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Lapatulllka [165]
3 years ago
12

Use image for both problems

Physics
1 answer:
Alchen [17]3 years ago
4 0

a potential

b kinetic

c potential

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V = 73.8 m/s

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An isotropic point source emits light at wavelength 510 nm, at the rate of 170 W. A light detector is positioned 410 m from the
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Answer:

\frac{dB}{dt} = 3.03 \times 10^6 T/s

Explanation:

As we know that the power emitted by the source is given as

P = 170 W

now we know that

P = \frac{N}{t} (\frac{hc}{\lambda})

now we know that energy density is given as

u = \frac{B^2}{2\mu_0} + \frac{\epsilon_0 E^2}{2}

now we have

E = B c

u = \frac{B^2}{2\mu_0}

intensity is defined as

I = \frac{P}{A}

now we have

\frac{I}{c} = u = \frac{B^2}{2\mu_0}[/tex]

now we have

\frac{dB}{dt} = \omega B

\frac{dB}{dt} = \frac{2\pi c B}{\lambda}

\frac{dB}{dt} = \frac{2\pi c \sqrt{2\mu_0 I}}{\lambda\sqrt c}

here we have

I = \frac{P}{4\pi r^2}

I = \frac{170}{4\pi (410)^2}

I = 8.05 \times 10^{-5}

now we have

\frac{dB}{dt} = \frac{2\pi\sqrt{2\mu_0 c (8.05 \times 10^{-5})}}{(510 nm)}

\frac{dB}{dt} = 3.03 \times 10^6 T/s

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Explanation:

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What did Thomson’s and Rutherford’s experiments have in common? They both used charged particles in their experiments. They both
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Both Thomson and Rutherford used charged particles in their experiments.

Explanation:

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3 years ago
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