<h2>
Answer:</h2>
4.2 C
<h2>
Explanation:</h2>
The charge (Q) moving is the product of the current(I) flowing through the torso of the person and the time taken (t) for the flow.
i.e
Q = I x t
Where;
I = current = 14.0A
t = time taken = 0.0300s
Substituting the values of I and t into the equation above gives
Q = 14.0 x 0.0300
Q = 4.2 C
Therefore quantity of charge moving is 4.2 C
Answer:
(a) Friction force = 50 N
(b) Work done by friction = 300 j
(c) Net work done = 0 j
Explanation:
We have given that the box is pulled by 6 meter so d = 6 m
Force applied on the box F = 60 N
We have have given that velocity is constant so acceleration will be zero
So to applied force will be utilized in balancing the friction force
So friction force 
Work done by friction force 
Work done by applied force 
So net work done = 300-300 = 0 j
In component form, the displacement vectors become
• 350 m [S] ==> (0, -350) m
• 400 m [E 20° N] ==> (400 cos(20°), 400 sin(20°)) m
(which I interpret to mean 20° north of east]
• 550 m [N 10° W] ==> (550 cos(100°), 550 sin(100°)) m
Then the student's total displacement is the sum of these:
(0 + 400 cos(20°) + 550 cos(100°), -350 + 400 sin(20°) + 550 sin(100°)) m
≈ (280.371, 328.452) m
which leaves the student a distance of about 431.8 m from their starting point in a direction of around arctan(328.452/280.371) ≈ 50° from the horizontal, i.e. approximately 431.8 m [E 50° N].
9 is D I believe, I don't know about 10
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.