Answer:
The Young's modulus for this alloy of aluminum is ![6.21\times10^{10}\ N/m^2](https://tex.z-dn.net/?f=6.21%5Ctimes10%5E%7B10%7D%5C%20N%2Fm%5E2)
Explanation:
Given that,
Diameter = 0.12 cm
Length = 2.4 m
Mass = 37 kg
Stretch length = 1.24 cm
Density = 2.7 g/cm³
We need to calculate the area
Using formula of area
![A=\pi\times r^2](https://tex.z-dn.net/?f=A%3D%5Cpi%5Ctimes%20r%5E2)
Put the value into the formula
![A=\pi\times(\dfrac{0.12\times10^{-2}}{2})^2](https://tex.z-dn.net/?f=A%3D%5Cpi%5Ctimes%28%5Cdfrac%7B0.12%5Ctimes10%5E%7B-2%7D%7D%7B2%7D%29%5E2)
![A=0.000001130\ m^2](https://tex.z-dn.net/?f=A%3D0.000001130%5C%20m%5E2)
![A=1.13\times10^{-6}\ m^2](https://tex.z-dn.net/?f=A%3D1.13%5Ctimes10%5E%7B-6%7D%5C%20m%5E2)
We need to calculate the Young's modulus
Using formula of Young's modulus
![Y=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta l}{l}}](https://tex.z-dn.net/?f=Y%3D%5Cdfrac%7B%5Cdfrac%7BF%7D%7BA%7D%7D%7B%5Cdfrac%7B%5CDelta%20l%7D%7Bl%7D%7D)
![Y=\dfrac{Fl}{A\Delta l}](https://tex.z-dn.net/?f=Y%3D%5Cdfrac%7BFl%7D%7BA%5CDelta%20l%7D)
Put the value into the formula
![Y=\dfrac{37\times9.8\times2.4}{1.13\times10^{-6}\times1.24\times10^{-2}}](https://tex.z-dn.net/?f=Y%3D%5Cdfrac%7B37%5Ctimes9.8%5Ctimes2.4%7D%7B1.13%5Ctimes10%5E%7B-6%7D%5Ctimes1.24%5Ctimes10%5E%7B-2%7D%7D)
![Y=6.21\times10^{10}\ N/m^2](https://tex.z-dn.net/?f=Y%3D6.21%5Ctimes10%5E%7B10%7D%5C%20N%2Fm%5E2)
Hence, The Young's modulus for this alloy of aluminum is ![6.21\times10^{10}\ N/m^2](https://tex.z-dn.net/?f=6.21%5Ctimes10%5E%7B10%7D%5C%20N%2Fm%5E2)
Answer:21.18 m
Explanation:
Given
initial speed u=10 m/s
height of building h=22 m
time taken to complete 22 m
![h=ut+\frac{1}{2}at^2](https://tex.z-dn.net/?f=h%3Dut%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2)
initial vertical velocity =0
![22=\frac{1}{2}gt^2](https://tex.z-dn.net/?f=22%3D%5Cfrac%7B1%7D%7B2%7Dgt%5E2)
![t=\sqrt{\frac{22\times 2}{g}}](https://tex.z-dn.net/?f=t%3D%5Csqrt%7B%5Cfrac%7B22%5Ctimes%202%7D%7Bg%7D%7D)
![t=2.11 s](https://tex.z-dn.net/?f=t%3D2.11%20s)
Horizontal Distance moved
![R=u_x\times t](https://tex.z-dn.net/?f=R%3Du_x%5Ctimes%20t)
![R=10\times 2.11](https://tex.z-dn.net/?f=R%3D10%5Ctimes%202.11)
![R=21.18 m](https://tex.z-dn.net/?f=R%3D21.18%20m)
Answer:
that initially the weather vane was at rest, by this load that remained on the pole it would begin to move.
Explanation:
Let us carefully analyze the situation, when the bar is facing the index post a load of equal magnitude, but opposite sign on its surface, these two charges are in balance; When the hand touches the pole, it creates a path to the ground where the charges that were induced on the pole can be balanced with the charge coming from the ground, leaving a zero charge on the pole.
Now if the hand is removed, there can be no exchange of charges with the earth. When the bar is removed, the induced loads are redistributed in the post, but the excess loads that came from the earth that have the same value and are of a sign opposite to the induced ones remain, you want to sign that they are of the same sign as the charges of the bar.
In summary, after the process, the post has a load of equal magnitude and sign (negative) that of the bar.
If we assume that initially the weather vane was at rest, by this load that remained on the pole it would begin to move.
Answer:
v = 2 m/s
Explanation:
Here we can use energy conservation to find the speed at the lowest point on its trajectory
As we know that by energy conservation
initial total gravitational potential energy = final total kinetic energy
now the height that is moved by the pendulum while it swing down is given as
![h = L(1 - cos30)](https://tex.z-dn.net/?f=h%20%3D%20L%281%20-%20cos30%29)
![h = 1.5(1 - cos30) = 0.200 m](https://tex.z-dn.net/?f=h%20%3D%201.5%281%20-%20cos30%29%20%3D%200.200%20m)
now we can use energy conservation as
![mgh = \frac{1}{2}mv^2](https://tex.z-dn.net/?f=mgh%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
![v = \sqrt{2gh}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B2gh%7D)
![v = \sqrt{2(9.8)(0.200)}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B2%289.8%29%280.200%29%7D)
![v = 2 m/s](https://tex.z-dn.net/?f=v%20%3D%202%20m%2Fs)