We can use the kinematic equation
where Vf is what we are looking for
Vi is 0 since we start from rest
a is acceleration
and d is the distance
we get
(Vf)^2 = (0)^2 + 2*(2)*(500)
(Vf)^2 = 2000
Vf = about 44.721
or 44.7 m/s [if you are rounding this by significant figures]
Answer:
b. electric potential energy.
Explanation:
The energy required to move a charge against the electric field is known as the electric potential energy. As in above case positively charged body is exerting an electric field on the positive charge. As the same charges repel so the charge tend to move away. In order to push it towards the body we need a work done. As it is hard to push the positive charged particle towards the positive electric field. So in the cases like these particle occupies the electric potential energy.
Answer:
D
Explanation:
We know the formula for Work to be:
W = f * d
Where W is work done
f is force
d is the distance
A)
Work = 50
Distance = 50
So, Force is:
Force = 50/50 = 1
B)
Work = 400
Distance = 80
Force = 400/80 = 5
C)
Work = 365
Distance = 73
Force = 365/73 = 5
D)
Work = 144
Distance = 16
Force = 144/16 = 9
Hence, D is the situation in which the force applied is the greatest.
The correct answer is 1.25 because it is 1/2 of 1 1/2 and that is 1.25.
