Answer:
(A) 88.92 cm
(B) 22.22 cm
Explanation:
distance (s) = 200 cm = 0.2 m
initial velocity (v) = 0 m/s
acceleration due to gravity (g) = 9.8 m/s^{2}
lets first find the time (T) it takes for the first drop to strike the floor
from s = ut + 
200 = 0 + 
200 = 
200 / 4.9 = 
T = 6.4
(A) When the first drop strikes the floor, how far below the nozzle is the second drop.
we can find how far the second drop was when the first drop hits the ground from the formula s = ut + 
where
- s = distance
- u = initial velocity = 0
- t = time, since the drops fall at regular (equal) intervals of time, the first drop striking the floor at the instant the fourth drop begins to fall there wold be 3 time intervals and this can be seen illustrated in the attached diagram. therefore the time of the second drop = 2/3 of the time it takes the first drop to strike the ground (
) - a = acceleration due to gravity = 9.8 m/s^{2}
substituting all required values we have
s = 0 + (
)
s = 0 + (
)
s = 88.92 cm
(B) When the first drop strikes the floor, how far below the nozzle is the third drop.
we can find how far the third drop was when the first drop hits the ground from the formula s = ut + 
where
- s = distance
- u = initial velocity = 0
- t = time, since the drops fall at regular (equal) intervals of time, the first drop striking the floor at the instant the fourth drop begins to fall there wold be 3 time intervals and this can be seen illustrated in the attached diagram. therefore the time of the third drop = 1/3 of the time it takes the first drop to strike the ground (
) - a = acceleration due to gravity = 9.8 m/s^{2}
substituting all required values we have
s = 0 + (
)
s = 0 + (
)
s = 22.22 cm
Answer:
Explanation:
Given
Four Point charges are Placed at the corner of a square with side length a
Length of diagonal 
Distance of charge from center 
Potential at the center due to four charges is given by

as all the four charges are equal and placed symmetrical about center
Work done required to bring a fifth charge q from infinity to the center of the square

The motion of the football that is being kicked follows a projectile path and is parabolic in shape. It has a maximum height of meters and the angle with respect to the ground is degrees. The maximum height is the highest position vertically that the ball could reach. From the equations of an object in a projectile motion, we use the equation <span>(V*sin20)^2 = 2gh in order to solve for velocity, V. We do as follows:
</span><span>((V) (sin(20)))^2 = 2gh
(V^2) (0.342^2) = 92.12
V = √ (92.12 / 0.342^2)
V = 28.06 m/s
</span>The initial velocity of the kick would be 28.06 m/s.