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the Earth from the question above. 1.89° 3. If the constellation Andromeda is overhead at midnight tonight, what time will it be overhead one month from now? 

          the Earth from the question above.
1.89°
3. If the constellation Andromeda is overhead at midnight tonight, what time will it be overhead one month
from now?
the Earth from the question above. 1.89° 3. If the constellation Andromeda is overhead at midnight tonight, what time will it be overhead one month from now?

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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'How was this answer gotten the Earth from the question above_ 1.890 3 . If the constellation Andromeda is overhead at midnight tonight; what time will it be overhead one month from now'
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00:01 High in the given problem the volume of the sphere is 80 centimeter cube and the tolerance of the volume which is dv is equal to 4 % so 4 % can be written as 0 .04 so this is the tolerance or we can say this is the change in volume so dv now we is the volume of the sphere is 4 over 3 pi r cube so when we differentiate both the side with respect to us so this will be dv and this is 4 pi r square d r so that's the change in the volume is related to the change in the radius now since we have volume as 80 so from here we get we have volume 80 so we can write this as 4 by 3 pi r cube which means that r is equal to root over cube root of 60 so 60 over pi to the power 1 over 3 that's the radius now dv is related to dr as so we can say write dr is equal to 1 over 100 pi and r square 100 5 r square so that is dr because a change in world dv is already 0 .04 so we'll put that as equal to…
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