OPTIONS :
A.) the force that the ball exerts on the wall
B.) the frictional force between the wall and the ball
C.) the acceleration of the ball as it approaches the wall
D.) the normal force that the wall exerts on the ball
Answer: D.) the normal force that the wall exerts on the ball
Explanation: The normal force acting on an object can be explained as a force experienced by an object when it comes in contact with a flat surface. The normal force acts perpendicular to the surface of contact.
In the scenario described above, Erica's tennis ball experiences an opposite reaction after hitting the wall.This is in relation to Newton's 3rd law of motion, which states that, For every action, there is an equal and opposite reaction.
The reaction force in this case is the normal force exerted on the ball by the wall perpendicular to the surface of contact.
<span>What we need to first do is split the ball's velocity into vertical and horizontal components. To do that multiply by the sin or cos depending upon if you're looking for the horizontal or vertical component. If you're uncertain as to which is which, look at the angle in relationship to 45 degrees. If the angle is less than 45 degrees, the larger value will be the horizontal speed, if the angle is greater than 45 degrees, the larger value will be the vertical speed. So let's calculate the velocities
sin(35)*18 m/s = 0.573576436 * 18 m/s = 10.32437585 m/s
cos(35)*18 m/s = 0.819152044 * 18 m/s = 14.7447368 m/s
Since our angle is less than 45 degrees, the higher velocity is our horizontal velocity which is 14.7447368 m/s.
To get the x positions for each moment in time, simply multiply the time by the horizontal speed. So
0.50 s * 14.7447368 m/s = 7.372368399 m
1.00 s * 14.7447368 m/s = 14.7447368 m
1.50 s * 14.7447368 m/s = 22.1171052 m
2.00 s * 14.7447368 m/s = 29.48947359 m
Rounding the results to 1 decimal place gives
0.50 s = 7.4 m
1.00 s = 14.7 m
1.50 s = 22.1 m
2.00 s = 29.5 m</span>
Answer:
a) 
b) 
Explanation:
Given:
height of water in one arm of the u-tube, 
a)
Gauge pressure at the water-mercury interface,:
we've the density of the water 
b)
Now the same pressure is balanced by the mercury column in the other arm of the tube:
<u>Now the difference in the column is :</u>
