Answer:
![7.3114*10^{-5}V](https://tex.z-dn.net/?f=7.3114%2A10%5E%7B-5%7DV)
Explanation:
To give a solution to the exercise, it is necessary to consider the concepts related to magnetic flux and Faraday's law of induction. Faraday's law states that the voltage induced in a closed circuit is directly proportional to the speed with which the magnetic flux that crosses any surface with the circuit as an edge changes over time.
It is represented under the equation,
![\epsilon = N \frac{\Delta\Phi}{\Delta t}](https://tex.z-dn.net/?f=%5Cepsilon%20%3D%20N%20%5Cfrac%7B%5CDelta%5CPhi%7D%7B%5CDelta%20t%7D)
Where,
is the induced electromotive force
N = Number of loops
= Time
= Magnetic Flux
For definition the change in magnetic flux is:
![\Delta \Phi = \Delta B A cos\phi](https://tex.z-dn.net/?f=%5CDelta%20%5CPhi%20%3D%20%5CDelta%20B%20A%20cos%5Cphi)
Where,
B= Magnetic field
Substituting at the first equation we have
![\epsilon = N \frac{\Delta B A Cos\phi}{\Delta t}](https://tex.z-dn.net/?f=%5Cepsilon%20%3D%20N%20%5Cfrac%7B%5CDelta%20B%20A%20Cos%5Cphi%7D%7B%5CDelta%20t%7D)
![\epsilon = N \frac{(B_2-B_1) (\pi r^2) Cos\phi}{\Delta t}](https://tex.z-dn.net/?f=%5Cepsilon%20%3D%20N%20%5Cfrac%7B%28B_2-B_1%29%20%28%5Cpi%20r%5E2%29%20Cos%5Cphi%7D%7B%5CDelta%20t%7D)
Our values are given by,
N = 1 turn
![B_2 = 1T](https://tex.z-dn.net/?f=B_2%20%3D%201T)
![B_1 = 0T](https://tex.z-dn.net/?f=B_1%20%3D%200T)
r = 1.6mm
![\phi = 0\°](https://tex.z-dn.net/?f=%5Cphi%20%3D%200%5C%C2%B0)
![\Delta t = 100ms](https://tex.z-dn.net/?f=%5CDelta%20t%20%3D%20100ms)
Replacing,
![\epsilon = (1) \frac{(1-0) (\pi (1.6*10^{-3})^2) Cos(0)}{110*10^{-3}}](https://tex.z-dn.net/?f=%5Cepsilon%20%3D%20%281%29%20%5Cfrac%7B%281-0%29%20%28%5Cpi%20%281.6%2A10%5E%7B-3%7D%29%5E2%29%20Cos%280%29%7D%7B110%2A10%5E%7B-3%7D%7D)
![\epsilon = 7.311*10^{-5}V](https://tex.z-dn.net/?f=%5Cepsilon%20%3D%207.311%2A10%5E%7B-5%7DV)
<em>Therefore the magnitud of the induced emf around a horizontal circle of tissue is
</em>
Answer:
The answer is 8.44*10^9 Newton
Explanation:
The equation for universal gravitation is:
![F=G*m_{1}*m_{2}/r^{2}](https://tex.z-dn.net/?f=F%3DG%2Am_%7B1%7D%2Am_%7B2%7D%2Fr%5E%7B2%7D)
Where:
G= 6.674×10^−11 N · (m/kg)^2
M(earth)=5.972 × 10^24 kg
M(asteroid)=5.0×10^13 kg
r=1.537*10^9 m (Calculated from average distance between moon and earth)
The force will be:
![F=6.674*10^{-11}*5.972*10^{24}*5*10^{13}/(1.537*10^9)^2\\F=8.44*10^9](https://tex.z-dn.net/?f=F%3D6.674%2A10%5E%7B-11%7D%2A5.972%2A10%5E%7B24%7D%2A5%2A10%5E%7B13%7D%2F%281.537%2A10%5E9%29%5E2%5C%5CF%3D8.44%2A10%5E9)
8.44*10^9 Newton
The gravity ( im not sure)