Answer:
1.04 s
Explanation:
The computation is shown below:
As we know that
t = t' × 1 ÷ (√(1 - (v/c)^2)
here
v = 0.5c
t = 1.20 -s
So,
1.20 = t' × 1 ÷ (√(1 - (0.5/c)^2)
1.20 = t' × 1 ÷ (√(1 - (0.5)^2)
1.20 = t' ÷ √0.75
1.20 = t' ÷ 0.866
t' = 0.866 × 1.20
= 1.04 s
The above formula should be applied
In order to persuade the electrons in the wire to flow, you need
a potential difference between the ends of the wire. Then the
electrons will want to get away from the more-negative end and
go to the more-positive end. If both ends of the wire are at the
same potential, then the electrons have no reason to go anywhere,
and they just stay where they are.
Choice-d says this.
This question is incomplete, the complete question is;
A weightlifter holds a 1,300 N barbell 1 meter above the ground. One end of a 2-meter-long chain hangs from the center of the barbell. The chain has a total weight of 400 N. How much work (in J) is required to lift the barbell to a height of 2 m?
What is the average force (average with respect to height of the barbell from the ground) exerted by the weightlifter in the process?
Answer: Average force exerted by the weightlifter in the process = 1600N
Explanation:
To find Work done to lift a barbell and half of the hanging chain we say;
W₁ = ( 1300N + (1/2 × 400N)) × 1m
W₁ = (1300 + 200) Nm
W₁ = 1500J
now work done to lift the upper half of the chain we say:
W₂ = (1/2 × 400N) × (1/2 × 1m)
W₂ = 200N × 0.5m
W₂ = 100J
So total work done will be
W = W₁ + W₂
W = 1500J + 100J
W = 1600J
To find the average force exerted by the weight lifter, we say;
F = W/D
F = (1600 / 1m) N
F = 1600N
∴Average force = 1600N
