The new magnitude of the force of attraction will be 6 times the original force of attraction
<h3>How to determine the initial force </h3>
- Mass 1 = m₁
- Mass 2 = m₂
- Gravitational constant = G
- Distance apart = r
- Initial force (F₁) = ?
F = Gm₁m₂ / r²
F₁ = Gm₁m₂ / r²
<h3>How to determine the new force </h3>
- Mass 1 = 2m₁
- Mass 2 = 3m₂
- Gravitational constant = G
- Distance apart (r) = r
- New force (F₂) =?
F = Gm₁m₂ / r²
F₂ = G × 2m₁ × 3m₂ / r²
F₂ = 6Gm₁m₂ / r²
But
F₁ = Gm₁m₂ / r²
Therefore
F₂ = 6Gm₁m₂ / r²
F₂ = 6F₁
Thus, the new magnitude of the force of attraction will be 6 times the original force of attraction
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Answer:
Vf = 29.4 m/s
h = 44.1 m
Explanation:
Data:
- Initial Velocity (Vo) = 0 m/s
- Gravity (g) = 9.8 m/s²
- Time (t) = 3 s
- Final Velocity (Vf) = ?
- Height (h) = ?
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Final Velocity
Use formula:
Replace:
Multiply:
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Height
Use formula:
Replace:
Multiply time squared:
Simplify the s², and multiply in the numerator:
It divides:
What is the velocity when falling to the ground?
The final velocity is <u>29.4 meters per seconds.</u>
How high is the building?
The height of the building is <u>44.1 meters.</u>
Answer:
900J
Explanation:
w =f×s
60×15
=900J
thus the k.e of the body is 900j
Answer:
Moving a unit "positive" test charge from A to B will result in a reduction in potential
V = K Q / R potential at a point
V2 - V1 = K Q (1 / .4 - 1 / .15) = = k Q (.15 - .4) / .06 = -4.17 K Q
V2 - V1 = -4.17 * 9 & 10E9 * 6.25 E-8
V2 - V1 = -4.17 * 562.5 J/C
V = - 2346 Volts
Answer:
(a). The initial velocity is 28.58m/s
(b). The speed when touching the ground is 33.3m/s.
Explanation:
The equations governing the position of the projectile are
where is the initial velocity.
(a).
When the projectile hits the 50m mark, ; therefore,
solving for we get:
Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that
which gives
(b).
The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,
the vertical component of the velocity is
which gives a speed of