In preparing to shoot an arrow, an archer pulls a bow string back 0.424 m by exerting a force that increases uniformly from 0 to
1 answer:
Answer:
525.94N/m
Explanation:
According to Hooke's law, the extension or compression of an elastic material is proportional to an applied force provided that the elastic limit of the material is not exceeded.
where F is the applied force or load, k is the elastic constant or stiffness of the material and e is the extension.
In this problem, the bow string is assumed to behave like an elastic material that is stretched not beyond the elastic limit.
Given;
F = 223N;
e = 0.424m
k = ?
We make substitutions into equation (1) and then solve for k.
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Note: I'm not sure what do you mean by "weight 0.05 kg/L". I assume it means the mass per unit of length, so it should be "0.05 kg/m".
Solution:
The fundamental frequency in a standing wave is given by
where L is the length of the string, T the tension and m its mass. If we plug the data of the problem into the equation, we find
The wavelength of the standing wave is instead twice the length of the string:
So the speed of the wave is
And the time the pulse takes to reach the shop is the distance covered divided by the speed:
Solution:
The fundamental frequency in a standing wave is given by
where L is the length of the string, T the tension and m its mass. If we plug the data of the problem into the equation, we find
The wavelength of the standing wave is instead twice the length of the string:
So the speed of the wave is
And the time the pulse takes to reach the shop is the distance covered divided by the speed: