The magnitude of the acceleration of the ball while coming to rest is 477.43 m/s²
The direction of the acceleration of the ball is downwards
The given parameters
initial velocity of the ball, u = 0
height above the ground, h = 2.2 m
time of motion of the ball, t = 96 ms = 0.096 s
The magnitude of the acceleration of the ball while coming to rest is calculated as;
let the downwards direction of the acceleration be positive

The direction of the acceleration of the ball is downwards
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Answer:
A. 181.24 N
Explanation:
The magnitude of hte electrostatic force between two charged objects is given by the equation

where
k is the Coulomb's constant
q1, q2 are the magnitudes of the two charges
r is the separation between the charges
In this problem, we have:
is the magnitude of the 1st charge
is the magnitude of the 2nd charge
r = 2.5 cm = 0.025 m is the separation between the charges
Therefore, the magnitude of the electric force is:

So, the closest answer is
A) 181.24 N
Answer : The wavelength of photon is, 
Explanation : Given,
Energy of photon = 
Formula used :

As, 
So, 
where,
= frequency of photon
h = Planck's constant = 
= wavelength of photon = ?
c = speed of light = 
Now put all the given values in the above formula, we get:


Conversion used : 
Therefore, the wavelength of photon is, 
In most cases the temperature must increase for thermal expansion to occur. Most substances expand as temperature increases because the atoms or molecules vibrate faster as temperature increases and experience greater separation.
Answer:
v = 120 m/s
Explanation:
We are given;
earth's radius; r = 6.37 × 10^(6) m
Angular speed; ω = 2π/(24 × 3600) = 7.27 × 10^(-5) rad/s
Now, we want to find the speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator.
The angle will be;
θ = ¾ × 90
θ = 67.5
¾ is multiplied by 90° because the angular distance from the pole is 90 degrees.
The speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator will be:
v = r(cos θ) × ω
v = 6.37 × 10^(6) × cos 67.5 × 7.27 × 10^(-5)
v = 117.22 m/s
Approximation to 2 sig. figures gives;
v = 120 m/s