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Cerrena [4.2K]
3 years ago
7

For safety, the National Electrical Code limits the allowable amount of current which such a wire may carry. When used in indoor

wiring, the limit is 20.0 A for rubber insulated wire of that size. How much power would be dissipated in the wire of the above problem when carrying the maximum allowable current
Physics
1 answer:
ioda3 years ago
4 0

Answer:

P=I^2*R=400R where R is the resistance of the wire

Explanation:

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The answer is D the rising of warm air pushing down cool air.

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A ball of mass M collides with a stick with moment of inertia I = βml2 (relative to its center, which is its center of mass). Th
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Answer:

Part a)

v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}

Part b)

v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})

Explanation:

Since the ball and rod is an isolated system and there is no external force on it so by momentum conservation we will have

Mv_o = M v_1 + m v_2

here we also use angular momentum conservation

so we have

M v_o d = M v_1 d + \beta mL^2 \omega

also we know that the collision is elastic collision so we have

v_o = (v_2 + d\omega) - v_1

so we have

\omega = \frac{v_o + v_1 - v_2}{d}

also we know

M v_o d - M v_1 d = \beta mL^2(\frac{v_o + v_1 - v_2}{d})

also we know

v_1 = v_o - \frac{m}{M}v_2

so we have

M v_o d - M(v_o - \frac{m}{M}v_2)d = \beta mL^2(\frac{v_o + v_o - \frac{m}{M}v_2 - v_2}{d})

mv_2 d = \beta mL^2\frac{2v_o}{d} - \beta mL^2(1 + \frac{m}{M})\frac{v_2}{d}

now we have

(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})v_2 = \frac{2\beta mL^2v_o}{d}

v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}

Part b)

Now we know that speed of the ball after collision is given as

v_1 = v_o - \frac{m}{M}v_2

so it is given as

v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})

3 0
3 years ago
Reread the introduction section of this activity. Write a brief conclusion for this activity
ki77a [65]

Answer:

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An ant travels 2.78 cm (West) and then turns and travels 6.25 cm (South 40 degrees East). What is the ant's total displacement?
andrey2020 [161]
From\ cosine\ theorem:\\\\
c^2=a^2+b^2-2abcos(\angle between\ a\ and\ b)\\\\
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3 years ago
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