Givens:
Acceleration (a) = 32.17 ft/s² as that is the only force acting on the stone in free fall assuming no air resistance
Initial velocity (V1) = 0 ft/s as that it the starting speed of the stone as all objects about to move from rest must start at 0 velocity at rest = 0 velocity
Final velocity (V2) = 120 ft/s
Displacement aka. the
height of the cliff (Δd) = ? ft
The kinematic formula we can use is: V2² = V1² + 2(a)(Δd)
Isolate to solve for Δd and just plug in the numbers!
V2² = V1² + 2(a)(Δd)
(120)² = (0)² + 2(32.17)(Δd)
14400 = 2(32.17)(Δd)
= Δd
223.9 = Δd
Therefore, the height of the cliff is 223.9 ft! Hope this helped! :)
Answer:
1.97 × 10⁻¹⁸J
Explanation:
Charge of an electron q = -1.9 × 10⁻¹⁹C
Length of one side of the equilateral triangle at whose corners electrons are placed d = 3.5 × 10⁻¹⁰m
Coulomb constant k = 8.99 × 10⁹Nm²/C²
electrostatic potential = (3 kq²) / d
= (3 × 8.99 × 10⁹ × (-1.6 × 10⁻¹⁹)²) / 3.5 × 10⁻¹⁰
= 1.97 × 10¹⁸J
this would be momentum i believe
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<h3>Explanation</h3>
Constant of universal gravitation:
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Apply Newton's law of universal gravitation:
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