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3 years ago

How do you find the speed of an object given its mass and kinetic energy (what is the formula)?

1 answer:

8 0

v = √ { 2*(KE) ] / m } ;

Now, plug in the known values for "KE" ["kinetic energy"] and "m" ["mass"] ;

and solve for "v".
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Explanation:
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The formula is: KE = (½) * (m) * (v²) ;
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"Kinetic energy" = (½) * (mass) * (velocity , "squared")
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Note: Velocity is similar to speed, in that velocity means "speed and direction"; however, if you "square" a negative number, you will get a "positive"; since:  a "negative" multiplied by a "negative" equals a "positive".
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So, we have the formula:
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KE = (½) * (m) * (v²) ; to solve for "(v)" ; velocity, which is very similar to the "speed";
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we arrange the formula ;
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(KE) = (½) * (m) * (v²) ;  ↔ (½)*(m)* (v²) = (KE) ;
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→ We have: (½)*(m)* (v²) = (KE) ; we isolate, "m" (mass) on one side of the equation:
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→ We divide each side of the equation by: "[(½)* (m)]" ;
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→ [ (½)*(m)*(v²) ] /  [(½)* (m)] = (KE) / [(½)* (m)]<span> ;
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to get:
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→ v² =   (KE) / [(½)* (m)]

→ v² = 2 KE / m
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Take the "square root" of each side of the equation ;
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  → √ (v²) = √ { 2*(KE) ] / m }
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→ v = √ { 2*(KE) ] / m } ;

Now, plug in the known values for "KE" ["kinetic energy"] and "m" ["mass"];

and solve for "v".
______________________________________________________

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On the way to the moon, the Apollo astronauts reach a point where the Moon’s gravitational pull is stronger than that of Earth’s

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Answer:

rm = 38280860.6[m]

Explanation:

We can solve this problem by using Newton's universal gravitation law.

In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m

$r_{e} = distance earth to the astronaut [m].\\r_{m} = distance moon to the astronaut [m]\\r_{t} = total distance = 3.84*10^8[m]$

Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.

Mathematically this equals:

$F_{e} = F_{m}\\F_{e} =G*\frac{m_{e} *m_{a}}{r_{e}^{2} } \\$

$F_{m} =G*\frac{m_{m}*m_{a} }{r_{m} ^{2} } \\where:\\G = gravity constant = 6.67*10^{-11}[\frac{N*m^{2} }{kg^{2} } ] \\m_{e}= earth's mass = 5.98*10^{24}[kg]\\ m_{a}= astronaut mass = 100[kg]\\m_{m}= moon's mass = 7.36*10^{22}[kg]$

When we match these equations the masses cancel out as the universal gravitational constant

$G*\frac{m_{e} *m_{a} }{r_{e}^{2} } = G*\frac{m_{m} *m_{a} }{r_{m}^{2} }\\\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} }$

To solve this equation we have to replace the first equation of related with the distances.

$\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} } \\\frac{5.98*10^{24} }{(3.84*10^{8}-r_{m} )^{2} } = \frac{7.36*10^{22} }{r_{m}^{2} }\\81.25*r_{m}^{2}=r_{m}^{2}-768*10^{6}* r_{m}+1.47*10^{17} \\80.25*r_{m}^{2}+768*10^{6}* r_{m}-1.47*10^{17} =0$

Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.

$r_{m1,2}=\frac{-b+- \sqrt{b^{2}-4*a*c } }{2*a}\\ where:\\a=80.25\\b=768*10^{6} \\c = -1.47*10^{17} \\replacing:\\r_{m1,2}=\frac{-768*10^{6}+- \sqrt{(768*10^{6})^{2}-4*80.25*(-1.47*10^{17}) } }{2*80.25}\\\\r_{m1}= 38280860.6[m] \\r_{m2}=-2.97*10^{17} [m]$

We work with positive value

rm = 38280860.6[m] = 38280.86[km]

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Explanation:

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