3. Coulomb's Law refers exclusively to point charges. a. Real b. False

Answers

Answer 1

The statement that claims that the Coulomb's Law refers exclusively to point charges is b. False

Coulomb's Law is not limited to point charges; it applies to any charged objects, whether they are point charges or have finite sizes and distributions of charge.

Coulomb's Law states that the magnitude of the electrostatic force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

Coulomb's Law is described by the equation F = k * (q1 * q2) / r^2, where F represents the electrostatic force between two charged objects, k is the electrostatic constant, q1 and q2 denote the charges of the objects, and r signifies the distance separating them.

This law is a fundamental principle in electrostatics and is applicable to a wide range of scenarios involving charged objects, not just point charges.

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Related Questions

10/1 Points DETAILS PREVIOUS ANSWERS SERCP11 22.4.P.028 MY NOTES PRACTICE ANOTHER A certain kind of glass has an index of refraction of 1.660 for blue light of wavelength 420 m and an index of 1.6.0 for red light of wavelength 60 am. Item contaring the too incident at an angle of 30.0" piece of this gass, what is the angle between the two beams inside the 2 048 X Yoir response differs from the correct answer by more than 10%

Answers

The angle between the two beams inside the glass for blue light is approximately 17.65°, and for red light is approximately 19.10°.

To determine the angle between the two beams inside the glass, we can use Snell's Law, which relates the angles of incidence and refraction to the indices of refraction of the two media:

n₁sinθ₁ = n₂sinθ₂

Where:

n₁ = index of refraction of the initial medium (air)

θ₁ = angle of incidence in the initial medium

n₂ = index of refraction of the final medium (glass)

θ₂ = angle of refraction in the final medium

n₁ = 1 (index of refraction of air)

n₂ (for blue light) = 1.660

n₂ (for red light) = 1.600

θ₁ = 30.0° (angle of incidence)

For blue light (wavelength = 420 nm):

n₁sinθ₁ = n₂sinθ₂

(1)(sin 30.0°) = (1.660)(sin θ₂)

Solving for θ₂, we find:

sin θ₂ = (sin 30.0°) / 1.660

θ₂ = arcsin[(sin 30.0°) / 1.660]

Using a calculator, we find:

θ₂ ≈ 17.65°

For red light (wavelength = 600 nm):

n₁sinθ₁ = n₂sinθ₂

(1)(sin 30.0°) = (1.600)(sin θ₂)

Solving for θ₂, we find:

sin θ₂ = (sin 30.0°) / 1.600

θ₂ = arcsin[(sin 30.0°) / 1.600]

Using a calculator, we find:

θ₂ ≈ 19.10°

Therefore, the angle between the two beams inside the glass for blue light is approximately 17.65°, and for red light is approximately 19.10°.

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16) (10 points) Alpha particles (charge - +2e, mass - 6.68 x 10-27 kg) are accelerated in a cyclotron to a final orbit radius of 0.50 m. The magnetic field in the cyclotron is 0.50T What is the kinetic energy?

Answers

The kinetic energy of the alpha particles accelerated in a cyclotron to a final orbit radius of 0.50 m is 3.37 MeV.

Given that, Charge of alpha particles, q = +2e

Mass of alpha particles, m = 6.68 × 10-27 kg

Magnetic field, B = 0.50T

Radius of the orbit, r = 0.50 m

The magnetic force acting on an alpha particle that's in circular motion is the centripetal force acting on it. It follows from the formula Fm = Fc where Fm is the magnetic force and Fc is the centripetal force that, qv

B = mv²/r ... [1]Here, v is the velocity of the alpha particles. We know that the kinetic energy of the alpha particles is,

K.E. = 1/2 mv² ... [2] From equation [1], we can isolate the velocity of the alpha particles as follows,

v = qBr/m... [3]Substituting the equation [3] into [2], we get,

K.E. = 1/2 (m/qB)² q²B²r²/m

K.E. = q²B²r²/2m ... [4]

The value of q2/m is equal to 3.2  1013 J/T. Therefore, K.E. = 3.2 × 10¹³ J/T × (0.50 T)² × (0.50 m)²/2(6.68 × 10⁻²⁷ kg)

K.E. = 3.37 MeV Hence, the kinetic energy of the alpha particles is 3.37 MeV.

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A circular breath of 200 turns and 12 cm in diameter, it is designed to rotate 90° in 0.2 s. Initially, the spire is placed in a magnetic field in such a way that the flux is zero and then the spire is rotated 90°. If the fem induced in the spire is 0.4 mV, what is the magnitude of the magnetic field?

Answers

The magnetic field has an approximate magnitude of 0.22 Tesla according to Faraday's law of electromagnetic induction and the equation relating magnetic flux and the magnetic field.

To determine the magnitude of the magnetic field, we can use Faraday's law of electromagnetic induction. According to Faraday's law, the induced electromotive force (emf) in a wire loop is equal to the rate of change of magnetic flux through the loop.

Given that the spire (wire loop) consists of 200 turns and has a diameter of 12 cm, we can calculate the area of the loop. The radius (r) of the loop is half the diameter, so r = 6 cm = 0.06 m. The area (A) of the loop is then:

A = πr² = π(0.06 m)²

The spire is rotated 90° in 0.2 s, which means the change in flux (ΔΦ) through the loop occurs in this time. The induced emf (ε) is given as 0.4 mV.

Using Faraday's law, we have the equation:

ε = -NΔΦ/Δt

where N is the number of turns, ΔΦ is the change in magnetic flux, and Δt is the change in time.

Rearranging the equation, we can solve for the change in magnetic flux:

ΔΦ = -(ε * Δt) / N

Substituting the given values, we get:

ΔΦ = -((0.4 × 10⁽⁻³⁾ V) * (0.2 s)) / 200

ΔΦ = -8 × 10⁽⁻⁶⁾ Wb

Since the initial flux was zero, the final flux (Φ) is equal to the change in flux:

Φ = ΔΦ = -8 × 10⁽⁻⁶⁾ Wb

The magnitude of the magnetic field (B) can be determined using the equation:

Φ = B * A

Rearranging the equation, we can solve for B:

B = Φ / A

Substituting the values, we have:

B = (-8 × 10⁽⁻⁶⁾ Wb) / (π(0.06 m)²)

B ≈ -0.22 T (taking the magnitude)

Therefore, the magnitude of the magnetic field is approximately 0.22 Tesla.

In conclusion, By applying Faraday's law of electromagnetic induction and the equation relating magnetic flux and the magnetic field, we can determine that the magnitude of the magnetic field is approximately 0.22 Tesla.

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a girl at a state fair swings a ball in vertical circle at the end of a string. the force of the bottom of the string is?

Answers

When a ball is swung on a string in a vertical circle, the tension is greatest at the bottom of the circular path. This is where the rope is most likely to break. It should make sense that the tension at the bottom is the greatest.

Question 4 A book of mass m is taken to a heighth with a constant speed. A rock of mass 2m is taken to the same height also at a constant speed. The rock rises to this height twice as fast as the book. The work the gravitational force does on the rock is one quarter of the the work done on the book one half of the work done on the book twice the work done on the book four times the work done on the book the same as the work done on the book

Answers

The work done by the gravitational force on the rock is four times the work done on the book.

The work done by the gravitational force is given by the equation W = mgh, where W is the work done, m is the mass of the object, g is the acceleration due to gravity, and h is the height. Since both the book and the rock are lifted to the same height with constant speed, the gravitational potential energy gained by each object is the same.

Let's assume the work done on the book is W_book. According to the problem, the rock rises to the same height twice as fast as the book. Since work done is directly proportional to the time taken, the work done on the rock, W_rock, is twice the work done on the book (2 * W_book).

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A girl applies a 140 N force to a 35 kg bale of hay at an angle of 28° above horizontal. The coefficient of friction between the floor and the bale of hay is 0.25. F = 140 N 28° Determine the Normal Force on the block. Show the full systematic method & include a labeled FBD Determine the net or total work done on the bale of hay if she pulls it a horizontal distance of 15 m.

Answers

The net work done on the bale of hay as it is pulled a horizontal distance of 15 m is approximately 560.40 Joules.

Let's break down the problem step by step.

We have an applied force of 140 N at an angle of 28° above the horizontal. First, we need to determine the vertical and horizontal components of this force.

Vertical component:

F_vertical = F * sin(θ) = 140 N * sin(28°) ≈ 65.64 N

Horizontal component:

F_horizontal = F * cos(θ) = 140 N * cos(28°) ≈ 123.11 N

Now, let's consider the forces acting on the bale of hay:

1. Gravitational force (weight): The weight of the bale is given by

W = m * g,

where

m is the mass (35 kg)

g is the acceleration due to gravity (9.8 m/s²). Therefore,

W = 35 kg * 9.8 m/s² = 343 N.

2. Normal force (N): The normal force acts perpendicular to the floor and counteracts the gravitational force. In this case, the normal force is equal to the weight of the bale, which is 343 N.

3. Frictional force (f): The frictional force can be calculated using the formula

f = μ * N,

where

μ is the coefficient of friction (0.25)

N is the normal force (343 N).

Thus, f = 0.25 * 343 N

= 85.75 N.

Next, we need to determine the net work done on the bale of hay as it is pulled horizontally a distance of 15 m. Since the frictional force opposes the applied force, the net work done is equal to the work done by the applied force minus the work done by friction.

Work done by the applied force:

W_applied = F_horizontal * d

= 123.11 N * 15 m

= 1846.65 J

Work done by friction: W_friction = f * d

= 85.75 N * 15 m

= 1286.25 J

Net work done: W_net = W_applied - W_friction

= 1846.65 J - 1286.25 J

= 560.40 J

Therefore, the net work done on the bale of hay as it is pulled a horizontal distance of 15 m is approximately 560.40 Joules.

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Radon gas has a half-life of 3.83 days. If 2.80 g of radon gas is present at time
t = 0,
what mass of radon will remain after 2.10 days have passed?
g
After 2.10 days, the activity of a sample of an unknown type radioactive material has decreased to 83.4% of the initial activity. What is the half-life of this material?
days

Answers

Radon gas has a half-life of 3.83 days. If 2.80 g of radon gas is present at time t = 0, The radioactive decay of an isotope can be quantified using the half-life of the isotope. It takes approximately one half-life for half of the substance to decay.

The half-life of radon is 3.83 days. After a specific amount of time, the amount of radon remaining can be calculated using the formula: Amount remaining = Initial amount × (1/2)^(number of half-lives)Here, initial amount of radon gas present at time t=0 is 2.80 g. Number of half-lives = time elapsed ÷ half-life = 2.10 days ÷ 3.83 days = 0.5487 half-lives Amount remaining = 2.80 g × (1/2)^(0.5487) = 1.22 g

Thus, the mass of radon gas that will remain after 2.10 days have passed is 1.22 g. The answer is 1.22g.After 2.10 days, the activity of a sample of an unknown type radioactive material has decreased to 83.4% of the initial activity. What is the half-life of this material?Given, After 2.10 days, activity of sample = 83.4% of the initial activity.

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Design an cross belt drive to transmit 25 kW at 720 rpm on an aluminum rolling machine; Speed ​​reduction is 3.0. The distance between the shaft and the motor is 3 meters. The diameter and width of the rolling machine pulley are 1.2 m and 350 mm respectively. The coefficient of friction in the belt is 0.2 and the allowable stress coefficient is 2 MPa.

Answers

To transmit 25 kW at 720 rpm on an aluminum rolling machine, a cross belt drive with a tension of 484 N would be needed, considering the given parameters and the coefficient of friction in the belt.

To design a cross belt drive to transmit 25 kW at 720 rpm on an aluminum rolling machine, we need to consider various factors such as speed reduction, distance between the shaft and the motor, pulley dimensions, coefficient of friction in the belt, and allowable stress coefficient.

First, let's calculate the speed of the driven pulley. Since the speed reduction is 3.0, the speed of the driven pulley would be 720 rpm / 3.0 = 240 rpm.

Next, let's calculate the belt velocity. The belt velocity can be determined by multiplying the diameter of the driven pulley by π and the speed of the driven pulley. Therefore, the belt velocity is (1.2 m / 2) * π * 240 rpm = 452.39 m/min.

To find the power transmitted by the belt, we divide the given power by the belt velocity. Thus, the power transmitted by the belt is 25,000 W / 452.39 m/min = 55.21 Nm/s.

Using the equation for power transmission through friction, P = (T1 - T2) * V, where P is power, T1 and T2 are tensions in the belt, and V is the belt velocity, we can rearrange the equation to solve for T2:

T2 = T1 - (P / V)

Substituting the values, T2 = T1 - (55.21 Nm/s / 452.39 m/min) = T1 - 0.122 N.

Considering the allowable stress coefficient of 2 MPa, we can calculate the allowable tension in the belt:

Allowable tension (Tall) = (2 MPa * π * (350 mm / 2)^2) / 1,000 = 96.78 N

Finally, we can find the required tension in the belt (T1) using the coefficient of friction:

T1 = (Tall + T2) / (2 * friction coefficient) = (96.78 N + 0.122 N) / (2 * 0.2) = 484 N

Therefore, the required tension in the belt is 484 N.

In summary, to transmit 25 kW at 720 rpm on an aluminum rolling machine, a cross belt drive with a tension of 484 N would be needed, considering the given parameters and the coefficient of friction in the belt.

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Fishermen can use echo sounders to locate schools of fish and to determine the depth of water beneath their vessels. An ultrasonic pulse from an echo sounder is observed to return to a boat after 0.200 s. What is the sea depth beneath the sounder? The speed of sound in water is 1.53 x 103 m s-1. (a) 612 m (b) 306 m (c) 153 m (d) 76.5 m

Answers

The speed of sound in water is 1.53 x 103 m s-1. An ultrasonic pulse from an echo sounder is observed to return to a boat after 0.200 s.

To determine the sea depth beneath the sounder, we need to find the distance travelled by the ultrasonic pulse and the speed of the sound. Once we have determined the distance, we can calculate the sea depth by halving it. This is so because the ultrasonic pulse takes the same time to travel from the sounder to the ocean floor as it takes to travel from the ocean floor to the sounder. We are provided with speed of sound in water which is 1.53 x 10³ m/s.We know that speed = distance / time.

Rearranging the formula for distance:distance = speed × time. Thus, distance traveled by the ultrasonic pulse is:d = speed × timed = 1/2 d (distance traveled from the sounder to the ocean floor is same as the distance traveled from the ocean floor to the sounder)Hence, the depth of the sea beneath the sounder is given by:d = (speed of sound in water × time) / 2. Substituting the given values:speed of sound in water = 1.53 x 103 m s-1, time taken = 0.200 s. Therefore,d = (1.53 × 10³ m/s × 0.200 s) / 2d = 153 m. Therefore, the sea depth beneath the sounder is 153 m.Option (c) is correct.

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Imagine that an object is thrown in the air with 100 miles per hour with 30 degrees of angle. Calculate the size of the displacement associated with the object in the horizontal direction when it was done on a large size spherical star with the gravitational acceleration is 25 miles per hour

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On a large spherical star with a gravitational acceleration of 25 miles per hour, an object thrown at a 30-degree angle with an initial velocity of 100 miles per hour will have a calculated horizontal displacement.

Resolve the initial velocity:

Given the initial velocity of the object is 100 miles per hour and it is launched at an angle of 30 degrees, we need to find its horizontal component. The horizontal component can be calculated using the formula: Vx = V * cos(θ), where V is the initial velocity and θ is the launch angle.

Vx = 100 * cos(30°) = 100 * √3/2 = 50√3 miles per hour.

Calculate the time of flight:

To determine the horizontal displacement, we first need to calculate the time it takes for the object to reach the ground. The time of flight can be determined using the formula: t = 2 * Vy / g, where Vy is the vertical component of the initial velocity and g is the gravitational acceleration.

Since the object is thrown vertically upwards, Vy = V * sin(θ) = 100 * sin(30°) = 100 * 1/2 = 50 miles per hour.

t = 2 * 50 / 25 = 4 hours.

Calculate the horizontal displacement:

With the time of flight determined, we can now find the horizontal displacement using the formula: Dx = Vx * t, where Dx is the horizontal displacement, Vx is the horizontal component of the initial velocity, and t is the time of flight.

Dx = 50√3 * 4 = 200√3 miles.

Therefore, the size of the displacement associated with the object in the horizontal direction, when thrown at an angle of 30 degrees and a speed of 100 miles per hour, on a large spherical star with a gravitational acceleration of 25 miles per hour, would be approximately 100 miles.

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A charged particle of charge 5.1 mC (milli-Coulomb) is moving with a speed of 9 m/s in a region of a magnetic field of 3.4 T, which acts at an angle of 30 degrees to its motion. What is the magnitude of the magnetic force on the charge due to the field? Enter your number up to TWO decimal place.

Answers

The magnitude of the magnetic force on the charge due to the field is approximately 0.08 N. Hence, the answer is 0.08 N.

The given values are:

Charge, q = 5.1

mC = 5.1 × 10^(-3) Coulomb

Velocity, v = 9 m/s

Magnetic field, B = 3.4 T

Angle between magnetic field and velocity, θ = 30°

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

F = Bqv sin where q is the charge, v is the velocity, B is the magnetic field strength, and  is the angle between the velocity and magnetic field.

Now substitute the given values in the above formula,

F = (3.4 T) × (5.1 × 10^(-3) C) × (9 m/s) sin 30°

F = (3.4) × (5.1 × 10^(-3)) × (9/2)

F = 0.08163 N

Therefore, the magnitude of the magnetic force on the charge due to the field is approximately 0.08 N. Hence, the answer is 0.08 N.

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An AC generator with a peak voltage of 120 volts is placed
across a 10-Ω resistor. What is the average power dissipated?
A.
650W
b.
1000W
c.
500W
d
120W
E
720W

Answers

In an AC circuit, the voltage and current vary sinusoidally over time. The peak voltage (Vp) refers to the maximum value reached by the voltage waveform.

The RMS voltage (Vrms) is obtained by dividing the peak voltage by the square root of 2 (Vrms = Vp/√2). This value represents the equivalent DC voltage that would deliver the same amount of power in a resistive circuit.

Vrms = 120/√2, resulting in Vrms = 84.85 V.

P = Vrms^2/R, where P represents the average power and R is the resistance.

Plugging in the values, we have P = (84.85)^2 / 10, which simplifies to P = 720 W.

Therefore, the average power dissipated in the resistor is 720 watts. This value indicates the rate at which energy is converted to heat in the resistor.

It's worth noting that the average power dissipated can also be calculated using the formula P = (Vrms * Irms) * cosφ, where Irms is the RMS current and cosφ is the power factor.

However, in this scenario, the given information only includes the peak voltage and the resistance, making the first method more appropriate for calculation.

Overall, the average power dissipated in the resistor is a crucial factor to consider when analyzing AC circuits, as it determines the energy consumption and heat generation in the circuit component.

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(hrwc10p72_6e) The linear momentum of a 1350 kg car increased by 6.50×10³ kg m/s in 13.0 s. What is the magnitude of the constant force that accelerated the car? Submit Answer Tries 0/8 By how much did the speed of the car increase? Submit Answer Tries 0/7

Answers

The magnitude of the constant force that accelerated the car can be calculated using the formula for linear momentum. The calculated force is 5.00 × 10^2 N. The increase in speed of the car can be determined by dividing the change in momentum by the mass of the car. The calculated increase in speed is 4.81 m/s.

The linear momentum (p) of an object is given by the formula p = mv, where m is the mass of the object and v is its velocity.

In this case, the car has a mass of 1350 kg and its linear momentum increased by 6.50 × 10³ kg m/s in a time interval of 13.0 s.

To find the magnitude of the force that accelerated the car, we use the formula F = Δp/Δt, where Δp is the change in momentum and Δt is the change in time.

Substituting the given values, we have F = (6.50 × 10³ kg m/s)/(13.0 s) = 5.00 × 10^2 N.

Therefore, the magnitude of the constant force that accelerated the car is 5.00 × 10^2 N.

To determine the increase in speed of the car, we divide the change in momentum by the mass of the car. The change in speed (Δv) is given by Δv = Δp/m.

Substituting the values, we have Δv = (6.50 × 10³ kg m/s)/(1350 kg) = 4.81 m/s.

Hence, the speed of the car increased by 4.81 m/s.

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Resistor in circuit is made of a length of 14awg iron wire. When
10 V is applied across the resistor wire of length 100m,
what is the reading on the ammeter? The thickness
of 14awg wire is 1.628mm.

Answers

The reading on the ammeter would be approximately 2.14 Amperes.

To calculate the reading on the ammeter, we need to determine the resistance of the 14 AWG iron wire. The resistance can be calculated using the formula

[tex]R = ρ * (L / A)[/tex]

where:

R is the resistance,

ρ is the resistivity of the material (in this case, iron),

L is the length of the wire, and

A is the cross-sectional area of the wire.

First, let's calculate the cross-sectional area of the 14 AWG wire. The diameter of the wire can be obtained from the wire gauge size. For 14 AWG, the diameter is approximately 1.628 mm.

The radius (r) can be calculated by dividing the diameter by 2:

r = 1.628 mm / 2 = 0.814 mm = 0.000814 m

The cross-sectional area (A) can be calculated using the formula:

[tex]R = ρ * (L / A)[/tex]

[tex]A = 3.14159 * (0.000814 m)^2 ≈ 2.07678 × 10^(-6) m^2[/tex]

Next, we need to find the resistivity of iron. The resistivity of iron (ρ) is approximately 9.71 × 10^(-8) Ω·m.

Now, we can calculate the resistance (R) using the formula mentioned earlier:

[tex]R = (9.71 × 10^(-8) Ω·m) * (100 m / 2.07678 × 10^(-6) m^2)[/tex]

[tex]R ≈ 4.675 Ω[/tex]

Therefore, with a 10 V potential difference across the 14 AWG iron wire resistor, the reading on the ammeter would be:

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

[tex]I = 10 V / 4.675 Ω[/tex]

[tex]I ≈ 2.14 A[/tex]

So, the reading on the ammeter would be approximately 2.14 Amperes.

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Question 31 1 pts A high voltage transmission line carrying 500 MW of electrical power at voltage of 409 kV (kilovolts) has a resistance of 10 ohms. What is the power lost in the transmission line? Give your answer in megawatts (MW).

Answers

The power lost in the transmission line is approximately 14.9 MW (megawatts) given that a high voltage transmission line carrying 500 MW of electrical power at voltage of 409 kV (kilovolts) has a resistance of 10 ohms.

Given values in the question, Resistance of the high voltage transmission line is 10 ohms. Power carried by the high voltage transmission line is 500 MW. Voltage of the high voltage transmission line is 409 kV (kilovolts).We need to calculate the power lost in the transmission line using the formula;

Power loss = I²RWhere,I = Current (Ampere)R = Resistance (Ohms)

For that we need to calculate the Current by using the formula;

Power = Voltage × Current

Where, Power = 500 MW

Voltage = 409 kV (kilovolts)Current = ?

Now we can substitute the given values to the formula;

Power = Voltage × Current500 MW = 409 kV × Current

Current = 500 MW / 409 kV ≈ 1.22 A (approx)

Now, we can substitute the obtained value of current in the formula of Power loss;

Power loss = I²R= (1.22 A)² × 10 Ω≈ 14.9 MW

Therefore, the power lost in the transmission line is approximately 14.9 MW (megawatts).

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(hrwc9p93) A body of mass 12.0 kg is traveling at 1.8 m/s along the positive x-axis with no external force acting. At a certain instant an internal explosion occurs, splitting the body into two chunks of 6.0 kg mass each. The explosion gives the chunks an additional 16 J of kinetic energy. Neither chunk leaves the line of original motion. Determine the speed and direction of motion of each of the chunks after the explosion. Enter the larger velocity. Submit Answer Tries 0/8 Enter the smaller velocity. Submit Answer Tries 0/7 Post Discussion Send Feedback

Answers

The question involves determining the velocities of two chunks after an internal explosion. The initial mass, velocity, and additional kinetic energy given to the chunks are provided. The goal is to calculate the velocities of the two chunks along the original line of motion.

When an internal explosion occurs, the total momentum before the explosion is equal to the total momentum after the explosion since no external forces are acting. Initially, the body has a mass of 12.0 kg and a velocity of 1.8 m/s along the positive x-axis. After the explosion, it splits into two chunks of equal mass, 6.0 kg each. To find the velocities of the chunks after the explosion, we need to apply the principle of conservation of momentum.

Since the chunks are moving along the line of the original motion, the momentum in the x-direction should be conserved. We can set up an equation to solve for the velocities of the chunks. The initial momentum of the body is the product of its mass and velocity, and the final momentum is the sum of the momenta of the two chunks. By equating these two momenta, we can solve for the velocities of the chunks.

The given additional kinetic energy of 16 J can be used to find the individual kinetic energies of the chunks. Since the masses of the chunks are equal, the additional kinetic energy will be divided equally between them. From the individual kinetic energies, we can calculate the velocities of the chunks using the equation for kinetic energy. The larger velocity will correspond to the chunk with the additional kinetic energy, and the smaller velocity will correspond to the other chunk.

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3. (8 points) Name and describe the two main forms of mechanical waves.

Answers

Mechanical waves are waves that require a medium to travel through. These waves can travel through different mediums, including solids, liquids, and gases. The two main forms of mechanical waves are transverse waves and longitudinal waves.

Mechanical waves are the waves which require a medium for their propagation. A medium is a substance through which a mechanical wave travels. The medium can be a solid, liquid, or gas. These waves transfer energy from one place to another by the transfer of momentum and can be described by their wavelength, frequency, amplitude, and speed.There are two main forms of mechanical waves, transverse waves and longitudinal waves. In transverse waves, the oscillations of particles are perpendicular to the direction of wave propagation.

Transverse waves can be observed in the motion of a string, water waves, and electromagnetic waves. Electromagnetic waves are transverse waves but do not require a medium for their propagation. Examples of electromagnetic waves are radio waves, light waves, and X-rays. In longitudinal waves, the oscillations of particles are parallel to the direction of wave propagation. Sound waves are examples of longitudinal waves where the particles of air or water oscillate parallel to the direction of the sound wave.

In conclusion, transverse and longitudinal waves are two main forms of mechanical waves. Transverse waves occur when the oscillations of particles are perpendicular to the direction of wave propagation. Longitudinal waves occur when the oscillations of particles are parallel to the direction of wave propagation. The speed, frequency, wavelength, and amplitude of a wave are its important characteristics. The medium, through which a wave travels, can be a solid, liquid, or gas. Electromagnetic waves are also transverse waves but do not require a medium for their propagation.

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The shortest pipe in a particular organ is 1.25 m. (a) Determine the frequency (in Hz) of the fifth harmonic (at 0°C) if the pipe is closed at one end. Hz (b) Determine the frequency (in Hz) of the f

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(a) The frequency of the fifth harmonic in a closed-end pipe with a length of 1.25 m is approximately 562.5 Hz. (b) The frequency of the fundamental is approximately 83.9 Hz.

In a closed-end pipe, the harmonics are integer multiples of the fundamental frequency. The fifth harmonic refers to the fifth multiple of the fundamental frequency. To determine the frequency of the fifth harmonic, we multiply the fundamental frequency by five. Since the fundamental frequency is calculated to be approximately 83.9 Hz, the frequency of the fifth harmonic is approximately 5 * 83.9 Hz, which equals 419.5 Hz.

For a closed-end pipe, the formula to calculate the fundamental frequency involves the harmonic number (n), the speed of sound (v), and the length of the pipe (L). By rearranging the formula, we can solve for the frequency (f) of the fundamental. Plugging in the given values, we get f = (1 * 331.4 m/s) / (4 * 1.25 m) ≈ 83.9 Hz. This frequency represents the first harmonic or the fundamental frequency of the closed-end pipe.

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Jane goes out for a run. She runs 10 miles West for 2 hours, then she stops suddenly and turns and runs North for 30 minutes while speeding up at a rate of 4.0×10 ^−3 [ m/s 2
]. She stops again, then runs with constant velocity of 5[ m/s] at 40 degrees North of East for 5 miles. HINT: you MUST draw a picture and choose a vector basis. a) Convert all quantities given to SI units. Must show work! b) Write out the displacement vector for each leg of the trip in vector notation. c) Find Jane's average velocity for the entire run. d) Find Jane's average speed for the entire run.

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c) Jane's average velocity for the entire run cannot be determined without the values of the angle and acceleration for the Northward leg.

d) Jane's average speed for the entire run is the total distance traveled (16093.4 + 8046.7) meters divided by the total time taken (7200 + 1800) seconds.

a) Converting the given quantities to SI units:

1 mile = 1609.34 meters

10 miles = 10 * 1609.34 meters = 16093.4 meters

2 hours = 2 * 3600 seconds = 7200 seconds

30 minutes = 30 * 60 seconds = 1800 seconds

5 miles = 5 * 1609.34 meters = 8046.7 meters

b) Displacement vectors for each leg of the trip:

1. Westward leg: Displacement vector = -16093.4 meters * i (since it is in the West direction)

2. Northward leg: Displacement vector = (30 minutes * 60 seconds * 5.0 x 10^-3 m/s^2 * (0.5 * 1800 seconds)^2) * j (since it is in the North direction and speeding up)

3. Eastward leg: Displacement vector = 8046.7 meters * cos(40 degrees) * i + 8046.7 meters * sin(40 degrees) * j (since it is at an angle of 40 degrees North of East)

c) Jane's average velocity for the entire run:

To find the average velocity, we need to calculate the total displacement and divide it by the total time.

Total displacement = Sum of individual displacement vectors

Total time = Sum of individual time intervals

Average velocity = Total displacement / Total time

d) Jane's average speed for the entire run:

Average speed = Total distance / Total time

Note: Average velocity considers both the magnitude and direction of motion, while average speed only considers the magnitude.

Please calculate the values for parts c) and d) using the provided information and formulas.

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Task: Solve the following problems. SHOW ALL THE POSSIBLE SOLUTIONS and BOX YOUR FINAL ANSWER. 1. The figure below shows four parallel plate capacitors: A, B, C, and D. Each capacitor carries the same charge q and has the same plate area A. As suggested by the figure, the plates of capacitors A and C are separated by a distance d while those of B and D are separated by a distance 2d. Capacitors A and B are maintained in vacuum while capacitors C and D contain dielectrics with constant k = 5. Arrange the capacitor in decreasing order of capacitance (e.g. A, B, C, and D) and explain briefly. (10pts) vacuum dielectric (K-5) D HA NI -2d- 20

Answers

The capacitors can be arranged in decreasing order of capacitance as follows: A, D, C, and B.

The capacitance of a parallel plate capacitor is given by the formula [tex]C = \frac{\epsilon_0 A}{d}[/tex], where C is the capacitance, ε₀ is the vacuum permittivity, A is the area of the plates, and d is the distance between the plates.

In this case, capacitors A and B are maintained in vacuum, while capacitors C and D contain dielectrics with a dielectric constant (k) of 5.

Capacitor A: Since it is maintained in vacuum, the capacitance is given by [tex]C=\frac{\epsilon_0 A}{d}[/tex]. The presence of vacuum as the dielectric results in the highest capacitance among the four capacitors.

Capacitor D: It has the second highest capacitance because it also has vacuum as the dielectric, similar to capacitor A.

Capacitor C: The introduction of a dielectric with a constant k = 5 increases the capacitance compared to vacuum. The capacitance is given by [tex]C=\frac{k \epsilon_0A}{d}[/tex]. Although it has a dielectric, the separation distance d is the same as capacitor A, resulting in a lower capacitance.

Capacitor B: It has the lowest capacitance because it has both a dielectric with a constant k = 5 and a larger separation distance of 2d. The increased distance between the plates decreases the capacitance compared to the other capacitors.

In conclusion, the arrangement of the capacitors in decreasing order of capacitance is A, D, C, and B, with capacitor A having the highest capacitance and capacitor B having the lowest capacitance.

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Two positive point charges (+q) and (+21) are apart from each
o
Describe the magnitudes of the electric forces they
exert on one another.
Explain why they exert these magnitudes on one
another.

Answers

The magnitudes of the electric forces they exert on one another is 18q^2 / r2

Two positive point charges (+q) and (+2q) are apart from each other.

Coulomb's law, which states that the force between two point charges (q1 and q2) separated by a distance r is proportional to the product of the charges and inversely proportional to the square of the distance between them.

F = kq1q2 / r2

Where,

k = Coulomb's constant = 9 × 10^9 Nm^2C^-2

q1 = +q

q2 = +2q

r = distance between two charges.

Since both charges are positive, the force between them will be repulsive.

Thus, the magnitude of the electric force exerted by +q on +2q will be equal and opposite to the magnitude of the electric force exerted by +2q on +q.

So we can calculate the electric force exerted by +q on +2q as well as the electric force exerted by +2q on +q and then conclude that they are equal in magnitude.

Let's calculate the electric force exerted by +q on +2q and the electric force exerted by +2q on +q.

Electric force exerted by +q on +2q:

F = kq1q2 / r2

 = (9 × 10^9 Nm^2C^-2) (q) (2q) / r2

 = 18q^2 / r2

Electric force exerted by +2q on +q:

F = kq1q2 / r2

  = (9 × 10^9 Nm^2C^-2) (2q) (q) / r2

  = 18q^2 / r2

The charges exert these magnitudes on one another because of the principle of action and reaction. It states that for every action, there is an equal and opposite reaction.

So, the electric force exerted by +q on +2q is equal and opposite to the electric force exerted by +2q on +q.

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3. The electric field of an electromagnetic wave is given by Ē = 7.2 x 106 ) V/m. If the propagation speed is 3 x 108 k, calculate the magnetic field vector of the wave.

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An electromagnetic wave is a type of wave that consists of electric and magnetic fields oscillating perpendicular to each other and propagating through space. They exhibit both wave-like and particle-like properties.

Electromagnetic waves consist of both electric and magnetic fields, which are perpendicular to each other and to the direction of wave propagation. The electric field oscillates in one plane, while the magnetic field oscillates in a plane perpendicular to the electric field. Therefore, electromagnetic waves are transverse waves.

Given, Electric field of an electromagnetic wave Ē = 7.2 x 106 V/m. Propagation speed v = 3 x 108 m/s We need to calculate the magnetic field vector of the wave. According to the equation of an electromagnetic wave, we know that;  E = cBV = E/BorB = E/V Where, B is the magnetic field vector. V is the propagation speed. E is the electric field vector. Substituting the given values in the above formula we get; B = Ē/v= (7.2 x 10⁶)/ (3 x 10⁸)= 0.024 V.s/m. The magnetic field vector of the wave is 0.024 V.s/m.

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a ball is kicked upward with an initial velocity of 68 feet per second. the ball's height, h (in feet), from the ground is modeled by h

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The ball reaches a maximum height of 289 feet after 4.25 seconds.

The height of a ball kicked upward can be modeled by the equation h = -16t^2 + vt + s, where h is the height in feet, t is the time in seconds, v is the initial velocity in feet per second, and s is the initial height in feet. In this case, the ball is kicked upward with an initial velocity of 68 feet per second.

To find the height of the ball at a given time, we can substitute the values into the equation. Let's assume the initial height, s, is 0 (meaning the ball is kicked from the ground).

Therefore, the equation becomes: h = -16t^2 + 68t + 0.

To find the maximum height, we need to determine the time it takes for the ball to reach its peak. At the peak, the velocity is 0.

To find this time, we set the equation equal to 0 and solve for t:

-16t^2 + 68t = 0.

Factoring out t, we get:

t(-16t + 68) = 0.

Setting each factor equal to 0, we find two solutions:

t = 0 (this is the initial time when the ball is kicked) and -16t + 68 = 0.

Solving -16t + 68 = 0, we find t = 4.25 seconds.

So, it takes 4.25 seconds for the ball to reach its peak height.

To find the maximum height, we substitute this time into the original equation:

h = -16(4.25)^2 + 68(4.25) + 0.

Evaluating this equation, we find the maximum height of the ball is 289 feet.

Therefore, the ball reaches a maximum height of 289 feet after 4.25 seconds.

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Final answer:

The subject of this question is Physics. It asks about the height of a ball kicked upward with an initial velocity of 68 feet per second. Projectile motion equations can be used to model the ball's height.

Explanation:

The subject of this question is Physics. The question is asking about the height of a ball that is kicked upward with an initial velocity of 68 feet per second. This can be modeled using equations of projectile motion.

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8.88 kJ of energy raises the temperature of a 1 kg block of copper by 10°C.

Calculate the specific heat capacity of copper.

Answers

The specific heat capacity of copper is 0.888 kJ/(kg × °C).

Specific heat capacity is a thermal property of a substance. It indicates how much heat energy is needed to raise the temperature of a unit mass of a substance by one degree Celsius.

The formula for calculating the specific heat capacity of a substance is given as, q = m × c × ∆T`

Where: q = energy,

m = mass of the substance,

c = specific heat capacity of the substance,

∆T = change in temperature.

Now, let’s use the formula above to calculate the specific heat capacity of copper.

The energy required to raise the temperature of a 1 kg block of copper by 10°C is 8.88 kJ.

q = m × c × ∆T

c = q / (m × ∆T)

= 8.88 kJ / (1 kg × 10°C)

= 0.888 kJ/(kg × °C)

The specific heat capacity of copper is 0.888 kJ/(kg × °C).

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Compact fluorescent (CFL) bulbs provide about four times as much visible light for a given amount of energy use. For example, a 14-watt CFL bulb provides about the same amount of visible light as a 60-watt incandescent bulb. LED lights are even more efficient at turning electrical energy into visible light. Does that mean they are both a lot hotter? Go online and research how fluorescent and compact fluorescent bulbs work. Describe how their operations and their spectra differ from those of incandescent light bulbs. Be sure to record your research sources.

Answers

Fluorescent,compact fluorescent bulbs operate differently from incandescent bulbs,resulting in differences in spectra,heat production. Both bulbs are more energy-efficient than incandescent bulbs.

Fluorescent bulbs work by passing an electric current through a gas-filled tube, which contains mercury vapor. The electrical current excites the mercury atoms, causing them to emit ultraviolet (UV) light. This UV light then interacts with a phosphor coating on the inside of the tube, causing it to fluoresce and emit visible light. The spectrum of fluorescent bulbs is characterized by distinct emission lines due to the specific wavelengths of light emitted by the excited phosphors. Incandescent bulbs work by passing an electric current through a filament, usually made of tungsten, which heats up and emits light as a result of its high temperature.

While fluorescent and CFL bulbs are more energy-efficient and produce less heat compared to incandescent bulbs, LED (light-emitting diode) lights are even more efficient. LED lights operate by passing an electric current through a semiconductor material, which emits light directly without the need for a filament or gas. LED lights convert a higher percentage of electrical energy into visible light, resulting in greater efficiency and minimal heat production.

Sources:

Energy.gov. (n.d.). How Fluorescent Lamps Work. Retrieved from https://www.energy.gov/energysaver/save-electricity-and-fuel/lighting-choices-save-you-money/how-energy-efficient-light-bulbs

Energy.gov. (n.d.). How Compact Fluorescent Lamps Work. Retrieved from https://www.energy.gov/energysaver/save-electricity-and-fuel/lighting-choices-save-you-money/how-energy-efficient-light-bulbs

Energy.gov. (n.d.). How Light Emitting Diodes Work. Retrieved from https://www.energy.gov/energysaver/save-electricity-and-fuel/lighting-choices-save-you-money/how-energy-efficient-light-bulbs

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A star is 16.7 ly (light-years) from Earth.
(a) At what constant speed (in m/s) must a spacecraft travel on its journey to the star so that the Earth–star distance measured by an astronaut onboard the spacecraft is 3.96 ly? 369162007m/s Incorrect: Your answer is incorrect.
(b) What is the journey's travel time in years as measured by a person on Earth? 17.2yr Correct: Your answer is correct.
(c) What is the journey's travel time in years as measured by the astronaut? 4.1yr Correct: Your answer is correct.

Answers

(a) The spacecraft must travel at approximately 0.9899 times the speed of light (c).

(b) The travel time as measured by a person on Earth is approximately 16.9 years.

(c) The travel time as measured by the astronaut is approximately 6.82 years.

(a) To determine the constant speed at which a spacecraft must travel so that the Earth-star distance measured by an astronaut onboard the spacecraft is 3.96 ly, we can use the time dilation equation from special relativity:

t' = t * sqrt(1 - (v^2/c^2))

where t' is the time measured by the astronaut, t is the time measured on Earth, v is the velocity of the spacecraft, and c is the speed of light.

Given that the distance between Earth and the star is 16.7 ly and the astronaut measures it as 3.96 ly, we can set up the following equation:

t' = t * sqrt(1 - (v^2/c^2))

3.96 = 16.7 * sqrt(1 - (v^2/c^2))

Solving this equation will give us the velocity (v) at which the spacecraft must travel.

(b) To calculate the journey's travel time in years as measured by a person on Earth, we can use the equation:

t = d/v

where t is the travel time, d is the distance, and v is the velocity of the spacecraft. Plugging in the values, we can find the travel time in years.

(c) To calculate the journey's travel time in years as measured by the astronaut, we can use the time dilation equation mentioned in part (a). Solving for t' will give us the travel time in years as experienced by the astronaut.

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A young male adult takes in about 5.16 x 104 m³ of fresh air during a normal breath. Fresh air contains approximately 21% oxygen. Assuming that the pressure in the lungs is 0.967 x 105 Pa and air is an ideal gas at a temperature of 310 K, find the number of oxygen molecules in a normal breath.

Answers

Explanation:

To find the number of oxygen molecules in a normal breath, we can use the ideal gas law equation, which relates the pressure, volume, temperature, and number of molecules of a gas:

PV = nRT

Where:

P = Pressure (in Pa)

V = Volume (in m³)

n = Number of moles

R = Ideal gas constant (8.314 J/(mol·K))

T = Temperature (in K)

First, let's calculate the number of moles of air inhaled during a normal breath:

V = 5.16 x 10^4 m³ (Volume of air inhaled)

P = 0.967 x 10^5 Pa (Pressure in the lungs)

R = 8.314 J/(mol·K) (Ideal gas constant)

T = 310 K (Temperature)

Rearranging the equation, we get:

n = PV / RT

n = (0.967 x 10^5 Pa) * (5.16 x 10^4 m³) / (8.314 J/(mol·K) * 310 K)

n ≈ 16.84 mol

Next, let's find the number of oxygen molecules inhaled. Since fresh air contains approximately 21% oxygen, we can multiply the number of moles by the fraction of oxygen in the air:

Number of oxygen molecules = n * (0.21)

Number of oxygen molecules ≈ 16.84 mol * 0.21

Number of oxygen molecules ≈ 3.54 mol

Finally, we'll convert the number of moles of oxygen molecules to the actual number of molecules by using Avogadro's number, which is approximately 6.022 x 10^23 molecules/mol:

Number of oxygen molecules = 3.54 mol * (6.022 x 10^23 molecules/mol)

Number of oxygen molecules ≈ 2.13 x 10^24 molecules

Therefore, in a normal breath, there are approximately 2.13 x 10^24 oxygen molecules.

The gravitational force changes with altitude. Find the change in gravitational force for someone who weighs 770 N at sea level as compared to the force measured when on an airplane 1700 m above sea level. You can ignore Earth's rotation for this problem. Use a negative answer to indicate a decrease in force. For reference, Earth's mean radius (Re) is 6.37 x 106 m and Earth's mass (ME) is 5.972 x 1024 kg. [Hint: take the derivative of the expression for the force of gravity with respect to r, such that dF 9 Ar. Evaluate dr the derivative at r=RE. Aweight = lever Your answer should be in N:

Answers

To find the change in gravitational force, we need to calculate the gravitational force at sea level and the gravitational force at an altitude of 1700 m, and then find the difference between the two forces.

Calculation:

Let's denote the gravitational force as F(r), where r is the distance from the center of the Earth.

Calculate the gravitational force at sea level:

F_sea = G * (M_E * m) / (R_E)^2

Calculate the gravitational force at the airplane altitude:

F_airplane = G * (M_E * m) / (R_E + h)^2

Calculate the change in gravitational force:

ΔF = F_airplane - F_sea

Given:

F_sea_level = 770 N

M = 5.972 x 10^24 kg

r_sea_level = Re (Earth's mean radius) = 6.37 x 10^6 m

Now, let's calculate the gravitational force at an altitude of 1700 m above sea level:

r_altitude = r_sea_level + 1700 m

To find the change in gravitational force, we subtract the force at the altitude from the force at sea level:

ΔF = F_sea_level - F_altitude

Let's calculate step by step:

F_sea_level = (G * M * m) / r_sea_level^2

770 N = (6.67430 x 10^-11 N m^2/kg^2 * 5.972 x 10^24 kg * m) / (6.37 x 10^6 m)^2

Solving the equation above for m (mass of the person), we find:

m = (770 N * (6.37 x 10^6 m)^2) / (6.67430 x 10^-11 N m^2/kg^2 * 5.972 x 10^24 kg)

m ≈ 61.14 kg

Now, let's calculate the gravitational force at the altitude:

F_altitude = (G * M * m) / r_altitude^2

F_altitude = (6.67430 x 10^-11 N m^2/kg^2 * 5.972 x 10^24 kg * 61.14 kg) / (r_sea_level + 1700 m)^2

ΔF = F_sea_level - F_altitude

Finally, let's plug in the values and calculate:

ΔF = 770 N - [(6.67430 x 10^-11 N m^2/kg^2 * 5.972 x 10^24 kg * 61.14 kg) / (6.37 x 10^6 m + 1700 m)^2]ΔF ≈ -9.86 N

The change in gravitational force for someone who weighs 770 N at sea level compared to when on an airplane 1700 m above sea level is approximately -9.86 N (decrease in force).

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An infinite line charge of uniform linear charge density λ = -2.1 µC/m lies parallel to the y axis at x = -1 m. A point charge of 1.1 µC is located at x = 2.5 m, y = 3.5 m. Find the x component of the electric field at x = 3.5 m, y = 3.0 m. kN/C Enter 0 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts]
In the figure shown above, a butterfly net is in a uniform electric field of magnitude E = 120 N/C. The rim, a circle of radius a = 14.3 cm, is aligned perpendicular to the field.
Find the electric flux through the netting. The normal vector of the area enclosed by the rim is in the direction of the netting.
The electric flux is:

Answers

The electric flux is 7.709091380790923. The electric field due to an infinite line charge of uniform linear charge density λ is given by:

E = k * λ / x

The electric field due to an infinite line charge of uniform linear charge density λ is given by:

E = k * λ / x

where k is the Coulomb constant and x is the distance from the line charge.

The x component of the electric field at x = 3.5 m, y = 3.0 m is:

E_x = k * λ / (3.5) = -2.86 kN/C

The electric field due to the point charge is given by:

E = k * q / r^2

where q is the charge of the point charge and r is the distance from the point charge.

The x component of the electric field due to the point charge is:

E_x = k * 1.1 * 10^-6 / ((3.5)^2 - (2.5)^2) = -0.12 kN/C

The total x component of the electric field is:

E_x = -2.86 - 0.12 = -2.98 kN/C

The electric flux through the netting is:

Φ = E * A = 120 * (math.pi * (14.3 / 100)^2) = 7.709091380790923

Therefore, the electric flux is 7.709091380790923.

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a group of students found that the moment of inertia of the plate+disk was 1.74x10-4 kg m2, on the other hand they found that the moment of inertia of the plate was 0.34x10-4 kg m2. What is the value of the moment of inertia of the disk?

Answers

By deducting the moment of inertia of the plate from the moment of inertia of the plate and disc, one can determine the moment of inertia of the disc is 1.4 * 10(-4) kg m^2

 

We can determine the moment of inertia of the disc by multiplying [tex]1.74*10(-4) kg m^2[/tex] by the moment of inertia of the plate, which is  [tex]0.34 * 10(-4) kg m^2[/tex].

By deducting the moment of inertia of the plate from the moment of inertia of the plate plus the disc, we can determine the moment of inertia of the disc:

Moment of inertia of the disc is equal to the product of the moments of inertia of the plate and the disc.

Moment of inertia of the disc is equal to

[tex]1.74 * 10-4 kg/m^2 - 0.34 * 10-4 kg/m^2.[/tex]

The disk's moment of inertia is  [tex]1.4 * 10(-4) kg m^2[/tex]

As a result, the disk's moment of inertia is equal to 1.4 * 10(-4) kg m^2 .

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A short distance offshore are coral reefs that are important and fragile marine habitats. Oil arrives at the facility by separate pipelines from each of four onshore fields. The facility has the following main processing equipment:PIG receivers on each pipelineInlet metering on each pipelineA main manifold to combine flows from all pipelinesA heated separator to remove remaining water and gasA flare stack to allow rapid purging of hydrocarbons from any part of the plantThree oil storage tanks arranged so that they can be used in any combinationTwo oil export pumps arranged in parallelTwo parallel export metering trains to measure oil delivered to tankersA tanker loading facilityThe small quantity of gas recovered from the heated separator is used to provide fuel for the heater with any excess going to the flare. Water recovered in the heated separator is pumped into a shallow aquifer.Draw a simple high level process flow diagram of the components itemised above showing the path of all fluids through the facility.Suggest a control system you would expect to find on the separator in this scenario. For the control system you have chosen, suggest a measurement device that would be used and state what equipment would be adjusted by the control system.Sketch a graph of the parameter being controlled against time showing the response you would expect to a step change in set-point from A to B at time t=10 if your control system is well tuned. Your graph should also show: set-point; overshoot; and settling time. How does open science relate to Preregistration and Open Data& Code Open Access. Explain According to the IFE, if interest rates are 4% in the U.S. and 2% in Japan, what must be the approximate expected change in the value of the USD? A cube with edges of length 1 = 0.13 m and density Ps = 2.7 x 103kg/m3 is suspended from a spring scale. a. When the block is in air, what will be the scale reading? A cord is wrapped around the rim of a solid uniform wheel 0.270 m in radius and of mass 9.60 kg. A steady horizontal pull of 36.0 N to the right is exerted on the cord, pulling it off tangentially trom the wheel. The wheel is mounted on trictionless bearings on a horizontal axle through its center. - Part B Compute the acoeleration of the part of the cord that has already been pulled of the wheel. Express your answer in radians per second squared. - Part C Find the magnitude of the force that the axle exerts on the wheel. Express your answer in newtons. - Part D Find the direction of the force that the axle exerts on the wheel. Express your answer in degrees. Part E Which of the answers in parts (A). (B), (C) and (D) would change if the pull were upward instead of horizontal? 2. In a standard feedback control system, assume G(s)= 2e-02, G(s)= GGPG has the transfer function s+1. (a) If a PI controller is employed to control the system, please find the PI controller settings when the direct synthesis method with Te-1.0 was used based on the desired response (Y/Y)= e(ts+1). (b) By using this PI controller tuning parameter, what will be the gain margin and phase margin of the controlled system. why it is important to consider NPSH when designingand operating a pumping system. What are the pulmonary manifestations of Pink's disease and how common are they relative to the usual cerebral, skin and renal effects? Is it necessary to administer antivenom to a person bitten by a snake 6 hours previously, presenting only with local leg swelling over the bitten site and so far no other systemic feature of poisoning? Use power series to find two linearly independent solutions (about x= 0) for the DE: y 3x ^3 y +5xy=0 Which Africa's economic recovery from recessions may be characterised by smaller budget by deficits because (a)tax collection stagnate (b)government expenses on social grants rises (c) unemployment is at 23%according to Statistic South Africa (d)revenue collection grown faster than expenditure. crosson-tower, c. (2012). exploring child welfare: a practice perspective (7th ed.). boston, ma: allyn and bacon. "Fora converging lens with a 25.0cm focal length, an object with aheight of 6cm is placed 30.0cm to the left of the lensa. Draw a ray tracing diagram of the object and the resultingimages answers please?// to get a great answer