Answer:

Negative sign shows that velocity of the car is decreases at a constant rate
Explanation:
We have given velocity of the car is decreases from 32 m /sec to 24 m/sec in 4 second
So initial velocity of the car u = 32 m /sec
And finally car reaches to a velocity of 24 m/sec
Time taken to change in velocity = 4 sec
So final velocity v = 24 m/sec
From first equation of motion v = u+at
So 

Negative sign shows that velocity of the car is decreases at a constant rate
Hi there!
Question - Which of the following statements is true?
Answer - C. You are exposed to nuclear radiation every day.
Why - "radiation in the form of elementary particles emitted by an atomic nucleus, as alpha rays or gamma rays, produced by decay of radioactive substances or by nuclear fission."
As an airplane moves through the air, its wings cause changes in the
speed and pressure of the air moving past them. These changes result in
the upward force called lift.
The Bernoulli principle states that an increase in the speed of a fluid
occurs simultaneously with a decrease in the pressure exerted by the
fluid.
A wing is shaped and tilted so the air moving over it moves faster than
the air moving under it. As air speeds up, its pressure goes down. So
the faster-moving air above exerts less pressure on the wing than the
slower-moving air below. The result is an upward push on the wing—lift!
I believe it would best represent Newton’s first law; an object tends to stay at rest and an object tends to stay in motion unless acted upon by an unbalanced force. When the dog stops walking, the doll will continue to go forward because there is no unbalanced force acting in it.
Answer:
A. The time taken for the car to stop is 3.14 secs
B. The initial velocity is 81.64 ft/s
Explanation:
Data obtained from the question include:
Acceleration (a) = 26ft/s2
Distance (s) = 256ft
Final velocity (V) = 0
Time (t) =?
Initial velocity (U) =?
A. Determination of the time taken for the car to stop.
Let us obtain an express for time (t)
Acceleration (a) = Velocity (V)/time(t)
a = V/t
Velocity (V) = distance (s) /time (t)
V = s/t
a = s/t^2
Cross multiply
a x t^2 = s
Divide both side by a
t^2 = s/a
Take the square root of both side
t = √(s/a)
Now we can obtain the time as follow
Acceleration (a) = 26ft/s2
Distance (s) = 256ft
Time (t) =..?
t = √(s/a)
t = √(256/26)
t = 3.14 secs
Therefore, the time taken for the car to stop is 3.14 secs
B. Determination of the initial speed of the car.
V = U + at
Final velocity (V) = 0
Deceleration (a) = –26ft/s2
Time (t) = 3.14 sec
Initial velocity (U) =.?
0 = U – 26x3.14
0 = U – 81.64
Collect like terms
U = 81.64 ft/s
Therefore, the initial velocity is 81.64 ft/s