Time= s/v
Speed =5km/h
Time=30min
Distance is required
Distance=time*speed
30min*5km/h=600m
Answer:
Given that,
- Power = 2000 W
- time = 60 seconds
- distance= 10m
Power = work done ÷ time
Here, since the movement is vertical, w = mgh
So,
Power = mgh÷t
2000 = (m × 9.8 ×10) ÷ 60
m = (2000 ×60) ÷98
m = 1224.5kg
Hi there!
Question - Which of the following statements is true?
Answer - C. You are exposed to nuclear radiation every day.
Why - "radiation in the form of elementary particles emitted by an atomic nucleus, as alpha rays or gamma rays, produced by decay of radioactive substances or by nuclear fission."
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
Sure what do u need help with
