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iren [92.7K]
3 years ago
10

How to solve number 10

Physics
1 answer:
Vinil7 [7]3 years ago
6 0

The 'formulas' to use are just the definitions of 'power' and 'work':

Power = (work done) / (time to do the work)

and  

Work = (force) x (distance) .

Combine these into one. Take the definition of 'Work', and write it in place of 'work' in the definition of power.

Power = (force x distance) / (time)

From the sheet, we know the power, the distance, and the time.  So we can use this one formula to find the force.

Power = (force x distance) / (time)

Multiply each side by (time):  (Power) x (time) = (force) x (distance)

Divide each side by (distance): Force = (power x time) / (distance).

Look how neat, clean, and simple that is !

Force = (13.3 watts) x (3 seconds) / (4 meters)

Force = (13.3 x 3 / 4) (watt-seconds / meter)

Force = 39.9/4 (joules/meter)

<em>Force = 9.975 Newtons</em>

Is that awesome or what !

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The solution to the questions are given as

  • t=40.39 \mathrm{sec}
  • \varepsilon &=(0.12v)e^{0.057t}
  • 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}

\quad$ and, $\phi_{B}=\int B \cdot d A=\int B \cdot d A \cdot \cos \theta$

\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}

\varepsilon &=(0.12v)e^{0.057t}

(b) Here, $\varepsilon_{0}=0.12 \mathrm{~V} \quad\left(a t_{2} t=0 \mathrm{sec}\right)$

\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}

c)

In conclusion, the direction of the induced current will be Counterclockwise.

Read more about current

brainly.com/question/13076734

#SPJ1

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