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RideAnS [48]
3 years ago
5

Using our understanding of the law of gravity. What happens to the gravity as we triple the distance between two objects?

Physics
1 answer:
MA_775_DIABLO [31]3 years ago
3 0

Answer:

1/9

Explanation:

<em>Newton’s Law of Universal Gravitation </em>

Objects with mass feel an attractive force that is proportional to their masses and inversely proportional to the square of the distance.

F = GMm/r²

where  

F - the gravitatioal force in Newtons,  

M   and m  -two masses in kilograms  

r  - the separation in meters.  

G  - the gravitational constant (6.674*10 ⁻¹¹ N (m/kg) ² )

Because of the magnitude of  G , gravitational force is very small unless large masses are involved.

So according to above equation , when the masses are not changing , force is inversely propotional to the square of distance

F1 ∝ 1/r² ---------------(1)

F2 ∝ 1(3r)²

F2 ∝ 1/9r²--------------(2)

(2)/(1)

\frac{F_2}{F_1} =\frac{1}{9}\\ F_2 =\frac{F_1}{9}

From their you get as the distance tripled, Force reduce by a factor of 9(3³)

for example , assume the distance get doubled ,Force reduce by a factor of 4 (2²)

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Calculate the average induced voltage between the tips of the wings of an airplane flying above East Lansing at a speed of 885 k
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Answer:

=0.855V

Explanation:

The induced voltage can be calculated using below expression

E =B x dA/dt

Where dA/dt = area

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speed = 885 km/h

speed = 885 x 10^3 m/hr

speed = 885 x 10^3/60 x60 m/s

speed = 245.8 m/s

If The aircraft wing sweep out" an area

at t= 50.4seconds then we have;

dA/dt = 50.4 x 245.8

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Then from the expression above

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2 years ago
Addition of a lubricant such as oil to a surface will __ friction ?
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Initial velocity vector vA has a magnitude of 3.00 meters per second and points 20.0o north of east, while final velocity vector
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Answer:

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70^{\circ} south of east

Explanation:

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v_{ay}=v_a\sin\theta\\\Rightarrow v_{ay}=3\times \sin20^{\circ}\\\Rightarrow v_{ay}=1.03\ \text{m/s}

x and y component of v_b

v_{bx}=v_b\cos \theta\\\Rightarrow v_{bx}=6\times \cos 320^{\circ}\\\Rightarrow v_{bx}=4.6\ \text{m/s}

v_{by}=v_b\sin\theta\\\Rightarrow v_{by}=6\times \sin320^{\circ}\\\Rightarrow v_{by}=-3.86\ \text{m/s}

\Delta v=v_b-v_a\\\Rightarrow \Delta v=(4.6-2.82)\hat{i}+(-3.86-1.03)\hat{j}\\\Rightarrow \Delta v=1.78\hat[i}-4.89\hat{j}

Magnitude

|\Delta v|=\sqrt{(-4.89)^2+1.78^2}\\\Rightarrow \Delta v=5.2\ \text{m/s}

Direction

\theta=\tan{-1}|\dfrac{-4.89}{1.78}|\\\Rightarrow \theta=70^{\circ}

The magnitude of the change in velocity vector is 5.2\ \text{m/s} and the direction is 70^{\circ} south of east.

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