Explanation:
Δy = v₀ t + ½ at²
Δy = (0 m/s) (7.0 s) + ½ (-9.8 m/s²)(7.0s)²
= -34.3
Answer:
The magnitude of the force the garbage exerts on the floor of the service elevator is 12 N
Explanation:
Given;
mass of the garbage, m = 10.0 kg
acceleration of the elevator, a = 1.2 m/s²
The magnitude of the force the garbage exerts on the floor of the service elevator is given by;
F = ma
F = 10 x 1.2
F = 12 N
Therefore, the magnitude of the force the garbage exerts on the floor of the service elevator is 12 N
Answer:
F = 1024 N
Explanation:
Given that,
Mass of the cart, m = 80 kg
Radius of the loop, r = 5 m
Speed of the cart, v = 8 m/s
To find,
the force exerted on the track on the cart.
Solution,
The force exerted on the track on the cart is called centripetal force. When an object moves in a circular path it possess centripetal force. It is given by :


F = 1024 N
So, the force exerted on the track on the cart is 1024 N.