Answer:
False
Explanation:
The second you let go its gonna release kinetic energy that's why it's potential
Answer:
The answer is: To accelerate an object <u>the force applied to the object</u> has to increase.
Explanation:
the acceleration of an object <u>increases with increased force</u> and <u>decreases with increased mass.</u>
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Answer:
![\gamma=6.07*10^5\frac{N}{m^2}](https://tex.z-dn.net/?f=%5Cgamma%3D6.07%2A10%5E5%5Cfrac%7BN%7D%7Bm%5E2%7D)
Explanation:
For a linear elastic material Young's modulus is a constant that is given by:
![\gamma=\frac{F/A}{\Delta L/L_0}](https://tex.z-dn.net/?f=%5Cgamma%3D%5Cfrac%7BF%2FA%7D%7B%5CDelta%20L%2FL_0%7D)
Here, F is the force exerted on an object under tensio, A is the area of the cross-section perpendicular to the applied force,
is the amount by which the length of the object changes and
is the original length of the object. In this case the force is the weight of the mass:
![F=mg\\F=55kg(9.8\frac{m}{s^2})\\F=539N](https://tex.z-dn.net/?f=F%3Dmg%5C%5CF%3D55kg%289.8%5Cfrac%7Bm%7D%7Bs%5E2%7D%29%5C%5CF%3D539N)
Replacing the given values in Young's modulus formula:
![\gamma=\frac{F/\pi r^2}{(L_1-L_0)/L_0}\\\gamma=\frac{539N/\pi(0.045m)^2}{(5.31m-4.66m)/4.66m}\\\gamma=6.07*10^5\frac{N}{m^2}](https://tex.z-dn.net/?f=%5Cgamma%3D%5Cfrac%7BF%2F%5Cpi%20r%5E2%7D%7B%28L_1-L_0%29%2FL_0%7D%5C%5C%5Cgamma%3D%5Cfrac%7B539N%2F%5Cpi%280.045m%29%5E2%7D%7B%285.31m-4.66m%29%2F4.66m%7D%5C%5C%5Cgamma%3D6.07%2A10%5E5%5Cfrac%7BN%7D%7Bm%5E2%7D)