The maximum height reached by the ball is 99.2 m
Explanation:
When the ball is thrown straight up, it follows a free fall motion, which is a uniformly accelerated motion with constant acceleration (
towards the ground). Therefore, we can use the following suvat equation:

where
v is the final velocity
u is the initial velocity
a is the acceleration
s is the displacement
In this problem, we have:
u = 44.1 m/s is the initial vertical velocity of the ball
v = 0 is the final velocity when the ball reaches the maximum height
s is the maximum height
is the acceleration of gravity (downward, so negative)
Solving for s, we find the maximum height reached by the ball:

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Answer: A
Explanation: We know that f=p*n
f=50*300=15000 Hz = 15kHz.
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Answer:
v = 20.31 m/s
Explanation:
p = mv -> v = p/m = 32,500 kg*m/s / 1,600 kg = 20.31 m/s
Answer:
Charge on each is 2 x 10⁻¹⁰.
Explanation:
We know that Force between two point charges is given b the Coulomb's law as:
F = kq₁q₂/r^2
k = 9 x 10^9
r = 3.00 cm
= 0.03 m
q₁ = q₂
F = 4.00 x 10^-7
Rearranging the formula, we get:
F = k q²/r²
q² = Fr²/k
q² = 4 x 10⁻⁷ x 0.03²/(9x10⁹)
q² = 4 x 10⁻²⁰
q = 2 x 10⁻¹⁰
As there is force of repulsion between the charges, the charges must be both positive or both negative.