A 2200 kilogram car is accelerating at 3.4 m/s/s. what is the NET force?
1 answer:
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Answer:
The velocity is 19.39 m/s
Solution:
As per the question:
Mass, m = 75 kg
Radius, R = 19.2 m
Now,
When the mass is at the top position in the loop, then the necessary centrifugal force is to keep the mass on the path is provided by the gravitational force acting downwards.
where
v = velocity
g = acceleration due to gravity
Answer:
the distance that the object is raised above its initial position is 5.625 m.
Explanation:
Given;
applied effort, E = 15 N
load lifted by the ideal pulley system, L = 16 N
distance moved by the effort, d₁ = 6 m
let the distance moved by the object = d₂
For an ideal machine, the mechanical advantage is equal to the velocity ratio of the machine.
M.A = V.R
Therefore, the distance that the object is raised above its initial position is 5.625 m.