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2 years ago

12

a flashlight operates with a current of 3.0 a and a power of 4.5 W. what is the voltage of the flashlights battery?

1 answer:

Dafna11 [192]2 years ago

4 0

Answer:

V=1.5V

Explanation:

p=I×V

4.5=3×V

$v = \frac{4.5}{3}$

$v = \frac{3}{2}$

v=1.5

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The direction of the magnetic field is parallel to the field line at any point in space. A small compass will point in a directi

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3 years ago

A 10-turn coil of wire having a diameter of 1.0 cm and a resistance of 0.50 Ω is in a 1.0 mT magnetic field, with the coil orien

n200080 [17]

Answer:

The voltage across the capacitor is 1.57 V.

Explanation:

Given that,

Number of turns = 10

Diameter = 1.0 cm

Resistance = 0.50 Ω

Capacitor = 1.0μ F

Magnetic field = 1.0 mT

We need to calculate the flux

Using formula of flux

$\phi=NBA$

Put the value into the formula

$\phi=10\times1.0\times10^{-3}\times\pi\times(0.5\times10^{-2})^2$

$\phi=7.85\times10^{-7}\ Tm^2$

We need to calculate the induced emf

Using formula of induced emf

$\epsilon=\dfrac{d\phi}{dt}$

Put the value into the formula

$\epsilon=\dfrac{7.85\times10^{-7}}{dt}$

Put the value of emf from ohm's law

$\epsilon =IR$

$IR=\dfrac{7.85\times10^{-7}}{dt}$

$Idt=\dfrac{7.85\times10^{-7}}{R}$

$Idt=\dfrac{7.85\times10^{-7}}{0.50}$

$Idt=0.00000157=1.57\times10^{-6}\ C$

We know that,

$Idt=dq$

$dq=1.57\times10^{-6}\ C$

We need to calculate the voltage across the capacitor

Using formula of charge

$dq=C dV$

$dV=\dfrac{dq}{C}$

Put the value into the formula

$dV=\dfrac{1.57\times10^{-6}}{1.0\times10^{-6}}$

$dV=1.57\ V$

Hence, The voltage across the capacitor is 1.57 V.

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3 years ago

An Olympic runner completes the 200-meter sprint in 23 seconds. What is the runner’s average speed? (Round your answer to the ne

LUCKY_DIMON [66]

The answer that is got 8.7 . I got that because if you divide 200 by 23 you get <span>8.69565217391 and if you round that you get 8.7</span>

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4 years ago

Read 2 more answers

What is the radius of a tightly wound solenoid of circular cross-section that has 180turns if a change in its internal magnetic

Nata [24]

Answer:

Radius of cross section, r = 0.24 m

Explanation:

It is given that,

Number of turns, N = 180

Change in magnetic field, $\dfrac{dB}{dt}=3\ T/s$

Current, I = 6 A

Resistance of the solenoid, R = 17 ohms

We need to find the radius of the solenoid (r). We know that emf is given by :

$E=N\dfrac{d\phi}{dt}$

$E=N\dfrac{d(BA)}{dt}$

Since, E = IR

$IR=NA\dfrac{dB}{dt}$

$A=\dfrac{IR}{N.\dfrac{dB}{dt}}$

$A=\dfrac{6\ A\times 17\ \Omega}{180\times 3\ T/s}$

$A=0.188\ m^2$

or

$A=0.19\ m^2$

Area of circular cross section is, $A=\pi r^2$

$r=\sqrt{\dfrac{A}{\pi}}$

$r=\sqrt{\dfrac{0.19}{\pi}}$

r = 0.24 m

So, the radius of a tightly wound solenoid of circular cross-section is 0.24 meters. Hence, this is the required solution.

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3 years ago

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Answer:

-963.93 m/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

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The acceleration of Superman would be -963.93 m/s² from Lois' perspective

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3 years ago