Answer:
A) 37 m
Explanation:
The car is moving of uniformly accelerated motion, so the distance it covers can be calculated by using the following SUVAT equation:
(1)
where
v = 0 m/s is the final velocity of the car
u = 24 m/s is the initial velocity
a is the acceleration
d is the length of the skid
We need to find the acceleration first. We know that the force responsible for the (de)celeration is the force of friction, so:

where
m = 1000 kg is the mass of the car
is the coefficient of friction
a is the deceleration of the car
g = 9.8 m/s^2 is the acceleration due to gravity
The negative sign is due to the fact that the force of friction is against the motion of the car, so the sign of the acceleration will be negative because the car is slowing down. From this equation, we find:

And we can substitute it into eq.(1) to find d:

Answer:
the velocity of the point P located on the horizontal diameter of the wheel at t = 1.4 s is 
Explanation:
The free-body diagram below shows the interpretation of the question; from the diagram , the wheel that is rolling in a clockwise directio will have two velocities at point P;
- the peripheral velocity that is directed downward
along the y-axis
- the linear velocity
that is directed along the x-axis
Now;


Also,

where
(angular velocity) = 

∴ the velocity of the point P located on the horizontal diameter of the wheel at t = 1.4 s is 
Answer:
Refractive index of unknown liquid = 1.56
Explanation:
Using Snell's law as:
Where,
is the angle of incidence ( 65.0° )
is the angle of refraction ( 53.0° )
is the refractive index of the refraction medium (unknown liquid, n=?)
is the refractive index of the incidence medium (oil, n=1.38)
Hence,
Solving for
,
Refractive index of unknown liquid = 1.56
Answer:
hmax=81ft
Explanation:
Maximum height of the object is the highest vertical position along its trajectory.
The vertical velocity is equal to 0 (Vy = 0)

we isolate th (needed to reach the maximum height hmax)

The formula describing vertical distance is:

So, given y = hmax and t = th, we can join those two equations together:


if we launch a projectile from some initial height h all you need to do is add this initial elevation

