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Ymorist [56]
3 years ago
6

In a circuit with one 12.0 resistor, a current of 0.500 A is flowing. This circuit is powered by a single

Physics
1 answer:
Juliette [100K]3 years ago
4 0

Answer:

6 V

Explanation:

In a circuit, the relationship between voltage, current and resistance is given by Ohm's law:

V=RI

where

V is the voltage

R is the resistance

I is the current

In this circuit, we have

R=12.0\Omega is the resistance

I = 0.500 A

so we can find the voltage of the battery by using Ohm's law:

V=(12.0 \Omega)(0.500 A)=6 V

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3 years ago
A piano wire with mass 2.95 g and length 79.0 cm is stretched with a tension of 29.0 N . A wave with frequency 105 Hz and amplit
Romashka-Z-Leto [24]

The concept needed to solve this problem is average power dissipated by a wave on a string. This expression ca be defined as

P = \frac{1}{2} \mu \omega^2 A^2 v

Here,

\mu = Linear mass density of the string

\omega =  Angular frequency of the wave on the string

A = Amplitude of the wave

v = Speed of the wave

At the same time each of this terms have its own definition, i.e,

v = \sqrt{\frac{T}{\mu}} \rightarrow Here T is the Period

For the linear mass density we have that

\mu = \frac{m}{l}

And the angular frequency can be written as

\omega = 2\pi f

Replacing this terms and the first equation we have that

P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2(\sqrt{\frac{T}{\mu}})

P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2 (\sqrt{\frac{T}{m/l}})

P = 2\pi^2 f^2A^2(\sqrt{T(m/l)})

PART A ) Replacing our values here we have that

P = 2\pi^2 (105)^2(1.8*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})

P = 0.2320W

PART B) The new amplitude A' that is half ot the wavelength of the wave is

A' = \frac{1.8*10^{-3}}{2}

A' = 0.9*10^{-3}

Replacing at the equation of power we have that

P = 2\pi^2 (105)^2(0.9*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})

P = 0.058W

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3 years ago
What is the voltage in a circuit if the current is 6.2 A and the resistance is 18 ohms? A) 2.9 V B) 11.8 V C) 24.2 V D) 111.6 V
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V = I · R  =  (6.2 A) · (18 Ω) =  111.6 volts  (D)
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