The initial kinetic energy of the car is

Then, the velocity of the car is decreased by half:

so, the new kinetic energy is

So, the new kinetic energy is 1/4 of the initial kinetic energy of the car. Numerically:
If we pull an object vertically upwards then we need to apply a force which is equal in the magnitude of the weight of the object

now when we pull the same object upwards along an inclined plane with angle then we require a force which will balance the component of weight along the inclined
so it is given as

so as if we compare the two forces we can say that since the value of sine is always less than 1 for an angle less than 90 degree
so in the 2nd case when we pull the object along the inclined plane it will require less effort
so correct answer is
<em>A. reduce effort</em>
The answer is C voltmeter
Answer:
a)3.5s
b)28.57m/S
c)34.33m/S
d)44.66m/S
Explanation:
Hello!
we will solve this exercise numeral by numeral
a) to find the time the ball takes in the air we must consider that vertically the ball experiences a movement with constant acceleration whose value is gravity (9.81m / S ^ 2), that the initial vertical velocity is zero, we use the following equation for a body that moves with constant acceleration

where
Vo = Initial speed
=0
T = time
g=gravity=9.81m/s^2
y = height=60m
solving for time

T=3.5s
b)The horizontal speed remains constant since there is no horizontal acceleration.
with the value of the distance traveled (100m) and the time that lasts in the air (3.5s) we estimate the horizontal speed

c)
to find the final vertical velocity we use the equations for motion with constant velocity as follows
Vf=Vo+g.t
Vf=0+(9.81 )(3.5)=34.335m/S
d)Finally, to find the resulting velocity, we add the horizontal and vertical velocities vectorially, this is achieved by finding the square root of the sum of its squares
