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goldfiish [28.3K]
4 years ago
12

Which one of the following is not equivalent to 2.50 miles?

Physics
1 answer:
lisabon 2012 [21]4 years ago
6 0

Answer:

Choice C is not equivalent to 2.50 miles.

Explanation:

The given data is now converted into feet, inches, kilometers, yards and centimeters:

mi - ft

x = (2.50\,mi)\cdot \left(5280\,\frac{ft}{mi} \right)

x = 13200\,ft

x = 1.320\times 10^{4}\,ft (Choice A)

mi - in

x = (2.50\,mi)\cdot \left(5280\,\frac{ft}{mi} \right)\cdot \left(12\,\frac{in}{ft} \right)

x = 158400\,in

x = 1.584\times 10^{5} (Choice B)

mi - km

x = (2.50\,mi)\cdot \left(1.61\,\frac{km}{mi} \right)

x = 4.025\,km (Different from Choice C)

mi - yd

x = (2.50\,mi)\cdot \left(5280\,\frac{ft}{mi}\right) \cdot \left(\frac{1}{3}\,\frac{yd}{ft}  \right)

x = 4400\,yd

x = 4.40\times 10^{3}\,yd (Choice D)

mi - cm

x = (2.50\,mi)\cdot \left(1.61\,\frac{km}{mi} \right)\cdot \left(100000\,\frac{cm}{km}\right)

x = 402500\,cm

x = 4.025\times 10^{5}\,cm (Choice E)

Choice C is not equivalent to 2.50 miles.

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Answer:

When like charges come together, they repel each other. For instance, when the north and south poles of a magnet come together, they push each other apart. The like poles in the magnet repel each other and unlike poles attract each other much. The same reaction occurs in like and unlike charges. Also, the repulsion acts along the line between the two charges.

4 0
3 years ago
A sound has a sound level of 30 dB. Its intensity is what multiple of the standard reference level for intensities?
arsen [322]
<h2>Answer:</h2>

1000th multiple of the standard reference level for intensities.

<h2>Explanation:</h2>

The sound intensity level (β), measured in decibels, of a sound with an intensity of I is defined as follows;

β = 10 log (I / I₀)       --------------------(i)

Where;

I₀ = reference intensity

Given from the question;

β = sound level = 30dB

Substitute this value into equation (i) as follows;

30 = 10 log (I / I₀)

Divide both sides by 3;

3 = log (I / I₀)

Take antilog of both sides;

10^(3) = (I / I₀)

1000 = I / I₀

Solve for I;

I = 1000I₀

Therefore the intensity of the sound is 1000 times the standard reference level for intensities (I₀)

7 0
3 years ago
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kolezko [41]
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If the phase of the vibrating sources was changed so that they were vibrating completely out of phase, what effect would this ha
Over [174]

Answer:

There would be complete destructive interference.

Explanation:

This is because since the waves are completely out of phase, the phase difference is half wavelength, that is the phase angle is 180°. The vibrating sources are 180° out of phase with each other.

Since this is the case, the crest of the one source meets the trough of the other, this causes the resultant vibrational wave to cancel out, thus producing a destructive interference pattern.

Since the vibrating sources are completely out of phase, every point they meet is completely out of phase, so the resultant interference pattern would produce a complete destructive interference pattern of no wave.

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In order to draw the free body diagram, first let's calculate the friction force acting on the crate:

\begin{gathered} F_f=N\cdot\mu \\ F_f=40\cdot9.8\cdot0.7 \\ F_f=274.4\text{ N} \end{gathered}

Since the friction force is greater than the force applied, the crate will not move, and the friction force will be equal to the force applied.

The weight force is equal to 40 * 9.8 = 392 N.

So, drawing the diagram, we have:

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