A hydraulic system is used to lift a 20,000 N vehicle in an auto garage. If the vehicle sits on

Physics

1 answer:

ollegr [7]2 years ago

7 0

Answer:

Explanation:

Area covered by the vehicle on the piston A1 = 0.5 m2 Weight of the vehicle, F1 = 2000 kg × 9.8 m s-2 Area on which force F2 is applied, A = 0.03 m2 P1 = P2 ; F 1 A 1 F1A1 = F 2 A 2 F2A2 and F2 = F 1 A 1 F1A1 A2 ; F2 = (2000 × 9.8) 0.03 0.5 0.030.5 = 1176 NRead more on Sarthaks.com - https://www.sarthaks.com/957542/hydraulic-system-used-lift-2000-vehicle-in-an-auto-garage-if-the-vehicle-sits-on-piston-area

You might be interested in

A cat dozes on a stationary merry-go-round, at a radius of 4.4 m from the center of the ride. The operator turns on the ride and

monitta

Answer:

The coefficient of static friction is 0.29

Explanation:

Given that,

Radius of the merry-go-round, r = 4.4 m

The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 7.7 s.

We need to find the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding. For this the centripetal force is balanced by the frictional force.

$\mu mg=\dfrac{mv^2}{r}$