Answer:
1.429*10^-5 m
Explanation:
From the question, we are given that
Diameter of the cable, d = 3 cm = 0.03 m
Force on the cable, F = 2 kN
Young Modulus, Y = 2*10^11 Pa
Area of the cable = πd²/4 = (3.142 * 0.03²) / 4 = 0.0028 / 4 = 0.0007 m²
The fractional length = Δl/l
Δl/l = F/AY
Δl/l = 2000 / 0.0007 * 2*10^11
Δl/l = 2000 / 1.4*10^8
Δl/l = 1.429*10^-5 m
Therefore, the fractional length is 1.429*10^-5 m long
Answer:
(a) I_A=1/12ML²
(b) I_B=1/3ML²
Explanation:
We know that the moment of inertia of a rod of mass M and lenght L about its center is 1/12ML².
(a) If the rod is bent exactly at its center, the distance from every point of the rod to the axis doesn't change. Since the moment of inertia depends on the distance of every mass to this axis, the moment of inertia remains the same. In other words, I_A=1/12ML².
(b) The two ends and the point where the two segments meet form an isorrectangle triangle. So the distance between the ends d can be calculated using the Pythagorean Theorem:
![d=\sqrt{(\frac{1}{2}L) ^{2}+(\frac{1}{2}L) ^{2} } =\sqrt{\frac{1}{2}L^{2} } =\frac{1}{\sqrt{2} } L=\frac{\sqrt{2} }{2} L](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%28%5Cfrac%7B1%7D%7B2%7DL%29%20%5E%7B2%7D%2B%28%5Cfrac%7B1%7D%7B2%7DL%29%20%5E%7B2%7D%20%7D%20%3D%5Csqrt%7B%5Cfrac%7B1%7D%7B2%7DL%5E%7B2%7D%20%7D%20%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D%20%7D%20L%3D%5Cfrac%7B%5Csqrt%7B2%7D%20%7D%7B2%7D%20L)
Next, the point where the two segments meet, the midpoint of the line connecting the two ends of the rod, and an end of the rod form another rectangle triangle, so we can calculate the distance between the two axis x using Pythagorean Theorem again:
![x=\sqrt{(\frac{1}{2}L)^{2}-(\frac{\sqrt{2}}{4}L) ^{2} } =\sqrt{\frac{1}{8} L^{2} } =\frac{1}{2\sqrt{2}} L=\frac{\sqrt{2}}{4} L](https://tex.z-dn.net/?f=x%3D%5Csqrt%7B%28%5Cfrac%7B1%7D%7B2%7DL%29%5E%7B2%7D-%28%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B4%7DL%29%20%20%5E%7B2%7D%20%7D%20%3D%5Csqrt%7B%5Cfrac%7B1%7D%7B8%7D%20L%5E%7B2%7D%20%7D%20%3D%5Cfrac%7B1%7D%7B2%5Csqrt%7B2%7D%7D%20L%3D%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B4%7D%20L)
Finally, using the Parallel Axis Theorem, we calculate I_B:
![I_B=I_A+Mx^{2} \\\\I_B=\frac{1}{12} ML^{2} +\frac{1}{4} ML^{2} =\frac{1}{3} ML^{2}](https://tex.z-dn.net/?f=I_B%3DI_A%2BMx%5E%7B2%7D%20%5C%5C%5C%5CI_B%3D%5Cfrac%7B1%7D%7B12%7D%20ML%5E%7B2%7D%20%2B%5Cfrac%7B1%7D%7B4%7D%20%20ML%5E%7B2%7D%20%3D%5Cfrac%7B1%7D%7B3%7D%20ML%5E%7B2%7D)
The lines can be traced out with a compass. The needle is like a permanent magnet and the north indicator is the north end of a magnet.
Answer:
F = 0.00156[N]
Explanation:
We can solve this problem by using Newton's proposed universal gravitation law.
![F=G*\frac{m_{1} *m_{2} }{r^{2} } \\](https://tex.z-dn.net/?f=F%3DG%2A%5Cfrac%7Bm_%7B1%7D%20%2Am_%7B2%7D%20%7D%7Br%5E%7B2%7D%20%7D%20%5C%5C)
Where:
F = gravitational force between the moon and Ellen; units [Newtos] or [N]
G = universal gravitational constant = 6.67 * 10^-11 [N^2*m^2/(kg^2)]
m1= Ellen's mass [kg]
m2= Moon's mass [kg]
r = distance from the moon to the earth [meters] or [m].
Data:
G = 6.67 * 10^-11 [N^2*m^2/(kg^2)]
m1 = 47 [kg]
m2 = 7.35 * 10^22 [kg]
r = 3.84 * 10^8 [m]
![F=6.67*10^{-11} * \frac{47*7.35*10^{22} }{(3.84*10^8)^{2} }\\ F= 0.00156 [N]](https://tex.z-dn.net/?f=F%3D6.67%2A10%5E%7B-11%7D%20%2A%20%5Cfrac%7B47%2A7.35%2A10%5E%7B22%7D%20%7D%7B%283.84%2A10%5E8%29%5E%7B2%7D%20%7D%5C%5C%20F%3D%200.00156%20%5BN%5D)
This force is very small compare with the force exerted by the earth to Ellen's body. That is the reason that her body does not float away.
My Response:
If the mass of an object increases, then its kinetic energy will increase proportionally because mass and kinetic energy have a linear relationship when graphed.
Sample Response:
If the speed of an object increases, then its kinetic energy will increase proportionally because speed and kinetic energy have a linear relationship when graphed.