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stira [4]
3 years ago
9

You have a large flashlight that takes 4 D-cell batteries, each with a voltage of 1.5 volts. If the current in the flashlight is

2.0 amps, what is the resistance?
Physics
2 answers:
olganol [36]3 years ago
8 0
Resistance = voltage/current
= 1.5v/2amp
= .75 ohm
Elden [556K]3 years ago
8 0

Answer:

3 ohm

Explanation:

Use the concept of Ohm's law

V = I × R

Voltage of each battery = 1.5 V

Voltage of 4 batteries = 4 × 1.5 = 6 V

Current I = 2 A

6 = 2 × R

R = 3 ohm

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Answer:

The magnetic field strength due to current flowing in the wire is9.322 x 10⁻⁶ T.

Explanation:

Given;

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The strength of the resulting magnetic field at the given distance is calculated as;

B = \frac{\mu_o I}{2\pi R}

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B = \frac{\mu_o I}{2\pi R}\\\\B = \frac{4\pi*10^{-7} *21.3}{2\pi(0.457)}\\\\B = 9.322 *10^{-6} \ T

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4 years ago
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ad-work [718]

Answer:

True A and B

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Let's apply this equation to the case presented. The index of refraction and airs is 1 (n1 = 1)

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Let's analyze the statements time

A. False. We saw that it deviates

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3 years ago
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Answer:

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