Which statement is true if the refractive index of medium A is greater than that of medium B?

Physics

1 answer:

allsm [11]3 years ago

7 0

Answer:

I think it is C

Explanation:

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A trailer mechanic pushes a 2500 kg car, home to a speed v, performing a job during the 5000 J. In this process, the car moves 2

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3 years ago

A packing crate slides down an inclined ramp at constant velocity. Thus we can deduce thatA) a frictional force is acting on itb

torisob [31]

Answer:

A) a frictional force is acting on it

Explanation:

The crate is sliding down the inclined ramp at constant velocity: constant velocity means zero acceleration, a = 0. According to Newton's second law:

F = ma

this also means that the net force acting along the direction of the slope is zero.

Without frictional force, there would be only one force acting along this direction: the component of the weight of the crane parallel to the slope, acting downward:

$W_p = mg sin \theta$

where $\theta$ is the angle of the incline. Therefore, the net force along this direction would not be zero. In order to have a net force of zero, there must be another force acting upward on the crate: and the only force that is possibly acting on the crate against its direction of motion is the frictional force, whose magnitude must be equal to $W_p$ , in order to produce a net force of zero (and therefore, a zero acceleration).

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4 years ago

What are some examples of solids, liquids, and gases? please help me itys my sons homework!!?

Sergio [31]

Solids wall cement chat door tree
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4 years ago

Where on this diagram does the ball have the highest point of gravitational potential energy?

mixer [17]

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3 years ago

Which formulas have been correctly rearranged to solve for radius? Check all that apply. r = GM central/v^2 r =fcm/v^2 r =ac/v^2

jek_recluse [69]

The orbital radius is: $r=\frac{GM}{v^2}$

Explanation:

The problem is asking to find the radius of the orbit of a satellite around a planet, given the orbital speed of the satellite.

For a satellite in orbit around a planet, the gravitational force provides the required centripetal force to keep it in circular motion, therefore we can write:

$\frac{GMm}{r^2}=m\frac{v^2}{r}$