To what i see, the answer is....
C.
Answer: a. F doubled
b. F reduced by one-quarter i.e
1/4*(F)
c. 1/9*(F)
d. F increased by a factor of 4 i.e 4*F
e. F reduces 3/4*(F)
Explanation: Coulombs law states the force F of attraction/repulsion experience by two charges qA and qB is directly proportional to thier product and inversely proportional to the square of distance d between them. That is
F = k*(qA*qB)/d²
a. If qA is doubled therefore the force is doubled since they are directly proportional.
b. If qA and qB are half, that means thier new product would be qA/2)*qB/2 =qA*qB/4
Which means the product of charge is divided by 4 so the force would be divided by 4 too since they are directly proportional.
c. If d is tripped that is multiplied by 3. From the formula new d would be (3*d)²=9d² but force is inversely proportional to d² so instead of multiplying by 9 the force will be divided by 9
d. If d is cut into half that is divided by 2. The new d would be (d/2)²=d²/4. So d² is divided by 4 so the force would be multiplied by 4
e. If qA is tripled that is multiplied by 3. F would be multiplied by 3 also, if at the same time d is doubled (2*d)²= 4*d² . Force would be divided by 4 at same time. So we have,
3/4*F
The net force acting on the car is 3 which slows the car down.
Explanation:
The given data is as follows.
F = 3.2 N, m = 18.2 kg,
t = 0.82 sec
(a) Formula for impulse is as follows.
I = Ft = 
Ft = 
or, 
Putting the given values into the above formula as follows.
= 
= 0.144 m/s
Therefore, final velocity of the mass if it is initially at rest is 0.144 m/s.
(b) When velocity is 1.85 m/s to the left then, final velocity of the mass will be calculated as follows.
Ft =
or,
= 
= -1.705 m/s
Hence, we can conclude that the final velocity of the mass if it is initially moving along the x-axis with a velocity of 1.85 m/s to the left is 1.705 m/s towards the left.
Answer:
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