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
Answer:
a)30.14 rad/s2
b)43.5 rad/s
c)60633 J
d)42 kW
e)84 kW
Explanation:
If we treat the propeller is a slender rod, then its moments of inertia is

a. The angular acceleration is Torque divided by moments of inertia:

b. 5 revolution would be equals to
rad, or 31.4 rad. Since the engine just got started


c. Work done during the first 5 revolution would be torque times angular displacement:

d. The time it takes to spin the first 5 revolutions is

The average power output is work per unit time
or 42 kW
e.The instantaneous power at the instant of 5 rev would be Torque times angular speed at that time:
or 84 kW
Answer:f^n=m(a+g)
Explanation:khan academy said
<u>Answer:</u>
<em>The moon doesn’t change shape on its own.</em>
<u>Explanation:</u>
Shapes of moon that we observe is based on the different perspectives of view from the earth and position of moon with respect to the sun. The changes arise due to the rotation of earth on its own axis as well as the revolution of moon on its orbit. The moon doesn’t have any light of its own.
It just reflects off the light from the sun. Due to tidal locking phenomenon one face of the moon permanently faces the sun. Because of the changes in position of moon with respect to the sun the moon is lighted up variably giving rise to various phases like new moon, full moon, crescent etc.