Answer:
C: equal to mg
Explanation:
in free-fall, gravity is always the net force on an object
Answer:
x = 0.775m
Explanation:
Conceptual analysis
In the attached figure we see the locations of the charges. We place the charge q₃ at a distance x from the origin. The forces F₂₃ and F₁₃ are attractive forces because the charges have an opposite sign, and these forces must be equal so that the net force on the charge q₃ is zero.
We apply Coulomb's law to calculate the electrical forces on q₃:
(Electric force of q₂ over q₃)
(Electric force of q₁ over q₃)
Known data
q₁ = 15 μC = 15*10⁻⁶ C
q₂ = 6 μC = 6*10⁻⁶ C
Problem development
F₂₃ = F₁₃
(We cancel k and q₃)

q₂(2-x)² = q₁x²
6×10⁻⁶(2-x)² = 15×10⁻⁶(x)² (We cancel 10⁻⁶)
6(2-x)² = 15(x)²
6(4-4x+x²) = 15x²
24 - 24x + 6x² = 15x²
9x² + 24x - 24 = 0
The solution of the quadratic equation is:
x₁ = 0.775m
x₂ = -3.44m
x₁ meets the conditions for the forces to cancel in q₃
x₂ does not meet the conditions because the forces would remain in the same direction and would not cancel
The negative charge q₃ must be placed on x = 0.775 so that the net force is equal to zero.
All of the following
involve waves of electromagnetic energy except the rumble of thunder during a storm.
Electromagnetic waves<span> <span>are
used to transmit long/short/FM wavelength radio </span>waves, and TV/telephone/wireless signals or energies. They are
also responsible for transmiting energy in the form of microwaves, infrared radiation<span> (IR), visible light (VIS),
ultraviolet light (UV), X-rays, and gamma rays.</span></span>
The correct answer between all
the choices given is the second choice or letter B. I am hoping that this
answer has satisfied your query and it will be able to help you in your
endeavor, and if you would like, feel free to ask another question.
(a) The ball’s maximum speed over the net is v(max) = √2gh.
(b) The maximum speed of the horizontally moving ball clearing the net is about 27 m/s.
(c) Speed of the ball is independent of its mass.
<h3>
Time of motion of the ball</h3>
The time of motion of the ball is calculated as follows;
h = vt + ¹/₂gt²
1 = 0 + ¹/₂(9.8)t²
1 = 4.9t²
t² = 1/4.9
t² = 0.204
t = 0.452 s
<h3>Horizontal speed of the ball</h3>
The horizontal speed of the ball is calculated as follows;
X = vt
v = X/t
v = (12 m)/(0.452)
v = 26.6 m/s ≈ 27 m/s (proved)
<h3>Conservation of energy</h3>
P.E = K.E
mgh = ¹/₂mv²
gh = ¹/₂v²
2gh = v²
√2gh = v(max)
Speed of the ball is independent of its mass.
Learn more about horizontal velocity here: brainly.com/question/24681896
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