A bomb falls with velocity v(t)=C(1−e−kt), where C and k are constants. What is the terminal velocity of the bomb? That is, what
is the maximum valocity the bomb can reach regardless of the height of the bomber?
1 answer:
Answer:
The terminal velocity is equal to C.
Explanation:
Making the assumption that k and t are positive, we then have that -kt is negative. The value of e^(-kt) will be equal to 1/(e^(kt))
If t increases, e^(kt) will increase exponentially and its reciprocal 1/(e^(kt)) will approach zero.
So, we have:
v(t) = C*(1-0)
v(t) = C*(1)
v(t) = C
Therefore, C is the terminal velocity.
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Answer:
L= 12 light years
Explanation:
for length dilation we use the formula
now calculating Lo
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now putting the values of v and Lo in the above equation we get
= 1.136×10^17 m
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m
so L= 12 light years
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