For the work-energy theorem, the work needed to stop the bus is equal to its variation of kinetic energy:
where
W is the work
Kf is the final kinetic energy of the bus
Ki is the initial kinetic energy of the bus
Since the bus comes at rest, its final kinetic energy is zero:
, so the work done by the brakes to stop the bus is
And the work done is negative, because the force applied by the brake is in the opposite direction to that of the bus motion.
<span>¿Qué estás pidiendo en esta situación. Hay entornos de diferencia de los animales. Algunos viven en entornos de tundra, otra veraniega en vivo, ambientes cálidos.</span>
At the time that I'll call ' Q ', the height of the stone that was
dropped from the tower is
H = 50 - (1/2 G Q²) ,
and the height of the stone that was tossed straight up
from the ground is
H = 20Q - (1/2 G Q²) .
The stones meet when them's heights are equal,
so that's the time when
<span>50 - (1/2 G Q²) = 20Q - (1/2 G Q²) .
This is looking like it's going to be easy.
Add </span><span>(1/2 G Q²) to each side.
Then it says
50 = 20Q
Divide each side by 20: 2.5 = Q .
And there we are. The stones pass each other
2.5 seconds
after they are simultaneously launched.
</span>
<span>if we assume the origin is at the dropping point and the object is merely dropped and not thrown up or down then y0 = 0 and v0 = 0. The equation reduces to </span>
<span>y = 0 + 0t + ½gt² </span>
<span>y = ½gt² </span>
<span>t = √(2y/g) </span>
<span>in the ft - lb - s system </span>
<span>y = -100 ft </span>
<span>g = -32.2 ft / s² </span>
<span>t = √(2y/g) </span>
<span>t = √(2(-100) / (-32.2)) </span>
<span>t = 2.5 s</span>
Answer:
The velocity of the man from the frame of reference of a stationary observer is, V₂ = 5 m/s
Explanation:
Given,
Your velocity, V₁ = 2 m/
The velocity of the person, V₂ =?
The velocity of the person relative to you, V₂₁ = 3 m/s
According to the relative velocity of two
V₂₁ = V₂ -V₁
∴ V₂ = V₂₁ + V₁
On substitution
V₂ = 3 + 2
= 5 m/s
Hence, the velocity of the man from the frame of reference of a stationary observe is, V₂ = 5 m/s
