Answer:
(a) ![L=2.6742\times 10^{40}\, kg.m^2.s^{-1}](https://tex.z-dn.net/?f=L%3D2.6742%5Ctimes%2010%5E%7B40%7D%5C%2C%20kg.m%5E2.s%5E%7B-1%7D)
(b) ![L_s=7.07 \times 10^{23} \, kg.m^2.s^{-1}](https://tex.z-dn.net/?f=L_s%3D7.07%20%5Ctimes%2010%5E%7B23%7D%20%5C%2C%20kg.m%5E2.s%5E%7B-1%7D)
Explanation:
<u>For Earth we have:</u>
- mass of earth,
![m_E=5.97\times 10^{24}\, kg](https://tex.z-dn.net/?f=%20m_E%3D5.97%5Ctimes%2010%5E%7B24%7D%5C%2C%20kg)
- radius of earth,
![R_E=6.38\times 10^6m](https://tex.z-dn.net/?f=%20R_E%3D6.38%5Ctimes%2010%5E6m)
- orbital radius,
![r=1.5 \times 10^{11}m](https://tex.z-dn.net/?f=r%3D1.5%20%5Ctimes%2010%5E%7B11%7Dm)
- period of rotation,
![t_{rot}=24h=86400\, s](https://tex.z-dn.net/?f=t_%7Brot%7D%3D24h%3D86400%5C%2C%20s)
- period of revolution,
![t_{rev}= 1 yr=3.156\times 10^7 s](https://tex.z-dn.net/?f=t_%7Brev%7D%3D%201%20yr%3D3.156%5Ctimes%2010%5E7%20s)
(a)
Angular momentum, ![L=?](https://tex.z-dn.net/?f=L%3D%3F)
∵
...............................(1)
For a particle of mass m moving in a circular path at a distance r from the axis,
![I=m.r^2](https://tex.z-dn.net/?f=I%3Dm.r%5E2)
& ![v=r.\omega](https://tex.z-dn.net/?f=v%3Dr.%5Comega)
Putting respecstive values in eq. (1)
![L=m_E\times r^{2}\times \omega](https://tex.z-dn.net/?f=L%3Dm_E%5Ctimes%20r%5E%7B2%7D%5Ctimes%20%5Comega)
![L=5.97\times 10^{24}\times (1.5 \times 10^{11})^2\times \frac{2\pi}{3.156\times 10^7}](https://tex.z-dn.net/?f=L%3D5.97%5Ctimes%2010%5E%7B24%7D%5Ctimes%20%281.5%20%5Ctimes%2010%5E%7B11%7D%29%5E2%5Ctimes%20%5Cfrac%7B2%5Cpi%7D%7B3.156%5Ctimes%2010%5E7%7D)
![L=2.6742\times 10^{40}\, kg.m^2.s^{-1}](https://tex.z-dn.net/?f=L%3D2.6742%5Ctimes%2010%5E%7B40%7D%5C%2C%20kg.m%5E2.s%5E%7B-1%7D)
To model earth as a particle is reasonable because the distance between the sun and the earth is very large s compared to the radius if the earth.
(b)
For a uniform sphere of mass M and radius R and an axis through its center, ![I=\frac{2}{5}M.R^2](https://tex.z-dn.net/?f=I%3D%5Cfrac%7B2%7D%7B5%7DM.R%5E2)
using eq. (1)
![L_s=(\frac{2}{5}m_E.R_E^2) \omega](https://tex.z-dn.net/?f=L_s%3D%28%5Cfrac%7B2%7D%7B5%7Dm_E.R_E%5E2%29%20%5Comega)
![L_s=\frac{2}{5} \times 5.97\times 10^{24} \times 6.38\times 10^6\times \frac{2\pi}{86400}](https://tex.z-dn.net/?f=L_s%3D%5Cfrac%7B2%7D%7B5%7D%20%5Ctimes%205.97%5Ctimes%2010%5E%7B24%7D%20%5Ctimes%206.38%5Ctimes%2010%5E6%5Ctimes%20%5Cfrac%7B2%5Cpi%7D%7B86400%7D)
![L_s=7.07 \times 10^{23} \, kg.m^2.s^{-1}](https://tex.z-dn.net/?f=L_s%3D7.07%20%5Ctimes%2010%5E%7B23%7D%20%5C%2C%20kg.m%5E2.s%5E%7B-1%7D)
Newtons approach was creative because he thought of the new idea passing a single color of light through a prism. His approach was also logical. He reasoned that comparing white light and a single color of light would allow him to compare the two
The sequence of nucleotides in DNA genes determines the order of amino acids in a protein. This is the direct connection between your genes and your traits.
Answer:
the sound intensity level for actual intensity ( without the earplugs ) is 103.8 dB
Explanation:
Given the data in the question;
sound intensity reduced by the factor, m = 305
the sound intensity level experienced by the crew members wearing protective earplugs, L = 79 dB
Now, using the expression of sound intensity level;
L = 10log(
)
where
is the intensity at L level
so we substitute
79 = 10log(
)
= ![10^{7.9](https://tex.z-dn.net/?f=10%5E%7B7.9)
Now, expression for actual intensity;
= m![I_0](https://tex.z-dn.net/?f=I_0)
where
is the actual intensity
so we substitute
= 305 × ![10^{7.9](https://tex.z-dn.net/?f=10%5E%7B7.9)
Next, we write the expression of sound intensity level for reduced intensity;
L' = 10log(
)
So we substitute
L' = 10log( 305 ×
)
L' = 10log( 24227011159.09058 )
L' = 103.8 dB
Therefore, the sound intensity level for actual intensity ( without the earplugs ) is 103.8 dB
The negative work done states that the work is done by the object and not on the object.
<u>Explanation:
</u>
According to work energy theorem, the work done is equal to change in kinetic energy exhibited by the body. As the mass of the object is constant, and the velocity is decreased from 10 m/s to 4 m/s, the work done will be
![W=\frac{1}{2} \times\left(u^{2}-v^{2}\right)](https://tex.z-dn.net/?f=W%3D%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%5Cleft%28u%5E%7B2%7D-v%5E%7B2%7D%5Cright%29)
Here W is the work done, m is the mass and v is the final and u is the initial velocity of the object. As the initial velocity is greater than the final velocity. So
![W=\frac{1}{2} \times\left(4^{2}-10^{2}\right)=\frac{1}{2} \times(16-100)=\frac{1}{2} \times(-84)= - 42 J](https://tex.z-dn.net/?f=W%3D%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%5Cleft%284%5E%7B2%7D-10%5E%7B2%7D%5Cright%29%3D%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%2816-100%29%3D%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%28-84%29%3D%20-%2042%20J)
So the work done is negative for the given situation. The negative work done states that the work is done by the object and not on the object.