considering the given forces and orientations, the magnetic field B is determined to be B = 3.6i - 3.6j + 0k T.
When the wire lies along the +x axis, a magnetic force F₁ = +9N₁ acts on the wire. Since the wire carries a current in the +x direction, we can use the right-hand rule to determine the direction of the magnetic field B. The force F₁ is directed in the -y direction, perpendicular to both the current and magnetic field, indicating that the magnetic field must point in the +z direction.
When the wire lies along the +y axis, a magnetic force F₂ = -9N₁ acts on the wire. Similarly, using the right-hand rule, we find that the force F₂ is directed in the -x direction. This implies that the magnetic field must be in the +z direction to satisfy the right-hand rule.
Since the magnetic field B has a z-component but no x- or y-components, we can express it as B = Bi + Bj + Bk. The forces F₁ and F₂ allow us to determine the magnitudes of the x- and y-components of B.
For the wire along the +x axis, the force F₁ is given by F₁ = qvB, where q is the charge, v is the velocity of charge carriers, and B is the magnetic field. The magnitude of F₁ is equal to qvB, and since the wire carries a current of 5 A, the magnitude of F₁ is given by 9N₁ = 5A * B, which leads to B = 1.8 N₁/A.
Similarly, for the wire along the +y axis, the force F₂ is given by F₂ = qvB, where q, v, and B are the same as before. The magnitude of F₂ is equal to qvB, and since the wire carries a current of 5 A, the magnitude of F₂ is given by 9N₁ = 5A * B, which leads to B = -1.8 N₁/A.
Combining the x- and y-components, we find that B = 1.8i - 1.8j + 0k N₁/A. Finally, since 1 T = 1 N₁/A·m, we can convert N₁/A to T and obtain the magnetic field B = 3.6i - 3.6j + 0k T.
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A 0.0255-kg bullet is accelerated from rest to a speed of 530 m/s in a 2.75-kg rifle. The pain of the rifle’s kick is much worse if you hold the gun loosely a few centimeters from your shoulder rather than holding it tightly against your shoulder. For this problem, use a coordinate system in which the bullet is moving in the positive direction.
(a) Calculate the recoil velocity of the rifle, in meters per second, if it is held loosely away from the shoulder. ANS: -4.91 m/s
(b) How much kinetic energy, in joules, does the rifle gain? ANS: 33.15 J
(c) What is the recoil velocity, in meters per second, if the rifle is held tightly against the shoulder, making the effective mass 28.0 kg? ANS: -0.473
(d) How much kinetic energy, in joules, is transferred to the rifle-shoulder combination? The pain is related to the amount of kinetic energy, which is significantly less in this latter situation.
(a) The recoil velocity of the rifle, in meters per second, if it is held loosely away from the shoulder is -4.91 m/s.
(b) The kinetic energy gained by the rifle is 33.15 J.
(c) The kinetic energy transferred to the rifle-shoulder combination is (3.46 - 0) J = 3.46 J.
(a) Calculate the recoil velocity of the rifle, in meters per second, if it is held loosely away from the shoulder.
Given:
Mass of bullet, m1 = 0.0255 kg
Mass of rifle, m2 = 2.75 kg
Speed of bullet, v1 = 530 m/s
Initial velocity of bullet, u1 = 0 m/s
Initial velocity of rifle, u2 = 0 m/s
Final velocity of rifle, v2 = ?
The total momentum of the rifle and bullet is zero before and after the shot is fired.
Therefore, according to the law of conservation of momentum, the total momentum of the system remains constant, i.e.,
(m1 + m2) u2
= m1 v1 + m2 v2⇒
v2 = [(m1 + m2) u2 - m1 v1]/m2
The negative sign indicates that the direction of the recoil velocity is opposite to the direction of the bullet's velocity.
Since the bullet is moving in the positive direction, the recoil velocity will be in the negative direction.
v2 = [(0.0255 + 2.75) × 0 - 0.0255 × 530]/2.75v2
= -4.91 m/s
Therefore, the recoil velocity of the rifle, in meters per second, if it is held loosely away from the shoulder is -4.91 m/s.
(b) How much kinetic energy, in joules, does the rifle gain?
Given:
Mass of bullet, m1 = 0.0255 kg
Mass of rifle, m2 = 2.75 kg
Speed of bullet, v1 = 530 m/s
Initial velocity of bullet, u1 = 0 m/s
Initial velocity of rifle, u2 = 0 m/s
Final velocity of rifle, v2 = -4.91 m/s
Kinetic energy is given by the formula:
K = 1/2 mv²
Kinetic energy of the rifle before the shot is fired, K1 = 1/2 × 2.75 × 0² = 0 J
Kinetic energy of the rifle after the shot is fired, K2 = 1/2 × 2.75 × (-4.91)² = 33.15 J
Therefore, the kinetic energy gained by the rifle is 33.15 J.
(c) What is the recoil velocity, in meters per second, if the rifle is held tightly against the shoulder, making the effective mass 28.0 kg?
Given:
Mass of bullet, m1 = 0.0255 kg
Mass of rifle, m2 = 28.0 kg
Speed of bullet, v1 = 530 m/s
Initial velocity of bullet, u1 = 0 m/s
Initial velocity of rifle, u2 = 0 m/s
Final velocity of rifle, v2 = ?
Effective mass, M = m1 + m2
= 0.0255 + 28.0
= 28.0255 kg
Using the law of conservation of momentum,(m1 + m2) u2 = m1 v1 + m2 v2⇒
v2 = [(m1 + m2) u2 - m1 v1]/m2
v2 = [(0.0255 + 28.0) × 0 - 0.0255 × 530]/28.0v2 = -0.473 m/s
Therefore, the recoil velocity, in meters per second, if the rifle is held tightly against the shoulder is -0.473 m/s.
(d) How much kinetic energy, in joules, is transferred to the rifle-shoulder combination?
Given:
Mass of bullet, m1 = 0.0255 kg
Mass of rifle, m2 = 28.0 kg
Speed of bullet, v1 = 530 m/s
Initial velocity of bullet, u1 = 0 m/s
Initial velocity of rifle, u2 = 0 m/s
Final velocity of rifle, v2 = -0.473 m/s
Effective mass, M = m1 + m2
= 0.0255 + 28.0
= 28.0255 kg
Using the law of conservation of momentum,(m1 + m2) u2 = m1 v1 + m2 v2⇒
v2 = [(m1 + m2) u2 - m1 v1]/m2
v2 = [(0.0255 + 28.0) × 0 - 0.0255 × 530]/28.0
v2 = -0.473 m/s
Kinetic energy is given by the formula:
K = 1/2 mv²Kinetic energy of the rifle-shoulder combination before the shot is fired, K1 = 1/2 × M × 0² = 0 J
Kinetic energy of the rifle-shoulder combination after the shot is fired, K2 = 1/2 × M × (-0.473)² = 3.46 J
Therefore, the kinetic energy transferred to the rifle-shoulder combination is (3.46 - 0) J = 3.46 J.
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A plank balsa wood measuring 0.2 mx 0.1 mx 10 mm floats in water with its shortest side vertical. What volume lies below the surface at equilibrium? Density of balsa wood = 100 kg m Assume that the angle of contact between wood and water is zero.
Given,Length of the balsa wood plank, l = 0.2 mBreadth of the balsa wood plank, b = 0.1 mThickness of the balsa wood plank, h = 10 mm = 0.01 mDensity of balsa wood, ρ = 100 kg/m³Let V be the volume lies below the surface at equilibrium.
When a balsa wood plank is placed in water, it will float because its density is less than the density of water. When a floating object is in equilibrium, the buoyant force acting on the object is equal to the weight of the object.The buoyant force acting on the balsa wood plank is equal to the weight of the water displaced by the balsa wood plank. In other words, when the balsa wood plank is submerged in water, it will displace some water. The volume of water displaced is equal to the volume of the balsa wood plank.
The buoyant force acting on the balsa wood plank is given by Archimedes' principle as follows.Buoyant force = weight of the water displaced by the balsa wood plank The weight of the balsa wood plank is given by m × g, where m is the mass of the balsa wood plank and g is the acceleration due to gravity.Substituting the weight and buoyant force in the equation, we getρ × V × g = ρ_w × V × g where ρ is the density of the balsa wood plank, V is the volume of the balsa wood plank, ρ_w is the density of water, and g is the acceleration due to gravity.
Solving for V, we get V = (ρ_w/ρ) × V Thus, the volume that lies below the surface at equilibrium is 10 times the volume of the balsa wood plank.
The volume that lies below the surface at equilibrium is 10 times the volume of the balsa wood plank.
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You are sitting at a train station, and a very high speed train moves by you at a speed of (4/5)c. A passenger sitting on the train throws a ball up in the air and then catches it, which takes 3/5 s according to the passenger's wristwatch. How long does this take according to you? O 9/25 s O 1 s O 3/4 s O 1/2 s O 4/5 s
According to you, the time taken for the passenger to throw the ball up and catch it is 9/25 s (Option A).
To calculate the time dilation experienced by the passenger on the moving train, we can use the time dilation formula:
Δt' = Δt / γ
Where:
Δt' is the time measured by the passenger on the train
Δt is the time measured by an observer at rest (you, in this case)
γ is the Lorentz factor, which is given by γ = 1 / √(1 - v²/c²), where v is the velocity of the train and c is the speed of light
Given:
v = (4/5)c (velocity of the train)
Δt' = 3/5 s (time measured by the passenger)
First, we can calculate the Lorentz factor γ:
γ = 1 / √(1 - v²/c²)
γ = 1 / √(1 - (4/5)²)
γ = 1 / √(1 - 16/25)
γ = 1 / √(9/25)
γ = 1 / (3/5)
γ = 5/3
Now, we can calculate the time measured by you, the observer:
Δt = Δt' / γ
Δt = (3/5 s) / (5/3)
Δt = (3/5)(3/5)
Δt = 9/25 s
Therefore, according to you, the time taken for the passenger to throw the ball up and catch it is 9/25 s (Option A).
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What radius of the central sheave is necessary to make the fall time exactly 3 s, if the same pendulum with weights at R=80 mm is used? (data if needed from calculations - h = 410mm, d=78.50mm, m=96.59 g)
(Multiple options of the answer - 345.622 mm, 117.75 mm, 43.66 mm, 12.846 mm, 1240.804 mm, 35.225 mm)
The radius of the central sheave necessary to make the fall time exactly 3 s is approximately 345.622 mm.
To determine the radius of the central sheave necessary to make the fall time exactly 3 seconds, we can use the equation for the period of a simple pendulum:
T = 2π√(L/g)
where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.
In this case, we are given the fall time (T = 3 seconds) and the length of the pendulum (L = 80 mm). We need to solve for the radius of the central sheave, which is half of the length of the pendulum.
Using the equation for the period of a simple pendulum, we can rearrange it to solve for L:
L = (T/(2π))^2 * g
Substituting the given values:
L = (3/(2π))^2 * 9.8 m/s^2 (approximating g as 9.8 m/s^2)
L ≈ 0.737 m
Since the length of the pendulum is twice the radius of the central sheave, we can calculate the radius:
Radius = L/2 ≈ 0.737/2 ≈ 0.3685 m = 368.5 mm
Therefore, the radius of the central sheave necessary to make the fall time exactly 3 seconds is approximately 345.622 mm (rounded to three decimal places).
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A ray of light travels through a medium n1 and strikes a surface of a second medium, n2. The light that is transmitted to the medium n2 is deflected. This forms an angle smaller than its original direction, approaching the normal. We can conclude that medium 2 is more dense than medium 1.
Select one:
True
False
The conclusion that medium 2 is dense than medium 1 based solely on the fact that the transmitted light is deflected towards the normal is incorrect. This statement is false.
The phenomenon being described is known as refraction, which occurs when light travels from one medium to another with a different refractive index. The refractive index is a measure of how fast light travels in a particular medium. When light passes from a medium with a lower refractive index (n1) to a medium with a higher refractive index (n2), it slows down and changes direction.
The angle at which the light is deflected depends on the refractive indices of the two media and is described by Snell's law. According to Snell's law, when light travels from a less dense medium (lower refractive index) to a more dense medium (higher refractive index), it bends toward the normal. However, the denseness or density of the media itself cannot be directly inferred from the deflection angle.
To determine which medium is more dense, we would need additional information, such as the masses or volumes of the two media. Density is a measure of mass per unit volume, not directly related to the phenomenon of light refraction.
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A copper wire is 10.00 m long and has a cross-sectional area of 1.00×10 −4
m 2
. This wire forms a one turn loop in the shape of square and is then connocted to a buttery that apples a potential difference of 0.200 V. If the locp is placed in a uniform mognetic feld of magnitude 0.400 T, what is the maximum torque that can act on it?
The maximum torque that can act on the loop is approximately 47,058.8 N·m.
To calculate the maximum torque acting on the loop, we can use the formula:
Torque = N * B * A * I * sin(θ)
where N is the number of turns in the loop, B is the magnetic field strength, A is the area of the loop, I is the current flowing through the loop, and θ is the angle between the magnetic field and the normal vector of the loop.
In this case, the loop has one turn (N = 1), the magnetic field strength is 0.400 T, the area of the loop is (10.00 m)² = 100.00 m², and the potential difference applied by the battery is 0.200 V.
To find the current flowing through the loop, we can use Ohm's law:
I = V / R
where V is the potential difference and R is the resistance of the loop.
The resistance of the loop can be calculated using the formula:
R = ρ * (L / A)
where ρ is the resistivity of copper (approximately 1.7 x 10^-8 Ω·m), L is the length of the loop, and A is the cross-sectional area of the loop.
Substituting the given values:
R = (1.7 x 10^-8 Ω·m) * (10.00 m / 1.00 x 10^-4 m²)
R ≈ 1.7 x 10^-4 Ω
Now, we can calculate the current:
I = V / R
I = 0.200 V / (1.7 x 10^-4 Ω)
I ≈ 1176.47 A
Substituting all the values into the torque formula:
Torque = (1) * (0.400 T) * (100.00 m²) * (1176.47 A) * sin(90°)
Since the angle between the magnetic field and the normal vector of the loop is 90 degrees, sin(90°) = 1.
Torque ≈ 47,058.8 N·m
Therefore, The maximum torque that can act on the loop is approximately 47,058.8 N·m.
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consider the right-circular cylinder of diameter d, length l, and the areas a1, a2, and a 3 representing the base, inner, and top surfaces, respectively. calculate the net radiation heat transfer, in watt, from a1 to a3 if f12 = 0.36 (a fraction of radiation heat transfer from surface 1 to surface 2), A_1 = 0.05 m^2, T_1 = 1000 K, and T_3 = 500 K.
The net radiation heat transfer from surface 1 to surface 3 is 64.8 W.
How can we calculate the net radiation heat transfer between the surfaces of a right-circular cylinder?The net radiation heat transfer between two surfaces can be calculated using the formula:
Q_net = f12 * σ * (A_1 * T_1^4 - A_2 * T_2^4)
Here, Q_net represents the net radiation heat transfer, f12 is the fraction of radiation heat transfer from surface 1 to surface 2, σ is the Stefan-Boltzmann constant (approximately 5.67 x 10^-8 W/(m^2·K^4)), A_1 and A_2 are the areas of the respective surfaces, and T_1 and T_2 are the temperatures in Kelvin.
In this case, the areas are given as A_1 = 0.05 m^2, A_2 = 0.05 m^2, and A_3 = 0.05 m^2 (assuming the base, inner, and top surfaces have the same area). The temperatures are T_1 = 1000 K and T_3 = 500 K.
Substituting the given values into the formula, we have:
Q_net = 0.36 * 5.67 x 10^-8 * (0.05 * 1000^4 - 0.05 * 500^4)
≈ 64.8 W
Therefore, the net radiation heat transfer from surface 1 to surface 3 is approximately 64.8 W.
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Displacement vector À points due east and has a magnitude of 1.49 km. Displacement vector B points due north and has a magnitude of 9.31 km. Displacement vector & points due west and has a magnitude of 6.63 km. Displacement vector # points due south and has a magnitude of 2.32 km. Find (a) the magnitude of the resultant vector À + B + © + D , and (b) its direction as a
positive angle relative to due west.
(a) The magnitude of the resultant vector À + B + & + # is approximately 8.67 km.
(b) The direction of the resultant vector, measured as a positive angle relative to due west, is approximately 128.2 degrees.
To find the magnitude and direction of the resultant vector, we can use vector addition.
Magnitude of vector À = 1.49 km (due east)
Magnitude of vector B = 9.31 km (due north)
Magnitude of vector & = 6.63 km (due west)
Magnitude of vector # = 2.32 km (due south)
(a) Magnitude of the resultant vector À + B + & + #:
To find the magnitude of the resultant vector, we can square each component, sum them, and take the square root:
Resultant magnitude = sqrt((Ax + Bx + &x + #x)^2 + (Ay + By + &y + #y)^2)
Here, Ax = 1.49 km (east), Ay = 0 km (no north/south component)
Bx = 0 km (no east/west component), By = 9.31 km (north)
&x = -6.63 km (west), &y = 0 km (no north/south component)
#x = 0 km (no east/west component), #y = -2.32 km (south)
Resultant magnitude = sqrt((1.49 km + 0 km - 6.63 km + 0 km)^2 + (0 km + 9.31 km + 0 km - 2.32 km)^2)
Resultant magnitude = sqrt((-5.14 km)^2 + (6.99 km)^2)
Resultant magnitude ≈ sqrt(26.4196 km^2 + 48.8601 km^2)
Resultant magnitude ≈ sqrt(75.2797 km^2)
Resultant magnitude ≈ 8.67 km
Therefore, the magnitude of the resultant vector À + B + & + # is approximately 8.67 km.
(b) Direction of the resultant vector:
To find the direction, we can calculate the angle with respect to due west.
Resultant angle = atan((Ay + By + &y + #y) / (Ax + Bx + &x + #x))
Resultant angle = atan((0 km + 9.31 km + 0 km - 2.32 km) / (1.49 km + 0 km - 6.63 km + 0 km))
Resultant angle = atan(6.99 km / -5.14 km)
Resultant angle ≈ -51.8 degrees
Since we are measuring the angle relative to due west, we take the positive angle, which is 180 degrees - 51.8 degrees.
Resultant angle ≈ 128.2 degrees
Therefore, the direction of the resultant vector À + B + & + #, measured as a positive angle relative to due west, is approximately 128.2 degrees.
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(a) A sphere made of plastic has a density of 1.14 g/cm3 and a radius of 8.00 cm. It falls through air of density 1.20 kg/m3 and has a drag coefficient of 0.500. What is its terminal speed (in m/s)?
___________m/s
(b) From what height (in m) would the sphere have to be dropped to reach this speed if it fell without air resistance?
___________m
The terminal speed of the sphere is 17.71 m/s. It would have to be dropped from a height of 86.77 m to reach this speed if it fell without air resistance.
The terminal velocity of an object is the maximum velocity it can reach when falling through a fluid. It is reached when the drag force on the object is equal to the force of gravity.
The drag force is proportional to the square of the velocity, so as the object falls faster, the drag force increases. Eventually, the drag force becomes equal to the force of gravity, and the object falls at a constant velocity.
The terminal velocity of the sphere can be calculated using the following formula:
v_t = sqrt((2 * m * g) / (C_d * A * rho_f))
where:
v_t is the terminal velocity in meters per second
m is the mass of the sphere in kilograms
g is the acceleration due to gravity (9.8 m/s^2)
C_d is the drag coefficient (0.500 in this case)
A is the cross-sectional area of the sphere in meters^2
rho_f is the density of the fluid (1.20 kg/m^3 in this case)
The mass of the sphere can be calculated using the following formula:
m = (4/3) * pi * r^3 * rho
where:
m is the mass of the sphere in kilograms
pi is a mathematical constant (3.14)
r is the radius of the sphere in meters
rho is the density of the sphere in kilograms per cubic meter
The cross-sectional area of the sphere can be calculated using the following formula:
A = pi * r^2
Plugging in the known values, we get the following terminal velocity for the sphere:
v_t = sqrt((2 * (4/3) * pi * (8.00 cm)^3 * (1.14 g/cm^3) * 9.8 m/s^2) / (0.500 * pi * (8.00 cm)^2 * 1.20 kg/m^3)) = 17.71 m/s
The height from which the sphere would have to be dropped to reach this speed if it fell without air resistance can be calculated using the following formula:
h = (v_t^2 * 2 / g)
where:
h is the height in meters
v_t is the terminal velocity in meters per second
g is the acceleration due to gravity (9.8 m/s^2)
Plugging in the known values, we get the following height:
h = (17.71 m/s)^2 * 2 / 9.8 m/s^2 = 86.77 m
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Question 23 1 pts Which of the following best describes the sizes of atoms? Atoms are so small that millions of them could fit across the period at the end of this sentence. Most atoms are about a millionth of a meter (1 micrometer) in diameter. Atoms are roughly the same size as typical bacteria. Atoms are too small to see by eye, but can be seen with a handheld magnifying glass.
The statement "Atoms are so small that millions of them could fit across the period at the end of this sentence" best describes the sizes of atoms
How is the size of an atomAtoms are the fundamental building blocks of matter and are incredibly tiny They consist of a nucleus at the center made up of protons and neutrons with electrons orbiting around it The size of an atom is typically measured in terms of its diameter
They are said to be smallest pasrticles that make up matter. Hence we have to conclude that toms are so small that millions of them could fit across the period at the end of this sentence" best describes the sizes of atoms
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Ancient pyramid builders are balancing a uniform rectangular stone slab of weight w, Part A tipped at an angle θ above the horizontal using a rope 1 The rope is held by five workers who share the force equally. If θ=14.0 ∘
, what force does each worker exert on the rope? Express your answer in terms of w (the weight of the slab). X Incorrect; Try Again; 4 attempts remaining Part B As θ increases, does each worker have to exert more or less force than in pa Figure Part C At what angle do the workers need to exert no force to balance the slab? Express your answer in degrees. θ * Incorrect; Try Again; 2 attempts remaining
The force that each worker exerts on the rope is 0.012w, where w is the weight of the slab. As θ increases, the force that each worker exerts decreases. At an angle of 45 degrees, the workers need to exert no force to balance the slab. Beyond this angle, the slab will tip over.
The force that each worker exerts on the rope is equal to the weight of the slab divided by the number of workers. This is because the force of each worker must be equal and opposite to the force of the other workers in order to keep the slab balanced.
The weight of the slab is w, and the number of workers is 5. Therefore, the force that each worker exerts is:
F = w / 5
The angle θ is the angle between the rope and the horizontal. As θ increases, the moment arm of the weight of the slab decreases. This is because the weight of the slab is acting perpendicular to the surface of the slab, and the surface of the slab is tilted at an angle.
The moment arm of the force exerted by the workers is the distance between the rope and the center of mass of the slab. This distance does not change as θ increases. Therefore, as θ increases, the torque exerted by the weight of the slab decreases.
In order to keep the slab balanced, the torque exerted by the workers must also decrease. This means that the force exerted by each worker must decrease.
At an angle of 45 degrees, the moment arm of the weight of the slab is zero. This means that the torque exerted by the weight of the slab is also zero. In order to keep the slab balanced, the torque exerted by the workers must also be zero. This means that the force exerted by each worker must be zero.
Beyond an angle of 45 degrees, the torque exerted by the weight of the slab will be greater than the torque exerted by the workers. This means that the slab will tip over.
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