Answer:
![\Delta E=2.87\times 10^6\ J](https://tex.z-dn.net/?f=%5CDelta%20E%3D2.87%5Ctimes%2010%5E6%5C%20J)
Explanation:
It is given that,
Depth of Death valley is 85 m below sea level, ![h_i=-85\ m](https://tex.z-dn.net/?f=h_i%3D-85%5C%20m)
The summit of nearby Mt. Whitney has an elevation of 4420 m, ![h_f=4420\ m](https://tex.z-dn.net/?f=h_f%3D4420%5C%20m)
Mass of the hiker, m = 65 kg
We need to find the change in potential energy. It is given by :
![\Delta E=mg(h_f-h_i)](https://tex.z-dn.net/?f=%5CDelta%20E%3Dmg%28h_f-h_i%29)
![\Delta E=65\times 9.8(4420-(-85))](https://tex.z-dn.net/?f=%5CDelta%20E%3D65%5Ctimes%209.8%284420-%28-85%29%29)
![\Delta E=2869685\ J](https://tex.z-dn.net/?f=%5CDelta%20E%3D2869685%5C%20J)
or
![\Delta E=2.87\times 10^6\ J](https://tex.z-dn.net/?f=%5CDelta%20E%3D2.87%5Ctimes%2010%5E6%5C%20J)
So, the change in potential energy of the hiker is
. Hence, this is the required solution.
Um this doesn't make since to me since you did not clearly state your awnser
Answer:
The maximum change in flux is ![\Delta \o = 0.1404 \ Wb](https://tex.z-dn.net/?f=%5CDelta%20%5Co%20%3D%200.1404%20%5C%20Wb)
The average induced emf ![\epsilon =0.11232 V](https://tex.z-dn.net/?f=%5Cepsilon%20%3D0.11232%20V)
Explanation:
From the question we are told that
The speed of the technician is ![v = 0.80 m/s](https://tex.z-dn.net/?f=v%20%3D%200.80%20m%2Fs)
The distance from the scanner is ![d = 1.0m](https://tex.z-dn.net/?f=d%20%3D%201.0m)
The initial magnetic field is ![B_i = 0T](https://tex.z-dn.net/?f=B_i%20%3D%200T)
The final magnetic field is ![B_f = 6.0T](https://tex.z-dn.net/?f=B_f%20%3D%206.0T)
The diameter of the loop is ![D = 19cm = \frac{19}{100} = 0.19 m](https://tex.z-dn.net/?f=D%20%3D%2019cm%20%3D%20%5Cfrac%7B19%7D%7B100%7D%20%3D%200.19%20m)
The area of the loop is mathematically represented as
![A = \pi [\frac{D}{2} ]^2](https://tex.z-dn.net/?f=A%20%20%3D%20%20%5Cpi%20%5B%5Cfrac%7BD%7D%7B2%7D%20%5D%5E2)
![= 3.142 \frac{0.19}{2}](https://tex.z-dn.net/?f=%3D%203.142%20%5Cfrac%7B0.19%7D%7B2%7D)
![= 0.02834 m^2](https://tex.z-dn.net/?f=%3D%200.02834%20m%5E2)
At maximum the change in magnetic field is mathematically represented as
![\Delta \o = (B_f - B_i)A](https://tex.z-dn.net/?f=%5CDelta%20%5Co%20%3D%20%28B_f%20-%20B_i%29A)
=> ![\Delta \o = (6 -0)(0.02834)](https://tex.z-dn.net/?f=%5CDelta%20%20%5Co%20%3D%20%286%20-0%29%280.02834%29)
![\Delta \o = 0.1404 \ Wb](https://tex.z-dn.net/?f=%5CDelta%20%5Co%20%3D%200.1404%20%5C%20Wb)
The average induced emf is mathematically represented as
![\epsilon = \Delta \o v](https://tex.z-dn.net/?f=%5Cepsilon%20%3D%20%20%5CDelta%20%5Co%20v)
![= 0.1404 * 0.80](https://tex.z-dn.net/?f=%3D%200.1404%20%2A%200.80)
![\epsilon =0.11232 V](https://tex.z-dn.net/?f=%5Cepsilon%20%3D0.11232%20V)
You are correct...amplitude will be the answer
In your question where the ask is to calculate the charge that the small sphere carries which is the mass of it is 441g moving at an acceleration of 13m/s^2 nad having and electric field of 5N/C. So the formula in getting the charge is mutliply the mass and the quotients of Acceleration and the Electric Field so the answer is 1,146.6