Answer:
-0.64525g
Explanation:
t = Time taken for the car to stop
u = Initial velocity = 95 km/h
v = Final velocity = 0 km/h
s = Displacement
a = Acceleration
Equation of motion
Converting to m/s²
g = Acceleration due to gravity = 9.81 m/s²
Dividing both the accelerations, we get
Hence, acceleration of the car is -0.64525g
Answer:
The answer to your question is : vf = 15.18 m/s
Explanation:
Data
vo = 24 m/s
d = 120 m
vf = ? when d = 60.0 m
Formula
vf² = vo² + 2ad
For d =100m
a = (vf² - vo²) / 2d
a = (0 -24²) / 2(100)
a = -576/200
a = 2.88 m/s²
Now, when d = 60
vf² = (24)² - 2(2.88)(60)
vf² = 576 - 345.6
vf² = 230.4
vf = 15.18 m/s
Explanation:
Both graphs show plotted points forming a curved line. Curved lines have changing slope; they may start with a very small slope and begin curving sharply (either upwards or downwards) towards a large slope. In either case, the curved line of changing slope is a sign of accelerated motion (i.e., changing velocity).
Answer:
OXYGEN
Explanation:brainlyist me
Answer:
192.08J
19.6m/s
Explanation:
Since there will be no potential energy when the ball is on the ground, the change in potential energy is equal to the potential energy at the start when the ball is 19.6m above the ground.
PE=mgh
=(1)(9.8)(19.6)
=192.08J
v²=u²+2as, where v is the final velocity, u is initial velocity, a is acceleration and s is distance. Initial velocity is 0 since it starts at rest.
v²=u²+2as
v²=0²+2(9.8)(19.6)
v=√384.16
=19.6m/s
