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3 years ago

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To complete a project, 200,000 Joules of work is needed. The time taken to complete the project is 20 seconds. How much power is

needed?

2 answers:

Leno4ka [110]3 years ago

5 0

Answer:

10,000

Explanation:

200,000/20 = power needed

200,000/20 = 10,000

Hope this Helps!

MArishka [77]3 years ago

3 0

Answer:

10,000 Watts

Explanation:

Power is the ratio of work to time, or work over time. It can be found using the following formula.

p=w/t

where p is the power, w is the work, and t is the time.

The project needed 200,000 Joules, and it took 20 seconds to complete the project. Therefore, w=200,000 and t=20. Substitute these values into the formula.

p=200,000/20

Divide

p=10,000

Add appropriate units. The units for power is Watts.

p=10,000 Watts

10,000 Watts of power is needed.

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A large crane consists of a 20 m, 3,000 kg arm that extends horizontally on top of a vertical tower. The arm extends 15 m toward

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Explanation:

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2 years ago

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

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

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

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

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

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

$\sqrt{3} :\sqrt{10}$

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