Asked by Kath
The temperature of the air is 20 Celsius when a tourist drops a coin into a wishing well. She hears the splash 5.30s after she drops the coin. If the air resistance is ignored, how deep is the well?
a) 344m b)332m c)137m d)120m e)0.60m
a) 344m b)332m c)137m d)120m e)0.60m
Answers
Answered by
bobpursley
time it takes the coin to fall: h=1/2 g t^2
t=sqrt(2h/g)
time it takes sound to come back: h/vs
given: those times add to 5.2 seconds
5.2=sqrt(2H/g)+h/v
a quadratic: let u=sqrt h so u^2=h
u^2/v+u(sqrt2/g)-5.2=0
u^2+u*vsqrt(2/g)-5.2v=0
calculate v for that temerature, then use the quadratic equation to solve for u (and square it for h)
t=sqrt(2h/g)
time it takes sound to come back: h/vs
given: those times add to 5.2 seconds
5.2=sqrt(2H/g)+h/v
a quadratic: let u=sqrt h so u^2=h
u^2/v+u(sqrt2/g)-5.2=0
u^2+u*vsqrt(2/g)-5.2v=0
calculate v for that temerature, then use the quadratic equation to solve for u (and square it for h)
Answered by
Kath
i dont really understand the explanation, so which would be the answer
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