I think the answer is: 6 inches
On average, the distance between Car's engine to the ground is around 8-10 inches.. But for small cars, the distance between car's engine to the ground could be as low as 5 inches - 6 inches from the ground.
When the flood reach the car's engine level, the water will got into the engine and disrupt it function, causing the car became unable to work properly
Answer: h = 0.30 m
Explanation:
A person jumping from height h would possess potential energy = m g h
which will convert completely into kinetic energy as person hits the ground. Now, the maximum energy absorbed by the person can be = 200 J
m = 67 kg
g = 9.8 m/s²
⇒ m g h = 200 J
⇒ h = 200 J / (67 kg × 9.8 m/s²) = 0.30 m
Hence, a person can land safely on both legs without breaking them from a height of 0.30 m only.
"Free fall" is the motion of an object when gravity is the ONLY force
acting on it.
In true 'free fall' the speed of an object increases at a constant rate
for the total duration of the fall. The rate of increase, on or near the
Earth's surface, is 9.8 meters per second for each second of fall.
True free fall is almost impossible to observe in everyday life, because
whenever we see anything falling, it's almost always falling through air,
so gravity is NOT the only force acting on it. The friction due to the
motion through air works against the gravitational force. In many cases,
the result is that the object's speed eventually stops increasing and
becomes constant, at a speed often described with the faux technical,
high-fallutin' sounding phrase "terminal velocity". It must be understood
that 'terminal velocity' is NOT a property of gravity or of free fall, but is
only a result of falling through some surrounding stuff that interferes with
the process of true 'free fall'.
Answer:
Can't answer without a picture
Explanation:
Answer:
Explanation:
Given that,
Initial angular velocity is 0
ωo=0rad/s
It has angular velocity of 11rev/sec
ωi=11rev/sec
1rev=2πrad
Then, wi=11rev/sec ×2πrad
wi=22πrad/sec
And after 30 revolution
θ=30revolution
θ=30×2πrad
θ=60πrad
Final angular velocity is
ωf=18rev/sec
ωf=18×2πrad/sec
ωf=36πrad/sec
a. Angular acceleration(α)
Then, angular acceleration is given as
wf²=wi²+2αθ
(36π)²=(22π)²+2α×60π
(36π)²-(22π)²=120πα
Then, 120πα = 8014.119
α=8014.119/120π
α=21.26 rad/s²
Let. convert to revolution /sec²
α=21.26/2π
α=3.38rev/sec
b. Time Taken to complete 30revolution
θ=60πrad
∆θ= ½(wf+wi)•t
60π=½(36π+22π)t
60π×2=58πt
Then, t=120π/58π
t=2.07seconds
c. Time to reach 11rev/sec
wf=wo+αt
22π=0+21.26t
22π=21.26t
Then, t=22π/21.26
t=3.251seconds
d. Number of revolution to get to 11rev/s
∆θ= ½(wf+wo)•t
∆θ= ½(0+11)•3.251
∆θ= ½(11)•3.251
∆θ= 17.88rev.
