Use the formular d = v x t
d = 2m
v= 100m/s
t= d / v
= 2 / 100
= 0.02sec
Answer:
b
Explanation:
because with no pole there is no role
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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It depends on the circuit. Sometimes it becomes a bit weaker, sometimes it stays the same.
Relative motion can best be defined as B<span> the motion of one object as it appears to another object.
An example is when you are in a car the car has the actual motion because it is the one moving but you are also moving because of relative motion.</span>
