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frez [133]
4 years ago
14

Which configuration would produce an electric current? A) Rotate a coil of copper wire between two magnets. B) Connect a wire be

tween a copper and zinc strip sitting in a beaker of water. C) Connect a wire to the (+) positive end of a battery and the other end to a light bulb. D) Connect a wire to the (-) negative end of a battery and the other end to a light bulb.
Physics
2 answers:
Lelu [443]4 years ago
8 0

Answer:

A.) Rotate a coil of copper wire between two magnets.

Explanation:

Rotating a coil of copper wire between two magnets would produce an electric current. This is an example of electromagnetic induction.

FrozenT [24]4 years ago
4 0

the answer is (A) the movement of the magnet relative to the coil

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Plz solve this question..
ivann1987 [24]

Answer:

Explanation:

Voltage = 12 (the two batteries are in series and the voltage adds).

Resistance = 3 ohms

I = V / R

I = 12/3

I = 4 amperes.

None of the answers are the right ones. If there is a fifth one that reads 4 amperes then it is the correct answer.

4 0
3 years ago
By what factor is the intensity of sound at a rock concert louder than that of a whisper when the two intensity levels are 120 d
lyudmila [28]

Answer:

d) 7.94\times 10^{9}

Explanation:

β₁ = sound level of sound at rock concert = 120 dB

β₂ = sound level of sound due to whisper = 21 dB

I₁ = Intensity of sound at rock concert

I₂ = Intensity of sound due to whisper

sound level of sound at rock concert is given as

\beta _{1} = 10 log\left ( \frac{I_{1}}{10^{-12}} \right )

120 = 10 log\left ( \frac{I_{1}}{10^{-12}} \right )

12 = log\left ( \frac{I_{1}}{10^{-12}} \right )               Eq-1

sound level due to whisper is given as

\beta _{2} = 10 log\left ( \frac{I_{2}}{10^{-12}} \right )

21 = 10 log\left ( \frac{I_{2}}{10^{-12}} \right )

2.1 = log\left ( \frac{I_{2}}{10^{-12}} \right )                          Eq-2

subtracting Eq-2 from Eq-1

12 - 2.1 = log\left ( \frac{I_{1}}{10^{-12}} \right ) - log\left ( \frac{I_{2}}{10^{-12}} \right )

9.9 = log\left ( \frac{I_{1}}{I_{2}} \right )

\left ( \frac{I_{1}}{I_{2}} \right ) = 7.94\times 10^{9}

6 0
3 years ago
"A 2000 kg car drives around a circular exit ramp with a radius of 30 meters. The car drives all the way around the ramp in 10 s
aev [14]

The speed of the car is 18.8 m/s

Explanation:

The variables that represent the givens are:

m = 2000 kg (mass of the car)

r = 30 m (radius of the curve)

t = 10 s (time elapsed to cover the curve)

The unknown is the speed of the car, which is given by:

v=\frac{d}{t}

where d is the distance covered by the car during the time t.

Here the car covers a circular trajectory of radius r, so the distance covered is

d=2\pi r=2\pi(30)=188m

Therefore, the speed of the car is

v=\frac{188}{10}=18.8 m/s

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

8 0
3 years ago
What is the Mechanical advantage of the inclined plane? A) 0.25 B) 4 C) 90 D) 150
Hunter-Best [27]

Answer:

it depends on the weight's ratio

(sorry)

3 0
3 years ago
If an object on a horizontal, frictionless surface is attached to a spring, displaced, and then released, it will oscillate. If
Rama09 [41]

Answer:

A) 0.120

B) 1.6s

C) 0.625 Hz

Explanation:

Here, X = 0.120 m

x = 0.120 m after t = 0.800 s

A- Amplitude = max displacement from equillibrium position = 0.120 m

B- Period = 2 * 0.800 = 1.6 s

C- Frequency = 1/peroid = 1/1.6

f = 0.625 Hz

3 0
3 years ago
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