The solution to the questions are given as
![t=40.39 \mathrm{sec}](https://tex.z-dn.net/?f=t%3D40.39%20%5Cmathrm%7Bsec%7D)
![\varepsilon &=(0.12v)e^{0.057t}](https://tex.z-dn.net/?f=%5Cvarepsilon%20%26%3D%280.12v%29e%5E%7B0.057t%7D)
- the direction of induced current will be Counterclock vise.
<h3>What is the direction of the
current induced in the loop, as viewed from above the loop.?</h3>
Given, $B(t)=(1.4 T) e^{-0.057 t}$
![$\varepsilon m f(\varepsilon)=-\frac{d \phi_{B}}{d t}](https://tex.z-dn.net/?f=%24%5Cvarepsilon%20m%20f%28%5Cvarepsilon%29%3D-%5Cfrac%7Bd%20%5Cphi_%7BB%7D%7D%7Bd%20t%7D)
![\quad$ and, $\phi_{B}=\int B \cdot d A=\int B \cdot d A \cdot \cos \theta$](https://tex.z-dn.net/?f=%5Cquad%24%20and%2C%20%24%5Cphi_%7BB%7D%3D%5Cint%20B%20%5Ccdot%20d%20A%3D%5Cint%20B%20%5Ccdot%20d%20A%20%5Ccdot%20%5Ccos%20%5Ctheta%24)
![\begin{aligned}\text { Here, } \theta &=30^{\circ} ; \\A &=\pi r^{2} \\a n \delta, R &=0.75 \mathrm{~m} \\\therefore \varepsilon &=-\frac{d}{d t}(B A \cdot \cos \theta)=-A \cdot \cos \theta \cdot \frac{d}{d t}(B(t)) \\\therefore \varepsilon &=-\pi R^{2} \cdot \cos \theta \cdot \frac{d}{d t}\left(e^{-0.057 t}\right)(1.4 T) \\\therefore \varepsilon &=+\pi(0.75)^{2} \cdot \cos 30 \cdot(0.057)(1.4) \cdot e^{-0.057 t}\left\{\because \frac{d}{d t} e^{-x}=-x \cdot e^{-x} .\right.\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%5Ctext%20%7B%20Here%2C%20%7D%20%5Ctheta%20%26%3D30%5E%7B%5Ccirc%7D%20%3B%20%5C%5CA%20%26%3D%5Cpi%20r%5E%7B2%7D%20%5C%5Ca%20n%20%5Cdelta%2C%20R%20%26%3D0.75%20%5Cmathrm%7B~m%7D%20%5C%5C%5Ctherefore%20%5Cvarepsilon%20%26%3D-%5Cfrac%7Bd%7D%7Bd%20t%7D%28B%20A%20%5Ccdot%20%5Ccos%20%5Ctheta%29%3D-A%20%5Ccdot%20%5Ccos%20%5Ctheta%20%5Ccdot%20%5Cfrac%7Bd%7D%7Bd%20t%7D%28B%28t%29%29%20%5C%5C%5Ctherefore%20%5Cvarepsilon%20%26%3D-%5Cpi%20R%5E%7B2%7D%20%5Ccdot%20%5Ccos%20%5Ctheta%20%5Ccdot%20%5Cfrac%7Bd%7D%7Bd%20t%7D%5Cleft%28e%5E%7B-0.057%20t%7D%5Cright%29%281.4%20T%29%20%5C%5C%5Ctherefore%20%5Cvarepsilon%20%26%3D%2B%5Cpi%280.75%29%5E%7B2%7D%20%5Ccdot%20%5Ccos%2030%20%5Ccdot%280.057%29%281.4%29%20%5Ccdot%20e%5E%7B-0.057%20t%7D%5Cleft%5C%7B%5Cbecause%20%5Cfrac%7Bd%7D%7Bd%20t%7D%20e%5E%7B-x%7D%3D-x%20%5Ccdot%20e%5E%7B-x%7D%20.%5Cright.%5Cend%7Baligned%7D)
![\varepsilon &=(0.12v)e^{0.057t}](https://tex.z-dn.net/?f=%5Cvarepsilon%20%26%3D%280.12v%29e%5E%7B0.057t%7D)
(b) ![Here, $\varepsilon_{0}=0.12 \mathrm{~V} \quad\left(a t_{2} t=0 \mathrm{sec}\right)$](https://tex.z-dn.net/?f=Here%2C%20%24%5Cvarepsilon_%7B0%7D%3D0.12%20%5Cmathrm%7B~V%7D%20%5Cquad%5Cleft%28a%20t_%7B2%7D%20t%3D0%20%5Cmathrm%7Bsec%7D%5Cright%29%24)
![\begin{aligned}&\therefore 1 . \varepsilon_{0}=\varepsilon_{0} \cdot e^{-e .057 t} \\&\therefore e^{0.057 t}=10 \quad \text { (taking log both thesides) } \\&\therefore 0.057 t=\ln (10)=2.303 \\&\therefore t=40.39 \mathrm{sec}\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%26%5Ctherefore%201%20.%20%5Cvarepsilon_%7B0%7D%3D%5Cvarepsilon_%7B0%7D%20%5Ccdot%20e%5E%7B-e%20.057%20t%7D%20%5C%5C%26%5Ctherefore%20e%5E%7B0.057%20t%7D%3D10%20%5Cquad%20%5Ctext%20%7B%20%28taking%20log%20both%20thesides%29%20%7D%20%5C%5C%26%5Ctherefore%200.057%20t%3D%5Cln%20%2810%29%3D2.303%20%5C%5C%26%5Ctherefore%20t%3D40.39%20%5Cmathrm%7Bsec%7D%5Cend%7Baligned%7D)
c)
In conclusion, the direction of the induced current will be Counterclockwise.
Read more about current
brainly.com/question/13076734
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Answer: P= mad/t or P=w/t so P= 300/6= 50 W
Answer:
I don't know why you are asking me?
Answer:
D = -4/7 = - 0.57
C = 17/7 = 2.43
Explanation:
We have the following two equations:
![3C + 4D = 5\ --------------- eqn (1)\\2C + 5D = 2\ --------------- eqn (2)](https://tex.z-dn.net/?f=3C%20%2B%204D%20%3D%205%5C%20---------------%20eqn%20%281%29%5C%5C2C%20%2B%205D%20%3D%202%5C%20---------------%20eqn%20%282%29)
First, we isolate C from equation (2):
![2C + 5D = 2\\2C = 2 - 5D\\C = \frac{2 - 5D}{2}\ -------------- eqn(3)](https://tex.z-dn.net/?f=2C%20%2B%205D%20%3D%202%5C%5C2C%20%3D%202%20-%205D%5C%5CC%20%3D%20%5Cfrac%7B2%20-%205D%7D%7B2%7D%5C%20--------------%20eqn%283%29)
using this value of C from equation (3) in equation (1):
![3(\frac{2-5D}{2}) + 4D = 5\\\\\frac{6-15D}{2} + 4D = 5\\\\\frac{6-15D+8D}{2} = 5\\\\6-7D = (5)(2)\\7D = 6-10\\\\D = -\frac{4}{7}](https://tex.z-dn.net/?f=3%28%5Cfrac%7B2-5D%7D%7B2%7D%29%20%2B%204D%20%3D%205%5C%5C%5C%5C%5Cfrac%7B6-15D%7D%7B2%7D%20%2B%204D%20%3D%205%5C%5C%5C%5C%5Cfrac%7B6-15D%2B8D%7D%7B2%7D%20%3D%205%5C%5C%5C%5C6-7D%20%3D%20%285%29%282%29%5C%5C7D%20%3D%206-10%5C%5C%5C%5CD%20%3D%20-%5Cfrac%7B4%7D%7B7%7D)
<u>D = - 0.57</u>
Put this value in equation (3), we get:
![C = \frac{2-(5)(\frac{-4}{7} )}{2}\\\\C = \frac{\frac{14+20}{7}}{2}\\\\C = \frac{34}{(7)(2)}\\\\C = \frac{17}{7}\\](https://tex.z-dn.net/?f=C%20%3D%20%5Cfrac%7B2-%285%29%28%5Cfrac%7B-4%7D%7B7%7D%20%29%7D%7B2%7D%5C%5C%5C%5CC%20%3D%20%5Cfrac%7B%5Cfrac%7B14%2B20%7D%7B7%7D%7D%7B2%7D%5C%5C%5C%5CC%20%3D%20%5Cfrac%7B34%7D%7B%287%29%282%29%7D%5C%5C%5C%5CC%20%3D%20%20%5Cfrac%7B17%7D%7B7%7D%5C%5C)
<u>C = 2.43</u>
Explanation:
When bullet is shot towards the monkey then let say the distance of monkey from the bullet is "d"
so we can find the time to reach the bullet to the monkey
![t = \frac{d}{vcos\theta}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7Bd%7D%7Bvcos%5Ctheta%7D)
Now similarly we can find the vertical displacement of the bullet in the same time
![\Delta y = vsin\theta t - \frac{1}{2}gt^2](https://tex.z-dn.net/?f=%5CDelta%20y%20%3D%20vsin%5Ctheta%20t%20-%20%5Cfrac%7B1%7D%7B2%7Dgt%5E2)
![\Delta y = v sin\theta (\frac{d}{vcos\theta}) - \frac{1}{2}gt^2](https://tex.z-dn.net/?f=%5CDelta%20y%20%3D%20v%20sin%5Ctheta%20%28%5Cfrac%7Bd%7D%7Bvcos%5Ctheta%7D%29%20-%20%5Cfrac%7B1%7D%7B2%7Dgt%5E2)
so it is given as
![\Delta y = d tan\theta - \frac{1}{2}gt^2](https://tex.z-dn.net/?f=%5CDelta%20y%20%3D%20d%20tan%5Ctheta%20-%20%5Cfrac%7B1%7D%7B2%7Dgt%5E2)
here if the monkey is initially at height H above the ground at given angle then we can say
![H = dtan\theta](https://tex.z-dn.net/?f=H%20%3D%20dtan%5Ctheta)
so we can say that
![\Delta y = H - \frac{1}{2}gt^2](https://tex.z-dn.net/?f=%5CDelta%20y%20%3D%20H%20-%20%5Cfrac%7B1%7D%7B2%7Dgt%5E2)
So if at the same time monkey will fall down then the height of monkey from ground after time "t" is given as
![\Delta y = H - \frac{1}{2}gt^2](https://tex.z-dn.net/?f=%5CDelta%20y%20%3D%20H%20-%20%5Cfrac%7B1%7D%7B2%7Dgt%5E2)
so here bullet will hit the monkey as both monkey and bullet are at same position.