What amount of heat is removed to lower the temperature of 80 grams of water from 75ºC to 45ºC? The specific heat of liquid wate
2 answers:
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Answer:
(a)
M = 1.898 x 10^27 kg
(b)
v = 13.74 km/s
(c) E = 0.28 N/kg
Explanation:
Time period, T = 3.55 days = 3.55 x 24 x 3600 second = 306720 s
Radius, r = 6.71 x 10^8 m
G = 6.67 x 10^-11 Nm^2/kg^2
(a)
M = 1.898 x 10^27 kg
(b) Let v be the orbital velocity
v = 13739.5 m/s
v = 13.74 km/s
(b) The gravitational field E is given by
E = 0.28 N/kg
Answer:
The final acceleration becomes (1/3) of the initial acceleration.
Explanation:
The second law of motion gives the relationship between the net force, mass and the acceleration of an object. It is given by :
m = mass
a = acceleration
According to given condition, if the mass of a sliding block is tripled while a constant net force is applied. We need to find how much does the acceleration decrease.
Let a' is the final acceleration,
m' = 3m
So, the final acceleration becomes (1/3) of the initial acceleration. Hence, this is the required solution.