Answer:
U = 80.91 J
Explanation:
In order to calculate the electric potential energy between the three charges you use the following formula:
(1)
k: Coulomb's constant = 8.98*10^9Nm^2/C^2
q1: q2 charge
r1,2: distance between charges 1 and 2.
For the three charges you have:
(2)
You use the fact that q1=q2=q3=q and that the distance between charges are equal. Then, in the equation (2) you have:
q = 1.45μC = 1.45*10^-6C
r = 0.700mm = 0.700*10^-3m

The electric potential energy between the three charges is 80.91 J
Answer: 2.92 s
Explanation:
Given
Mass of ball is 
The initial velocity of the ball is 
Velocity after the rebound is 
Force during the contact is 
We know, change in momentum is Impulse


Thus, the force is applied for 2.92 s
To solve the problem it is necessary to use Newton's second law and statistical equilibrium equations.
According to Newton's second law we have to

where,
m= mass
g = gravitational acceleration
For the balance to break, there must be a mass M located at the right end.
We will define the mass m as the mass of the body, located in an equidistant center of the corners equal to 4m.
In this way, applying the static equilibrium equations, we have to sum up torques at point B,

Regarding the forces we have,

Re-arrange to find M,



Therefore the maximum additional mass you could place on the right hand end of the plank and have the plank still be at rest is 16.67Kg
6. Since we are not sure if the person in the question is actively lifting the crate, we have to determine the downwards force of the crate due to gravity and compare it to the normal force.
F = ma
F = (15.3)(-9.8)
F = -150N
Since the downwards force of the crate is equivalent to the normal force, it means the person is applying no force in picking up the object. So to pick up a 150N object from scratch, you would have to exert more force than the weight of the object, so the answer is 294N.
7. Same idea as question 2.
First determine the weight of the object:
F = ma
F = (30)(-9.8)
F = -294N
The crate in question is not moving, so the magnitudes of the forces in the upwards and downwards direction has to equal to 0.
-294 + 150N + x = 0
x = 144N
So the person is exerting 144 N.
10. First find the force of block B to the right due to its acceleration:
F = ma
F = (24)(0.5)
F = 12N
So block B is moving 12N to the right relative to block A due to block A's movement to the left. However, block A is being applied a much greater force and is moving quicker to the left than block B is moving to the right of bock A. The force that is causing block B to experience the lower relative force to the right is because of the friction. To find the friction:
The sum of the forces in the leftward and rightward direction for block B must equal 12N.
75 - x = 12
x = 63N
So the force of friction of block A on block B is 63N to the left.
Answer:
work done is -150 kJ
Explanation:
given data
volume v1 = 2 m³
pressure p1 = 100 kPa
pressure p2 = 200 kPa
internal energy = 10 kJ
heat is transferred = 150 kJ
solution
we know from 1st law of thermodynamic is
Q = du +W ............1
put here value and we get
-140 = 10 + W
W = -150 kJ
as here work done is -ve so we can say work is being done on system