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Andre45 [30]
3 years ago
13

F = 3e8 divided by 6.56E-9 NO ANSWERS?

Physics
1 answer:
julsineya [31]3 years ago
7 0

Answer:

0.461\times10^{17}

Explanation:

The expression can be written as:

\frac{3\times10^8}{6.5\times10^{-9}}

That can be further simplified as:

\frac{3}{6.5}\times \frac{10^8}{10^{-9}}

Using \frac{1}{a^{-x}} = a^x\\\\ \frac{1}{10^{-9}}=10^9

Also, using the law of indices:

a^x\times a^y= a^{(x+y)}\\\\10^8\times10^9= 10^{(8+9)}= 10^{17}

Also, \frac{3}{6.5}=0.465

Now, the expression becomes:

=0.465\times10^8\times10^9\\=0.465\times10^{(8+9)}\\\\=0.465\times10^{17}

Therefore, the answer is 0.465\times10^{17}.

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Answer:

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3 years ago
What must be the magnitude of a uniform electric field if it is to have the same energy density as that possessed by a 0.50 T ma
Sindrei [870]

Answer:

E = 1.50 × 10^{8} V/m

Explanation:

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B = 0.50 T

solution

we know that energy density by the magnetic field is express as

\mu _b = \frac{B}{2\mu _o}   ...............1

and

energy density due to electric filed is

\mu _e = \frac{\epsilon _o E^2}{2}     ...............2

and here \mu _b = \mu _ e

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E = \frac{B}{\sqrt{\mu _o \times \epsilon _o}}      ...................3

put here value and we get

E = \frac{0.50}{\sqrt{4\pi \times 10^{-7} \times 8.852 \times 10^{-12}}}  

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