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Leni [432]
2 years ago
15

How do the accelerations of the rifle and bullet compare?

Physics
1 answer:
Rufina [12.5K]2 years ago
6 0

Answer:

The acceleration of the rifle equals the force that the bullet exerts on the rifle divided by the mass of the rifle. The two forces are equal, but since the mass of the rifle is much greater than the mass of the bullet, the acceleration of the rifle is much less than the bullet's acceleration.

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A ladder 7.85 m long leans against the side of a building. If the ladder is inclined at an angle of 79.5° to the horizontal, wha
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lanet R47A is a spherical planet where the gravitational acceleration on the surface is 3.45 m/s2. A satellite orbitsPlanet R47A
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2.6×10^6\:\text{m}

Explanation:

The acceleration due to gravity g is defined as

g = G\dfrac{M}{R^2}

and solving for R, we find that

R = \sqrt{\dfrac{GM}{g}}\:\:\:\:\:\:\:(1)

We need the mass M of the planet first and we can do that by noting that the centripetal acceleration F_c experienced by the satellite is equal to the gravitational force F_G or

F_c = F_G \Rightarrow m\dfrac{v^2}{r} = G\dfrac{mM}{r^2}\:\:\:\:\:(2)

The orbital velocity <em>v</em> is the velocity of the satellite around the planet defined as

v = \dfrac{2\pi r}{T}

where <em>r</em><em> </em>is the radius of the satellite's orbit in meters and <em>T</em> is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

\dfrac{4\pi^2 r}{T^2} = G\dfrac{M}{r^2}

Solving for <em>M</em>, we get

M = \dfrac{4\pi^2 r^3}{GT^2}

Putting this expression back into Eqn(1), we get

R = \sqrt{\dfrac{G}{g}\left(\dfrac{4\pi^2 r^3}{GT^2}\right)}

\:\:\:\:=\dfrac{2\pi}{T}\sqrt{\dfrac{r^3}{g}}

\:\:\:\:=\dfrac{2\pi}{(1.44×10^4\:\text{s})}\sqrt{\dfrac{(5×10^6\:\text{m})^3}{(3.45\:\text{m/s}^2)}}

\:\:\:\:= 2.6×10^6\:\text{m}

5 0
3 years ago
Two objects have a force of gravity of 100 N. If the distance between both objects decreases by a factor of 7, while the masses
Shkiper50 [21]

The gravitational force <em>F</em> between two masses <em>M</em> and <em>m</em> a distance <em>r</em> apart is

<em>F</em> = <em>G M m</em> / <em>r</em> ²

Decrease the distance by a factor of 7 by replacing <em>r</em> with <em>r</em> / 7, and decrease both masses by a factor of 8 by replacing <em>M</em> and <em>m</em> with <em>M</em> / 8 and <em>m</em> / 8, respectively. Then the new force <em>F*</em> is

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Answer: THIS IS THE EXACT ANSWER

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Explanation:

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