Answer:
See answer
Explanation:
The area of the circular loop is given by:
![A = \pi r^2](https://tex.z-dn.net/?f=A%20%3D%20%5Cpi%20r%5E2)
The magnetic flux is given by:
![\phi = \int \vec{B} \cdot d\vec{A}](https://tex.z-dn.net/?f=%5Cphi%20%3D%20%5Cint%20%5Cvec%7BB%7D%20%5Ccdot%20d%5Cvec%7BA%7D)
is parallel to
and
is constant in magnitude and direction therefore:
![\phi = \int \vec{B} \cdot d\vec{A}= \int BdAcos(0)= B\int dA= B*(\pi r^2)= \pi Br^2](https://tex.z-dn.net/?f=%5Cphi%20%3D%20%5Cint%20%5Cvec%7BB%7D%20%5Ccdot%20d%5Cvec%7BA%7D%3D%20%5Cint%20BdAcos%280%29%3D%20B%5Cint%20dA%3D%20B%2A%28%5Cpi%20r%5E2%29%3D%20%5Cpi%20Br%5E2)
Part A)
initially the flux is ![\phi =\pi B r^2](https://tex.z-dn.net/?f=%5Cphi%20%3D%5Cpi%20B%20r%5E2)
after the interval
the flux is
![\phi = 0](https://tex.z-dn.net/?f=%5Cphi%20%3D%200)
now, the EMF is defined as:
,
if we consider
very small then we can re-write it as:
![\epsilon =- \frac{\Delta \phi}{\Delta t}](https://tex.z-dn.net/?f=%5Cepsilon%20%3D-%20%5Cfrac%7B%5CDelta%20%5Cphi%7D%7B%5CDelta%20t%7D)
then:
![\epsilon =- \frac{-0.12}{0.0024} = 50 [V]](https://tex.z-dn.net/?f=%5Cepsilon%20%3D-%20%5Cfrac%7B-0.12%7D%7B0.0024%7D%20%3D%2050%20%5BV%5D)
Part B)
When looked down from above, the current flows counter clockwise, according to the right hand rule, if you place your thumb upwards (the direction of the magnetic field) and close your fingers, then the current will flow in the direction of your fingers.
Answer:
b. cannot be effectively tested
Explanation:
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)
Answer:
Option b, pothographs from drones.
Explanation:
the USGS (U.S. Geological Survey) decided to make photographic captures from drones to the volcanic surfaces, which allowed through observations to understand things like the characteristics of the lava, the height of the volcanic plumes (among others).
Podemos ver en el siguiente enlace un ejemplo de fotografía tomada desde un dron al Kilauea.
https://www.usgs.gov/media/images/k-lauea-volcano-drone-over-lava-channel