Newtons First Law of Motion:
An object at rest stays at rest and an object in motion<span> stays in </span>motion <span>with the same speed and in the same direction unless acted upon by an unbalanced force.</span>
Therefore, the relationship between force and motion is that it takes force to change the speed or direction of any object in motion.
Answer:
84.82N/C.
Explanation:
The x-components of the electric field cancel; therefore, we only care about the y-components.
The y-component of the differential electric field at the center is
.
Now, let us call 
the charge per unit length, then we know that
;
therefore,
Integrating
![$E = \frac{k \lambda }{R}*[-cos(\pi )+cos(0) ]$](https://tex.z-dn.net/?f=%24E%20%3D%20%5Cfrac%7Bk%20%5Clambda%20%20%20%7D%7BR%7D%2A%5B-cos%28%5Cpi%20%29%2Bcos%280%29%20%5D%24)
Now, we know that
and the radius of the semicircle is
therefore,
They have thick body coverings
To answer this problem, we will use the equations of motions.
Part (a):
For the ball to start falling back to the ground, it has to reach its highest position where its final velocity will be zero.
The equation that we will use here is:
v = u + at where
v is the final velocity = 0 m/sec
u is the initial velocity = 160 m/sec
a is acceleration due to gravity = -9.8 m/sec^2 (the negative sign is because the ball is moving upwards, thus, its moving against gravity)
t is the time that we want to find.
Substitute in the equation to get the time as follows:
v = u + at
0 = 160 - 9.8t
9.8t = 160
t = 160/9.8 = 16.3265 sec
Therefore, the ball would take 16.3265 seconds before it starts falling back to the ground
Part (b):
First, we will get the total distance traveled by the ball as follows:
s = 0.5 (u+v)*t
s = 0.5(160+0)*16.3265
s = 1306.12 meters
The equation that we will use to solve this part is:
v^2 = u^2 + 2as where
v is the final velocity we want to calculate
u is the initial velocity of falling = 0 m/sec (ball starting falling when it reached the highest position, So, the final velocity in part a became the initial velocity here)
a is acceleration due to gravity = 9.8 m/sec^2 (positive as ball is moving downwards)
s is the distance covered = 1306.12 meters
Substitute in the above equation to get the final velocity as follows:
v^2 = u^2 + 2as
v^2 = (0)^2 + 2(9.8)(1306.12)
v^2 = 25599.952 m^2/sec^2
v = 159.99985 m/sec
Therefore, the velocity of the ball would be 159.99985 m/sec when it hits the ground.
