A person stands on a spring balance at the equator.
(a) By what friction is the dinbalance reag less than hits true weight? (b) If the speed of earth's rotation is increased by such an amount that the balance reading is half the true weight, what will be the length of the day in this case?
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(a) Friction = 3.45 X
(b) The length of the day will be 2 hours
Explanation:
(a)
Step 1:
We know the Formula
Divide both side by g
is the fraction of weight which is less than its true weight
Step 2:
Fraction of weight is given by =
(b). For this situation the outward power because of pivot of earth= Half the genuine weight
Here
π = 3.14
Radius ( R ) = 6400 km
Converting km to m
Radius ( R ) = 6400×1000
g = 9.8 m/s2 (The gravity of Earth)
On substituting the values,
Square root both side
Converting Second to hour - divide by 3600
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