Answer:
318 Earth-pull
Explanation:
Using Newton's law of gravitational force
F = GMm / r²
F is directly proportional to mass
let earth = m
then m .......... Earth-pull
318 m = 318 m × Earth-pull / m = 318 Earth-pull
Answer:
ΔT = 0.02412 s
Explanation:
We will simply calculate the time for both the waves to travel through rail distance.
FOR THE TRAVELING THROUGH RAIL:

FOR THE WAVE TRAVELING THROUGH AIR:

The separation in time between two pulses can now be given as follows:

<u>ΔT = 0.02412 s</u>
Answer: P = 40 W
Explanation: Power is defined as the amount of work per unit time or expressed in the formula of P = W /t
Work is the product of force and distance so W = Fd.
So the working equation will be:
P = Fd / t
= 20 N ( 2.0 m) / 1 s
= 40 W
Is the amount of power exerted.
Answer:
a) t1 = v0/a0
b) t2 = v0/a0
c) v0^2/a0
Explanation:
A)
How much time does it take for the car to come to a full stop? Express your answer in terms of v0 and a0
Vf = 0
Vf = v0 - a0*t
0 = v0 - a0*t
a0*t = v0
t1 = v0/a0
B)
How much time does it take for the car to accelerate from the full stop to its original cruising speed? Express your answer in terms of v0 and a0.
at this point
U = 0
v0 = u + a0*t
v0 = 0 + a0*t
v0 = a0*t
t2 = v0/a0
C)
The train does not stop at the stoplight. How far behind the train is the car when the car reaches its original speed v0 again? Express the separation distance in terms of v0 and a0 . Your answer should be positive.
t1 = t2 = t
Distance covered by the train = v0 (2t) = 2v0t
and we know t = v0/a0
so distanced covered = 2v0 (v0/a0) = (2v0^2)/a0
now distance covered by car before coming to full stop
Vf2 = v0^2- 2a0s1
2a0s1 = v0^2
s1 = v0^2 / 2a0
After the full stop;
V0^2 = 2a0s2
s2 = v0^2/2a0
Snet = 2v0^2 /2a0 = v0^2/a0
Now the separation between train and car
= (2v0^2)/a0 - v0^2/a0
= v0^2/a0