Answer:
b) True. the force of air drag on him is equal to his weight.
Explanation:
Let us propose the solution of the problem in order to analyze the given statements.
The problem must be solved with Newton's second law.
When he jumps off the plane
fr - w = ma
Where the friction force has some form of type.
fr = G v + H v²
Let's replace
(G v + H v²) - mg = m dv / dt
We can see that the friction force increases as the speed increases
At the equilibrium point
fr - w = 0
fr = mg
(G v + H v2) = mg
For low speeds the quadratic depended is not important, so we can reduce the equation to
G v = mg
v = mg / G
This is the terminal speed.
Now let's analyze the claims
a) False is g between the friction force constant
b) True.
c) False. It is equal to the weight
d) False. In the terminal speed the acceleration is zero
e) False. The friction force is equal to the weight
Assuming you are looking for the acceleration a:
1.
2.
where T is the tension and a is the acceleration of the blocks. The acceleration of the two blocks and the acceleration of the pulley must be equal.
The torque on the pulley is given by:
3.
where
and
.
Combining the three equations:
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Answer:
5 ms-2
Explanation:
F = ma
F = 100N
m = 20kg ( you should make sure the unit is kg before you answer the question)
100 = 20a
a = 100÷ 20
a = 5 ms-2
Answer:
Explanation:
When two forces acting on a line of action and they are equal in magnitude but opposite it direction, it forms a couple.
Torque is defined as the product of either force and the perpendicular distance between the two forces.
It is a vector quantity.
The net torque is zero, it means the anticlockwise torque is equal to the clockwise torque.
It means they balances each other.
