Answer:
0.88752 kgm²
0.02236 Nm
Explanation:
m = Mass of ball = 1.2 kg
L = Length of rod = 0.86 m
= Angle = 90°
Rotational inertia is given by
![I=mL^2\\\Rightarrow I=1.2\times 0.86^2\\\Rightarrow I=0.88752\ kgm^2](https://tex.z-dn.net/?f=I%3DmL%5E2%5C%5C%5CRightarrow%20I%3D1.2%5Ctimes%200.86%5E2%5C%5C%5CRightarrow%20I%3D0.88752%5C%20kgm%5E2)
The rotational inertia is 0.88752 kgm²
Torque is given by
![\tau=FLsin\theta\\\Rightarrow \tau=2.6\times 10^{-2}\times 0.86sin90\\\Rightarrow \tau=0.02236\ Nm](https://tex.z-dn.net/?f=%5Ctau%3DFLsin%5Ctheta%5C%5C%5CRightarrow%20%5Ctau%3D2.6%5Ctimes%2010%5E%7B-2%7D%5Ctimes%200.86sin90%5C%5C%5CRightarrow%20%5Ctau%3D0.02236%5C%20Nm)
The torque is 0.02236 Nm
Some sort of magnetic metal
Metals are heavier per cubic unit than other materials such as air or water, and also are much more magnetic than other materials
I believe the correct response would be B. It would decrease.
Answer:
W = 55.12 J
Explanation:
Given,
Natural length = 6 in
Force = 4 lb, stretched length = 8.4 in
We know,
F = k x
k is spring constant
4 = k (8.4-6)
k = 1.67 lb/in
Work done to stretch the spring to 10.1 in.
![W =k\int_{6}^{10.1} x](https://tex.z-dn.net/?f=W%20%3Dk%5Cint_%7B6%7D%5E%7B10.1%7D%20x)
![W = \dfrac{k}{2}[x^2]_6^{10.1}](https://tex.z-dn.net/?f=W%20%3D%20%5Cdfrac%7Bk%7D%7B2%7D%5Bx%5E2%5D_6%5E%7B10.1%7D)
![W = \dfrac{1}{2}\times 1.67\times (10.1^2-6.0^2)](https://tex.z-dn.net/?f=W%20%3D%20%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%201.67%5Ctimes%20%2810.1%5E2-6.0%5E2%29)
W = 55.12 J
Work done in stretching spring from 6 in to 10.1 in is equal to 55.12 J.