Answer: (b)
Explanation:
Given
Mass of box
Applied force
Friction force
Box travels a distance of
time taken
Net unbalanced force
Work done by the unbalanced force
Power developed by unbalanced force
Thus, option (b) is correct.
Two in-phase loudspeakers that emit sound with the same frequency are placed along a wall and are separated by a distance of 5.0
1 answer:
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Answer:
The answer to the question is
At an altitude of 1413 km or 1.222·R above the north pole the weight of an object reduced to 67% of its earth-surface value
Explanation:
We make use of the gravitational formula as follows
F = G where
m₁ = mass of the object
m₂ = mass of the earth
d = distance between the two objects and
G = gravitational constant
if at the altitude the weight is reduced to 67 % of its weight on earth then
with all other variables remaining constant, we have
67% F = G =0.67× G×
cancelleing like ternss from both sides we have
1/R₂² =0.67/R₁² or r₁²/R₂² =0.67 and R₁/R₂ = 0.8185
or R₂ = R₁/0.8185 or 1.222·R = (6.37×10⁶ m)/0.8185 = 7.783×10⁶ m
Hence at an altitude = 1.413×10⁶ m the weight of the object will reduce to 67% its earth-surface value
(a) 120.8 m/s^2
The gravitational acceleration at a generic distance r from the centre of the planet is
where
G is the gravitational constant
M' is the mass enclosed by the spherical surface of radius r
r is the distance from the centre
For this part of the problem,
so the mass enclosed is just the mass of the core:
So the gravitational acceleration is
(b) 67.1 m/s^2
In this part of the problem,
and the mass enclosed here is the sum of the mass of the core and the mass of the shell, so
so the gravitational acceleration is