Answer:
The ball impact velocity i.e(velocity right before landing) is 6.359 m/s
Explanation:
This problem is related to parabolic motion and can be solved by the following equations:
----------------------(1)
---------(2)
----------------------- (3)
Where:
x = m is the horizontal distance travelled by the golf ball
is the golf ball's initial velocity
is the angle (it was a horizontal shot)
t is the time
y is the final height of the ball
is the initial height of the ball
g is the acceleration due gravity
V is the final velocity of the ball
Step 1: finding t
Let use the equation(2)


s
Substituting (6) in (1):
-------------------(4)
Step 2: Finding
:
From equation(4)


m/s (8)
Substituting
in (3):
v =42 .01 - 15.3566
V=26.359 m/s
Answer:
three point charge positioned one x-axis if the charge and corresponding positions are +32Mc x=0 +20Mc x=40cm - 60Mc x=60cm find force 32Mc
Explanation:

If the separation between the charges is increased then the magnitude of the force will increase in fact how the distance is being used in that formula.
Answer:
a) 6.1 m
b) 4.6 s
c) 1.326 m/s
d) 0.325 m
Explanation:
a) The wave length is the distance between 2 crests λ = 6.1m
b) The period of the wave is the time it takes from the lowest point to the next lowest point, which is twice the time it takes from the lowest point to the highest point = 2*2.3 = 4.6 s
c) The speed of the wave is the distance per unit of time, or wave length over period = 6.1 / 4.6 = 1.326 m/s
d)The amplitude A is half the distance from the highest point to the lowest point = 0.65 / 2 = 0.325 m
Answer:
Waves transfer energy, not motion