Asmall block slides without friction along a track toward a circular loop. the block has more than enough speed to remain firmly in contact with the track as it goes around the loop. the magnitude of the block’s acceleration at the top of the loop isa) zero b) greater than zero but less than g c) equal to g d) greater than g

d) greater than g

Explanation:

You have to apply Newton's Second Law to have an idea about the value of the block's aceleration.

∑F = ma

Where m is the mass, a is the acceleration and F represents all the forces involved.

You have to draw a free body diagram for the block at the top of the loop. (See the attachment)

Applying Newton's Second Law:

y: N+mg=ma

Where N is the normal force (the contact force that the loop's surface applies to the block) and mg is the weight.

Therefore a= + g

So, you can notice that the magnitude of the block's acceleration is greater than the acceleration of gravity because of the effect of the magnitude N/m.

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.