Ice at 0 C is kept in a cubical ice box of thickness 3cms and side 60 cms each. How much of ice will melt in one minute if the thermal conductivity of material is 0.2 W/mK and the external temperature is 300 C.
Please answer fast
Answers
Answer:
Radius of moon's orbit will double.
GIVEN:
Half of the earth were taken off by the impulse of a comet.
TO FIND:
The chnge produced in the moon's orbit.
EXPLANATION:
\boxed{ \bold{ \large{ \gray{Total \ Energy = K.E + P.E}}}}
Total Energy=K.E+P.E
BEFORE IMPACT:
Let mass of moon be m and mass of earth be M.
\sf \leadsto K.E = \dfrac{1}{2} mv^2⇝K.E=
2
1
mv
2
Let the distance between earth and moon be r.
\sf \leadsto P.E = - \dfrac{GMm}{r}⇝P.E=−
r
GMm
\sf \leadsto T. E= \dfrac{1}{2} mv^2 - \dfrac{GMm}{r}⇝T.E=
2
1
mv
2
−
r
GMm
AFTER IMPACT:
Now mass of earth will be M/2 and mass of moon will be m.
\sf \leadsto K.E = \dfrac{1}{2} mv^2⇝K.E=
2
1
mv
2
Now let the distance between earth and moon be r'.
\sf \leadsto P.E = - \dfrac{GMm}{2r'}⇝P.E=−
2r
′
GMm
\sf \leadsto T. E= \dfrac{1}{2} mv^2 - \dfrac{GMm}{2r'}⇝T.E=
2
1
mv
2
−
2r
′
GMm
By law of conservation of energy, energy before and after impact will be the same.
\sf \leadsto \dfrac{1}{2} mv^2 - \dfrac{GMm}{r} = \dfrac{1}{2} mv^2 - \dfrac{GMm}{2r'}⇝
2
1
mv
2
−
r
GMm
=
2
1
mv
2
−
2r
′
GMm
\sf \leadsto \dfrac{GMm}{r} = \dfrac{GMm}{2r'}⇝
r
GMm
=
2r
′
GMm
\sf \leadsto \dfrac{1}{r} = \dfrac{1}{2r'}⇝
r
1
=
2r
′
1
\sf \leadsto r'= \dfrac{r}{2}⇝r
′
=
2
r
Answer:
25 cm³ ice will melt in 6o sec