then solve for v
so
so v = (3m/s^2)(6s)
so v = 18m/s
A force F=0.12N is aplied on spring and spring elongates by 3cm . specific constant of spring
1 answer:
The spring constant is 4 N/m
Explanation:
When a spring is stretched/compressed by the application of a force, the relationship between the magnitude of the force applied and the elongation of the spring is given by Hooke's law:
where
F is the magnitude of the spring applied
k is the spring constant
x is the elongation of the spring, relative to its equilibrium position
For the spring in this problem, we have:
F = 0.12 N (force applied)
x = 3 cm = 0.03 m (elongation of the spring)
Therefore, we can solve the formula for k to find the spring constant:
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then solve for v
so
so v = (3m/s^2)(6s)
so v = 18m/s
Answer:
The work done by a particle from x = 0 to x = 2 m is 20 J.
Explanation:
A force on a particle depends on position constrained to move along the x-axis, is given by,
We need to find the work done on a particle that moves from x = 0.00 m to x = 2.00 m.
We know that the work done by a particle is given by the formula as follows :
So, the work done by a particle from x = 0 to x = 2 m is 20 J. Hence, this is the required solution.
Explanation:
Since, it is mentioned the there occurs no change in the temperature. This also means that there will occur no change in thermal energy of the system.
Hence, = 0. And, as
= 0 then there will be no work involved. This means that total energy added to the house will return to the outside air as heat.
Therefore,
Q = -(19000 J + 2000 J)
= -21000 J
or, |Q| = 21000 J
Thus, we can conclude that the magnitude of the energy transfer between the house and the outside air is 21000 J.