Answer:
The time taken for the race is 17.20 s.
Explanation:
It is given in the problem that a 62.0 kg sprinter starts a race with an acceleration of 1.44 meter per second square.The initial speed of the sprinter is zero as it starts from the rest.
Calculate the final speed of the sprinter.
The expression for the equation of the motion is as follows;

Here, u is the initial speed, v is the final speed, a is the acceleration and s is the distance.
Put u= 0, s=30 m and
.


Calculate time taken to cover 30 m distance.
The expression for the equation of motion is as follows;

Put u= 0, s=30 m and
.

t=6.45 s
Calculate the time taken to complete his race.
T= t+t'
Here, t is the time taken to cover 30 m distance and t' is the time taken to cover 100 m distance.

Put s= 30 m,
and s'= 100 m.

T= 17.20 s
Therefore, the time taken for the race is 17.20 s.
Golf ball because it weighs less so it has more power to yeah
Answer:
the normal force that the wall exerts on the ball
Explanation:
As Newton's third law states:
"when an object A exerts a force on object B, then object B exerts an equal and opposite force on object A".
If we apply this law to this problem, we can identify the ball as object A, and the wall as object B. As the ball hits the wall, the ball exerts a force on the wall (toward the direction of motion of the ball), so the wall exerts an equal and opposite force on the ball (in the opposite direction). This force is the normal force of the wall, and it is responsible for pushing the ball back towards Erica.
Answer:
n = 1.42
Explanation:
The refractive index for a medium is given by the ratio of the speed of light in vacuum to the speed of light in a medium.

So, the refractive index of the medium is 1.42.
Answer:
3.9 m/s
Explanation:
We are given that
Mass of car,m=
Initial velocity,u=0
Distance,s=5.9 m

Average friction force,f=
We have to find the speed of the car at the bottom of the driveway.
Net force,
Where 
Acceleration,


v=3.9 m/s