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Ierofanga [76]
2 years ago
8

HELP HELP HELP! Brainiest to the correct answer.

Physics
2 answers:
Trava [24]2 years ago
8 0

Answer:

V = W = 12 J/C * 150 C = 1800 J      work in moving 150 C of charge

P = W / t = 1800 J / 60 s  = 30 watts

Goshia [24]2 years ago
8 0

Answer:

30 Watts

Explanation:

We're provided with the following information:

  • Amount of Charge flown (Q) = 150 C
  • Potential Difference (V) = 12 V
  • Time (T) = 1 min

= 60 seconds

<h3>Current:</h3>

Current is the rate of flow of charge ,i. e., The amount of charge flowing "per second":

\boxed{\mathfrak{Current(I)=\frac{Charge(Q)}{Time(T)} }}

We have:

  • Charge flown (Q) = 150 C
  • Time (T) = 60 seconds

Finding the current flown using the above mentioned formula:

\implies \mathsf{I=\frac{150}{60} }

\implies \mathsf{I=\frac{5}{2}  }

<u>=> I = 2.5 A</u>

<em>A(Amperes) is the SI unit of Current</em>

<em />

<h3>Relation between Power, Voltage and Current:</h3>

We've a relation between Power, Current and Voltage, derived with the help of  "<em>Joule's Law of Heating"</em>, and that is:

\boxed{\mathfrak{Power(P)=Voltage(V) \times Current(I)}}

<em />

We can finally find Power, Since we have:

  • Current (I) = 2.5 A
  • Voltage (V) = 12 V

=> P = 12 × 2.5

<u>=> P = 30 W</u>

<em>W, Watts, is the Si unit of Power</em>

<em />

<em />

<em />

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A copper wire and a tungsten wire of the same length have the same resistance. What is the ratio of the diameter of the copper w
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Answer:

Therefore the ratio of diameter of the copper to that of the tungsten is

\sqrt{3} :\sqrt{10}

Explanation:

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R=\rho\frac{l}{A}

ρ is the resistivity.

The unit of resistance is ohm (Ω).

The resistivity of copper(ρ₁) is 1.68×10⁻⁸ ohm-m

The resistivity of tungsten(ρ₂) is 5.6×10⁻⁸ ohm-m

For copper:

A=\pi r_1^2 =\pi (\frac{d_1}{2} )^2

R_1=\rho_1\frac{l_1}{\pi(\frac{d_1}{2})^2 }

\Rightarrow (\frac{d_1}{2})^2=\rho_1\frac{l_1}{\pi R_1 }......(1)

Again for tungsten:

R_2=\rho_2\frac{l_2}{\pi(\frac{d_2}{2})^2 }

\Rightarrow (\frac{d_2}{2})^2=\rho_2\frac{l_2}{\pi R_2 }........(2)

Given that R_1=R_2   and    l_1=l_2

Dividing the equation (1) and (2)

\Rightarrow\frac{ (\frac{d_1}{2})^2}{ (\frac{d_2}{2})^2}=\frac{\rho_1\frac{l_1}{\pi R_1 }}{\rho_2\frac{l_2}{\pi R_2 }}

\Rightarrow( \frac{d_1}{d_2} )^2=\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}   [since R_1=R_2   and    l_1=l_2]

\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}}

\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{3}{10}}

\Rightarrow d_1:d_2=\sqrt{3} :\sqrt{10}

Therefore the ratio of diameter of the copper to that of the tungsten is

\sqrt{3} :\sqrt{10}

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