Answer
given,
initial speed of the car (v₁)= 19.8 mi/h
final speed of the car (v₂)= 59.9 mi/h
a) initial momentum = m v₁
P₁ = 19.8 m
final final momentum = m v₂
P₂ = 59.9 m
ratio =
=
ratio of momentum=
b) initial kinetic energy= 1/2 m v₁²
K₁ = 196.02 m
final kinetic energy= 1/2 m v₂²
K₂ = 1794.005 m
ratio =
=
ratio of Kinetic energy=
Answer:
Explanation:
Step one:
Given data
work-done in dragging the trash= 236J
applied force= 18.9N
distance moved= 24.4m
Required
The angle of the applied force
Step two:
We know that work done is
WD= F * distance
<em>The work is the product of the horizontal component of the force and the distance.
</em>
Horizontal force = 236 ÷ 24.4
= 9.67 N
Cos θ = Horizontal force ÷ Actual force
Cos θ = (236 ÷ 24.4) ÷ 18.9 = 236 ÷ 461.16
The angle is approximately 59˚
Answer B is the correct answer
We know that kinetic energy , where m is the mass of object and v is the velocity of object.
In this case only velocity is the variable, mass remains constant.
So point having higher velocity has higher kinetic energy.
When it leaves the racket, the ball will be having a certain height, but just before it reaches the ground it will not having any height. So maximum velocity of ball is at that time when it reaches just above the ground.
So option B is the correct answer.