A4 kg object hangs below a 6 kg object by a string of negligible mass. if the 6 kg object is pulled upward by a force of 440 n, what is the tension in the string?

176 N

Explanation:

Sum of the forces on the 6 kg object:

∑F = ma

F − Mg − T = Ma

Sum of the forces on the 4 kg object:

∑F = ma

T − mg = ma

Notice that since the objects are connected, they have the same acceleration.

Solving for acceleration in both equations:

a = (F − Mg − T) / M

a = (T − mg) / m

Set equal:

(F − Mg − T) / M = (T − mg) / m

Cross multiply and solve for T:

m (F − Mg − T) = M (T − mg)

mF − mMg − mT = MT − mMg

mF − mT = MT

mF = mT + MT

mF = (m + M) T

T = mF / (m + M)

Given F = 440 N, m = 4 kg, and M = 6 kg:

T = (4) (440) / (4 + 6)

T = 176 N

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.