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ANEK [815]
3 years ago
7

How is refraction different from reflection?

Physics
1 answer:
olga_2 [115]3 years ago
5 0
<span>The angle of refraction is not necessarily equal to the angle
of incidence, whereas the angle of reflection always is.</span>
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A

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accuracy

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You are trying to measure the mass of several different objects when you realize that there is a large wad of gum stuck to the underside of the balance pan. Removing the gum will improve the <u>accuracy</u> of your measurements.

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Town A lies 15 km north of town B. Town C lies 10 km west of town A. A small plane flies directly from town B to town C. What is
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In calisthenics, the resistance is ultimately provided by __________. A. free weights B. springs and elastic bands C. gravity D.
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Starting from zero, the electric current takes 2 seconds to reach half its maximum possible value in an RL circuit with a resist
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Answer:

time=4s

Explanation:

we know that in a RL circuit with a resistance R, an inductance L and a battery of emf E, the current (i) will vary in following fashion

i(t)=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}}), where imax=\frac{E}{R}

Given that, at i(2)=\frac{imax}{2} =\frac{E}{2R}

⇒\frac{E}{2R}=\frac{E}{R}(1-e^\frac{-2}{\frac{L}{R}})

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⇒\frac{1}{2}=e^\frac{-2}{\frac{L}{R}}

Applying logarithm on both sides,

⇒log(\frac{1}{2})=\frac{-2}{\frac{L}{R}}

⇒log(2)=\frac{2}{\frac{L}{R}}

⇒\frac{L}{R}=\frac{2}{log2}

Now substitute i(t)=\frac{3}{4}imax=\frac{3E}{4R}

⇒\frac{3E}{4R}=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})

⇒\frac{3}{4}=1-e^\frac{-t}{\frac{L}{R}}

⇒\frac{1}{4}=e^\frac{-t}{\frac{L}{R}}

Applying logarithm on both sides,

⇒log(\frac{1}{4})=\frac{-t}{\frac{L}{R}}

⇒log(4)=\frac{t}{\frac{L}{R}}

⇒t=log4\frac{L}{R}

now subs. \frac{L}{R}=\frac{2}{log2}

⇒t=log4\frac{2}{log2}

also log4=log2^{2}=2log2

⇒t=2log2\frac{2}{log2}

⇒t=4

5 0
3 years ago
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