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2 years ago

When a battery does 24 j of work on 10 c of charge, the voltage it supplies is?

2 answers:

7 0

W = 24 J ( W = work done)
C = 10C (C = charge)
therefore,
V(voltage) = W/C

V = 24/10
V = 2.4V
therefore
the voltage supplied is 2.4 Volts
<span> or 2.4 V </span>

Readme [11.4K]2 years ago

6 0

1 volt = 1 Joule per Coulomb.

24 J / 10 C = 2.4 J/C = 2.4 volts

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2 years ago

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2 years ago

An engine has a hot-reservoir temperature of 980 K and a cold-reservoir temperature of 570 K. The engine operates at three-fifth

hram777 [196]

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efficiency of engine is 0.21

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7 0

3 years ago

Read 2 more answers

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2 years ago

In a 200-turn automobile alternator, the magnetic flux in each turn is ΦB = 2.50 10-4 cos ωt, where ΦB is in webers, ω is the an

n200080 [17]

Answer:

$10.63952sin(209.43951t)$

10.63952 V

Explanation:

$N_t$ = Number of turns = 200

$\phi_B$ = Magnetic flux = $2.5\times 10^{-4}cos(\omega t)$

$\omega_e$ = Engine angular speed = $1\times 10^{3}\ rpm$

Alternator angular speed is given by

$N_a=2\times \omega_e\\\Rightarrow N_a=2\times 1\times 10^{3}\\\Rightarrow N_a=2\times 10^{3}\ rpm$

$\omega=N_a\dfrac{2\pi}{60}\\\Rightarrow \omega=2\times 10^{3}\dfrac{2\pi}{60}\\\Rightarrow \omega=209.43951\ rad/s$

Induced emf is given by

$\epsilon=-N_t\dfrac{d\phi_B}{dt}\\\Rightarrow \epsilon=-200\dfrac{d}{dt}2.54\times 10^{-4}cos(209.43951 t)\\\Rightarrow \epsilon=200\times 2.54\times 10^{-4}\times 209.43951 sin(209.43951t)\\\Rightarrow \epsilon=10.63952sin(209.43951t)$

The function is $10.63952sin(209.43951t)$

The induced maximum emf is 10.63952 V

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2 years ago