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3 years ago

What will most likely happen if a sound wave moves from the air through a solid

1 answer:

6 0

Depends on the structural integrity of the solid. We can use a subwoofer speaker for example; technically its a solid that could influence sound waves.

So basically, sound could travel through solids but, you need to know the thickness, mass and density of that solid; in-order to make conclusions.

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What is escape velocity

Bogdan [553]

The minimum speed needed a non-propelled object to escape from the gravitational influence of a massive body, that is to achieve an infinite distance from it. I think

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3 years ago

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An ideal step-down transformer is needed to reduce a primary voltage of 120 V to 6.0 V. What must be the ratio of the number of

Julli [10]

Answer:

$N_s : N_p = 20 : 1$

Explanation:

From the question we are told that

The primary voltage is $V_p = 120 \ V$

The secondary voltage is $V_s = 6 \ V$

Generally from the transformer equation we have that

$\frac{V_p}{V_s} = \frac{N_p}{N_s}$

$\frac{120}{6} = \frac{N_p}{N_s}$

=> $\frac{N_p}{N_s} = 20$

Therefore the ratio of the number of turns in the secondary to the number of turns in the primary is

$N_s : N_p = 20 : 1$

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3 years ago

How many changes of state are shown?<br> 0<br> 1<br> 2<br> 3

ivann1987 [24]

Answer: 2

I just took the test.

You’re welcome

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3 years ago

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Two small conducting point charges, separated by 0.4 m, carry a total charge of 200 C. They repel one another with a force of 12

Lunna [17]

Answer:

200 C

Explanation:

Let C1 and C2 be their charges. According to Coulomb's law

$F_C = k\frac{C_1C_2}{R^2}$

where k = $8.99\times10^9 nm^2/C^2$ is the constant, R = 0.4m is the distance between them, F = 120 N is their resulting charge force

$120 = 8.99\times10^9\frac{C_1C_2}{0.4^2}$

$C_1C_2 = \frac{120*0.4^2}{8.99\times10^9} = 2.13\times10^{-9}$

Since their total charge is 200C:

$C_1 + C_2 = 200$ or $C_1 = 200 - C_2$

We can substitute the above equation

$C_1C_2 = (200 - C_2)C_2 = 2.13\times10^{-9}$

$-C_2^2 +200C_2 - 2.13\times10^{-9} = 0$

$C= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$C= \frac{-200\pm \sqrt{(200)^2 - 4*(-1)*(-0.00000000213)}}{2*(-1)}$

$C= \frac{-200\pm200}{-2}$

$C = 1.06 \times 10^{-11}$ or $C \approx 200$

So the larger charge is C = 200 C

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3 years ago

On a distant planet, the GPE of a 65 kg astronaut is 11,115 j when they are on a 46 m tall cliff. What is the acceleration due t

Ainat [17]

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Explanation:

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