Agymnast of mass 66.0 kg hangs from a vertical rope attached to the ceiling. you can ignore the weight of the rope and assume that the rope does not stretch. use the value 9.81m/s2 for the acceleration of gravity.

The question is incomplete.

These are the parts missing.

Part A:

Calculate the tension T in the rope if the gymnast hangs motionless on the rope.

Express your answer in newtons.

Part B:

Calculate the tension T in the rope if the gymnast climbs the rope at a constant rate.

Express your answer in newtons.

Part C:

Calculate the tension T in the rope if the gymnast climbs up the rope with an upward acceleration of magnitude 1.50 m/s2 .

Express your answer in newtons.

Part D:

Calculate the tension T in the rope if the gymnast slides down the rope with a downward acceleration of magnitude 1.50 m/s2 .

Express your answer in newtons.

This is how you answer each part.

Part A:

Calculate the tension T in the rope if the gymnast hangs motionless on the rope.

The acceleration is zero, so the net force is 0:

T - mg = 0

T = mg = 66kg × 9.81 m/s² = 647.46 N

Part B:

Calculate the tension T in the rope if the gymnast climbs the rope at a constant rate.

The acceleration is zero too, so yet the tension is T = 647.46 N

Part C:

Calculate the tension T in the rope if the gymnast climbs up the rope with an upward acceleration of magnitude 1.50 m/s2 .

Now the acceleration of the gymnast is 1.50m/s², so she is adding a force of ma, which turns the equation into:

T - mg = ma

T = mg + ma = m (g + a) = 66kg (9.81 + 1.5)m/s² = 746.46N

So, the new tension is T = 746.46N

Part D:

Calculate the tension T in the rope if the gymnast slides down the rope with a downward acceleration of magnitude 1.50 m/s2 .

Now, the acceleration is downward so it has the opposite sign of that of the part C.

T = m(g -a) = 66 kg (9.81 - 1.5)m/s² = 548.46

The new tension is T = 548.46N

answer; how far out of place which the object finished relative to the starting location;

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.