There is a displacement. Just because the ball is thrown up,
and not crossways, doesn't mean its location is not moving. Remember, positive
displacement is together a displacement in the direction east, right, and up.
The velocity is the distance over time. To compute that, you must look how high
the ball moved before falling back down. Acceleration is expected to be
constant at 9.80m/s^2. That is the force of gravity. But remember that you are disregarding
air friction when you are computing the acceleration.
Answer:
7 m/s^2
Explanation:
Given that the jet is traveling 37.6 m/s when the pilot receives the message.
And it takes the pilot 5.37 s to bring the plane to a halt.
Acceleration of the plane can be calculated by using first equation of motion
V = U - at
Since the plane is going to stop, the final velocity V = zero.
And the acceleration will be negative
Substitute all the parameters into the formula
0 = 37.6 - 5.37a
5.37a = 37.6
Make a the subject of formula
a = 37.6 / 5.37
a = 7.0 m/s^2
Therefore, the acceleration of the plane to bring the plane to a halt is 7 m/s^2
Answer:
0.167m/s
Explanation:
According to law of conservation of momentum which States that the sum of momentum of bodies before collision is equal to the sum of the bodies after collision. The bodies move with a common velocity after collision.
Given momentum = Maas × velocity.
Momentum of glider A = 1kg×1m/s
Momentum of glider = 1kgm/s
Momentum of glider B = 5kg × 0m/s
The initial velocity of glider B is zero since it is at rest.
Momentum of glider B = 0kgm/s
Momentum of the bodies after collision = (mA+mB)v where;
mA and mB are the masses of the gliders
v is their common velocity after collision.
Momentum = (1+5)v
Momentum after collision = 6v
According to the law of conservation of momentum;
1kgm/s + 0kgm/s = 6v
1 =6v
V =1/6m/s
Their speed after collision will be 0.167m/s
Answer:
t = 0.33h = 1200s
x = 18.33 km
Explanation:
If the origin of coordinates is at the second car, you can write the following equations for both cars:
car 1:
(1)
xo = 10 km
v1 = 55km/h
car 2:
(2)
v2 = 85km/h
For a specific value of time t the positions of both cars are equal, that is, x=x'. You equal equations (1) and (2) and solve for t:
The position in which both cars coincides is:
Answer:
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