The elastic potential energy increases by a factor of 9
Explanation:
The elastic potential energy of a bowstring is given by
(1)
where
k is the spring constant
x is the elongation of the bowstring
Hooke's law states the relationship between the force applied and the elongation of an elastic object:
where
F is the force applied
x is the elongation
We can rewrite it as
And substituting into (1),
In this problem, the force applied to the bowstring is tripled,
F' = 3F
So the final elastic potential energy is:
So, the elastic potential energy increases by a factor of 9.
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Answer:
book on flat table
Explanation: the table pushes upward on the book while gravity pulls downward.
F = m.a
F = 3(2 I + 5 j)
F = 6i + 15j is resultant force
Magnitude force = √6+15 = √21
Answer:
Total pressure =1.01*10^5 Pa
Explanation:
given data:
atmospheric pressure = 1.013 *10^5 Pa\
height of water column = 60 cm
Total pressure will be sum of atmospheric pressure and pressure due to water column
P = Atmospheric pressure + pressure due to water column
pressure due to water =
\rho of water = 1420 kg/m^{3}
height of water column = 60 cm =0.60 m
Total pressure = 109649.6 Pa = 1.10*10^5 Pa
Total pressure =1.01*10^5 Pa
