Answer:
Moment of the force is 20 N-m.
Explanation:
Given:
Force exerted by the person is, 
Distance of application of force from the point about which moment is needed is, 
Now, we know that, moment of a force 'F' about a point at a perpendicular distance of 'd' from the same point is given as the product of the force and the perpendicular distance.
Therefore, the moment of the force about the end of the claw hammer is given as:

Hence, the moment of the force exerted by the person about the end of the claw hammer is 20 N-m.
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
Answer:
1. Acceleration
Explanation:
Newtons Second law gives the measure of acceleration
The universe is 13.8 billion years old.
Answer:
The force is the same
Explanation:
The force per meter exerted between two wires carrying a current is given by the formula

where
is the vacuum permeability
is the current in the 1st wire
is the current in the 2nd wire
r is the separation between the wires
In this problem

Substituting, we find the force per unit length on the two wires:

However, the formula is the same for the two wires: this means that the force per meter exerted on the two wires is the same.
The same conclusion comes out from Newton's third law of motion, which states that when an object A exerts a force on an object B, then object B exerts an equal and opposite force on object A (action-reaction). If we apply the law to this situation, we see that the force exerted by wire 1 on wire 2 is the same as the force exerted by wire 2 on wire 1 (however the direction is opposite).