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gavmur [86]
3 years ago
13

Is there any electric field inside a conductor carrying an electric current ??

Physics
1 answer:
denis-greek [22]3 years ago
3 0
I believe so not 100% sure but i am about 50% sure 
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Instructions:Select all the correct answers.
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The correct answer is 1 and 4.

Burning of fossil fuels releases harmful gases to the atmosphere which leads to climate change. Burning of hydrogen gas releases water which has no negative effect on climate.

Burning of both fossil fuels and hydrogen takes place inside expensive fuel cells.

Fossil fuels are widely available and easy to locate whereas hydrogen is available in the combined form with other elements. Hydrogen are not found in pure form on earth.

Fossil fuels are non-renewable source of energy which means they cannot be replaced in short time.

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Water (2510 g ) is heated until it just begins to boil. if the water absorbs 5.01×105 j of heat in the process, what was the ini
natka813 [3]
E=energy=5.09x10^5J = 509KJ 
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<span>E=mcΔT </span>
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5 0
3 years ago
A radar for tracking aircraft broadcasts a 12 GHz microwave beam from a 2.0-m-diameter circular radar antenna. From a wave persp
NISA [10]

A) 750 m

First of all, let's find the wavelength of the microwave. We have

f=12GHz=12\cdot 10^9 Hz is the frequency

c=3.0\cdot 10^8 m/s is the speed of light

So the wavelength of the beam is

\lambda=\frac{c}{f}=\frac{3\cdot 10^8 m/s}{12\cdot 10^9 Hz}=0.025 m

Now we can use the formula of the single-slit diffraction to find the radius of aperture of the beam:

y=\frac{m\lambda D}{a}

where

m = 1 since we are interested only in the central fringe

D = 30 km = 30,000 m

a = 2.0 m is the aperture of the antenna (which corresponds to the width of the slit)

Substituting, we find

y=\frac{(1)(0.025 m)(30000 m)}{2.0 m}=375 m

and so, the diameter is

d=2y = 750 m

B) 0.23 W/m^2

First we calculate the area of the surface of the microwave at a distance of 30 km. Since the diameter of the circle is 750 m, the radius is

r=\frac{750 m}{2}=375 m

So the area is

A=\pi r^2 = \pi (375 m)^2=4.42\cdot 10^5 m^2

And since the power is

P=100 kW = 1\cdot 10^5 W

The average intensity is

I=\frac{P}{A}=\frac{1\cdot 10^5 W}{4.42\cdot 10^5 m^2}=0.23 W/m^2

4 0
3 years ago
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