An open train car, with a mass of 2130 kg, coasts along a horizontal track at the speed 2.93 m/s. The car passes under a loading chute and, as it does so, gravel falls vertically into it for 3.03 s at the rate of 465 kg/s. What is the car's speed v f after the loading is completed

The final speed after the loading is completed is 1.764 m/s

Explanation:

Given;

mass of open train car, m₁ = 2130 kg

initial speed of the car, u₁ = 2.93 m/s

loading rate of gravel = 465 kg/s

time of loading = 3.03 s

mass of the gravel, m₂ = 465 kg/s x 3.03 s = 1408.95 kg

From the principle of conservation of linear momentum

m₁u₁ + m₂u₂ = v(m₁+m₂)

where;

v is the final speed after the loading is completed

2130 x 2.93 + 0 = v (2130 + 1408.95)

6240.9 = v (3538.95)

v = 6240.9/3538.95

v = 1.764 m/s

Therefore, the final speed after the loading is completed is 1.764 m/s

305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures