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Juli2301 [7.4K]
3 years ago
10

The centers of two 15.0 kg spheres are separated by 3.00 m. The magnitude of the gravitational force between the two spheres is

approximately what?
A) 1.11 × 10^-10 N
B) 3.34 × 10^-10 N
C) 1.67 × 10^-9 N
D) 5.00 × 10^-9 N
Physics
1 answer:
kompoz [17]3 years ago
7 0
 we have to use newtons law of gravitation which is
F=GMm/r^2 
G=6.67 x 10^<span>-11N kg^2/m^2
</span>M=<span>(15kg)
</span>m=15 kg
r=(3.0m)^2<span> 
</span>putting values we have 
<span>=(6.67 x 10^-11N kg^2/m^2)(15kg)(15kg)/(3.0m)^2 </span>
=1.67 x 10^-9N 
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A spacecraft on its way to Mars has small rocket engines mounted on its hull; one on its left surface and one on its back surfac
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Answer:

v=545.41 \frac{m}{s}

β=-25.93

Explanation:

Give the acceleration in 'x' and 'y' also the time can find the initial velocity using equation of uniform acceleration motion

For axis x

a_{x}=5.10\frac{m}{s^{2}} \\v_{fx}=3780\frac{m}{s} \\v_{fx}=v_{ox}+a_{x}*t\\v_{ox}=v_{fx}-a_{x}*t\\v_{ox}=3780 \frac{m}{s}-5.1\frac{m}{s^{2} }*645s\\v_{ox}=490.5\frac{m}{s}

For axis y

a_{y}=7.30\frac{m}{s^{2}} \\v_{fy}=4470\frac{m}{s} \\v_{fy}=v_{oy}+a_{y}*t\\v_{oy}=v_{fy}-a_{y}*t\\v_{oy}=4470\frac{m}{s}-7.30\frac{m}{s^{2} }*645s\\v_{oy}=-238.5\frac{m}{s}

Maginuted

v=\sqrt{v_{xo}^{2} +v_{yo}^{2} } \\v=\sqrt{490.5x^{2} +(-238.5)^{2} }\\v=545.41 \frac{m}{s}

The  direction is knowing when find the angle so

\beta =tan^{-1}*\frac{v_{yo} }{v_{xo}}\\\beta =tan^{-1}*\frac{-238.5}{490.5}\\\beta =tan^{-1}*-0.48\\\beta =-25.93

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