<h3><u>Given</u><u>:</u><u>-</u></h3>
Acceleration,a = 3 m/s²
Initial velocity,u = 0 m/s
Final velocity,v = 12 m/s
<h3><u>To</u><u> </u><u>be</u><u> </u><u>calculated:-</u><u> </u></h3>
Calculate the time take by a car.
<h3><u>Solution:-</u><u> </u></h3>
According to the first equation of motion:
v = u + at
★ Substituting the values in the above formula,we get:
⇒ 12 = 0 + 3 × t
⇒ 12 = 3t
⇒ 3t = 12
⇒ t = 12/3
⇒ t = 4 sec
A pair of equal gravitational forces ... one in each direction ...
exists between every speck of mass in the universe and every
other speck of mass.
Answer:
19.3m/s
Explanation:
Use third equation of motion

where v is the velocity at halfway, u is the initial velocity, g is gravity (9.81m/s^2) and h is the height at which you'd want to find the velocity
insert values to get answer
![v^2-0^2=2(9.81m/s^2)(38/2)\\v^2=9.81m/s^2 *38\\v^2=372.78\\v=\sqrt[]{372.78} \\v=19.3m/s](https://tex.z-dn.net/?f=v%5E2-0%5E2%3D2%289.81m%2Fs%5E2%29%2838%2F2%29%5C%5Cv%5E2%3D9.81m%2Fs%5E2%20%2A38%5C%5Cv%5E2%3D372.78%5C%5Cv%3D%5Csqrt%5B%5D%7B372.78%7D%20%5C%5Cv%3D19.3m%2Fs)
Answer:
715 N
Explanation:
Since the system is moving at a constant velocity, the net force must be 0. The tension on the road is equal and opposite direction with the kinetic friction force created by the road and the stuntman.
Let g = 9.8 m/s2
Gravity and equalized normal force is:
N = P = mg = 107*9.8 = 1048.6 N
Kinetic friction force and equalized tension force on the rope is

I think it was Isaac Newton