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dexar [7]
3 years ago
14

If you place a free electron and a free proton in the same electric field, how will the forces acting on them compare? 82. How w

ill the accelerations of the proton and the electron in the preceding problem compare? 83. How will the directions of travel compare for the electron and the proton in the preceding problem?
Physics
1 answer:
Paha777 [63]3 years ago
3 0

Answer:

the force experienced by both particles will be of same magnitude as the magnitude of charge is same for both particles but with different sign, the acceleration of both particles will be same in magnitude but opposite in direction, the direction of free electron will be opposite to the direction of electric field whereas the direction of proton will be in the direction of electric field,

Explanation:

the force experienced by both particles can be calculated by the formula

F=|q|E

the acceleration can be calculated by Newton's 2nd law i.e F=ma

the direction of electric field is always towards the positive charge so proton will be directed towards electric field and free electron will be directed away from the electric field.

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Assuming the incline to be frictionless and the zero of gravitational potential energy to be at the elevation of the horizontal
Leya [2.2K]

The energy in the system is given by the law of conservation of energy.

The energy stored in the spring plus the gravitational potential energy of the block when it has fully compressed the spring will be equal to m·g·h.

Reason:

The given parameters are;

Surface of the inclined plane = Frictionless

Potential energy at the horizontal line = Zero of gravitational potential

By the law of Conservation of Energy, we have;

The spring will be compressed by a distance, <em>x</em>

The energy stored in the compressed spring, K.E. = (1/2)·k·x²

Energy in the block when the block comes to rest at a height, h₁, will be, P.E. = m·g·h₁

Therefore, by conservation of energy, we have;

The initial potential of the block = The stored energy in the compressed string + The gravitational potential energy of the block when it has compressed.

Therefore, the correct option is; <u>The energy stored in the spring plus the gravitational potential energy of the block when it has fully compressed the spring will be equal to m·g·h</u>

Learn more here;

brainly.com/question/17713698

4 0
2 years ago
Two identical freight cars roll without friction (one at 1 m/s, the other at 2 m/s) toward each other on a level track. They col
Kazeer [188]

Answer:

They collide, couple together, and roll away in the direction that <u>the 2m/s car was rolling in.</u>

Explanation:

We should start off with stating that the conservation of momentum is used here.

Momentum = mass * speed

Since, mass of both freight cars is the same, the speed determines which has more momentum.

Thus, the momentum of the 2 m/s freight car is twice that of the 1 m/s freight car.

The final speed is calculated as below:

mass * (velocity of first freight car) + mass * (velocity of second freight car) = (mass of both freight cars) * final velocity

(m * V1) + (m * V2) = (2m * V)

Let's substitute the velocities 1m/s for the first car, and - 2m/s for the second. (since the second is opposite in direction)

We get:

m*1 + m*(-2) = 2m*V

solving this we get:

V = - 0.5 m/s

Thus we can see that both cars will roll away in the direction that the 2 m/s car was going in. (because of the negative sign in the answer)

7 0
3 years ago
A 26.0 kg child plays on a swing having support ropes that are 2.40 m long. A friend pulls her back until the ropes are 45.0 ∘ f
Sloan [31]
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8 0
3 years ago
A 34-m length of wire is stretched horizontally between two vertical posts. The wire carries a current of 68 A and experiences a
il63 [147K]

Answer:

7.28×10⁻⁵ T

Explanation:

Applying,

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make B the subject of the equation

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From the question,

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Substitute these values into equation 2

B = 0.16/(68×34×sin72°)

B = 0.16/(68×34×0.95)

B = 0.16/2196.4

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2 years ago
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Most earthquakes occur along or near the edges of the earth's lithospheric<span> plate. </span>
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