Answer:
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F=ma
for a Velocity/time
22/20
1.1
F=1.1. * 1100
F=1210newton
Answer:
w = 0.189 rad/ s
Explanation:
This exercise we work with the conservation of the moment, the system is made up of the merry-go-round and the child, for which we write the moment of two instants
Initial
L₀ = I₀ w₀
Final
= I w
L₀ = L_{f}
I₀ w₀ = I_{f} w
.w = I₀/I_{f} w₀
The initial moment of inertia is
I₀ = 500 kg. m2
The final moment of inertia
= 500 + m r²
I_{f} = 500 + 20 1.5
I_{f} = 530 kg m²
Initial angular velocity
w₀ = 0.20 rad / s
Let's calculate
w = 500/530 0.20
w = 0.189 rad / s
Answer:
110 N
Explanation:
When a force is applied on a body and body does not move, it means the body remains at rest.
In this condition, there is a contact force between the body and the floor which is called static friction.
Th static friction force is a self adjusting force and comes into play when the body is at rest.
Here, the applied force is 110 N and the chest is not moving, that means a static friction force is acting between the chest and the floor. This static friction force is the force of contact between the chest and the floor. The static friction force is equal to the applied force when the body does not move.
So, the contact force between the chest and the floor is 100 N.