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
Learn more about gravitational force:
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G is the answer for apex vs / Chehhhh
Answer:
Velocity = 3.25[m/s]
Explanation:
This problem can be solved if we use the Bernoulli equation: In the attached image we can see the conditions of the water inside the container.
In point 1, (surface of the water) we have the atmospheric pressure and at point 2 the water is coming out also at atmospheric pressure, therefore this members in the Bernoulli equation could be cancelled.
The velocity in the point 1 is zero because we have this conditional statement "The water surface drops very slowly and its speed is approximately zero"
h2 is located at point 2 and it will be zero.
![(P_{1} +\frac{v_{1}^{2} }{2g} +h_{1} )=(P_{2} +\frac{v_{2}^{2} }{2g} +h_{2} )\\P_{1} =P_{2} \\v_{1}=0\\h_{2} =0\\v_{2}=\sqrt{0.54*9.81*2}\\v_{2}=3.25[m/s]](https://tex.z-dn.net/?f=%28P_%7B1%7D%20%2B%5Cfrac%7Bv_%7B1%7D%5E%7B2%7D%20%7D%7B2g%7D%20%2Bh_%7B1%7D%20%29%3D%28P_%7B2%7D%20%2B%5Cfrac%7Bv_%7B2%7D%5E%7B2%7D%20%7D%7B2g%7D%20%2Bh_%7B2%7D%20%29%5C%5CP_%7B1%7D%20%3DP_%7B2%7D%20%5C%5Cv_%7B1%7D%3D0%5C%5Ch_%7B2%7D%20%3D0%5C%5Cv_%7B2%7D%3D%5Csqrt%7B0.54%2A9.81%2A2%7D%5C%5Cv_%7B2%7D%3D3.25%5Bm%2Fs%5D)
Answer:
(a) I_A=1/12ML²
(b) I_B=1/3ML²
Explanation:
We know that the moment of inertia of a rod of mass M and lenght L about its center is 1/12ML².
(a) If the rod is bent exactly at its center, the distance from every point of the rod to the axis doesn't change. Since the moment of inertia depends on the distance of every mass to this axis, the moment of inertia remains the same. In other words, I_A=1/12ML².
(b) The two ends and the point where the two segments meet form an isorrectangle triangle. So the distance between the ends d can be calculated using the Pythagorean Theorem:

Next, the point where the two segments meet, the midpoint of the line connecting the two ends of the rod, and an end of the rod form another rectangle triangle, so we can calculate the distance between the two axis x using Pythagorean Theorem again:

Finally, using the Parallel Axis Theorem, we calculate I_B:

<span>a) 1960 m
b) 960 m
Assumptions.
1. Ignore air resistance.
2. Gravity is 9.80 m/s^2
For the situation where the balloon was stationary, the equation for the distance the bottle fell is
d = 1/2 AT^2
d = 1/2 9.80 m/s^2 (20s)^2
d = 4.9 m/s^2 * 400 s^2
d = 4.9 * 400 m
d = 1960 m
For situation b, the equation is quite similar except we need to account for the initial velocity of the bottle. We can either assume that the acceleration for gravity is negative, or that the initial velocity is negative. We just need to make certain that the two effects (falling due to acceleration from gravity) and (climbing due to initial acceleration) counteract each other. So the formula becomes
d = 1/2 9.80 m/s^2 (20s)^2 - 50 m/s * T
d = 1/2 9.80 m/s^2 (20s)^2 - 50m/s *20s
d = 4.9 m/s^2 * 400 s^2 - 1000 m
d = 4.9 * 400 m - 1000 m
d = 1960 m - 1000 m
d = 960 m</span>