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Ronch [10]
3 years ago
6

Through which one of the following mediums is the velocity of a sound wave the greatest?

Physics
2 answers:
Scorpion4ik [409]3 years ago
7 0

Answer: C. Steel

Explanation: 100% correct on pf test. Hope this help everyone.

tino4ka555 [31]3 years ago
4 0

Answer: C. Steel

Explanation: When a sound wave travels through a solid body consisting

of an elastic material, the velocity of the wave is relatively

high. For instance, the velocity of a sound wave traveling

through steel (which is almost perfectly elastic) is about

5,060 meters per second. On the other hand, the velocity

of a sound wave traveling through an inelastic solid is

relatively low. So, for example, the velocity of a sound wave

traveling through lead (which is inelastic) is approximately

1,402 meters per second.

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What is the difference between a theory and a law?
algol [13]

A theory is a system of ideas that isn't exactly proven to be true fully. A law is a description of whatever scientific phenomena you're studying. All you need to know is a law describes, and a theory explains.

5 0
4 years ago
O How are electricity and magnetism related?
Nataly [62]
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4 years ago
How would the moon appear from Earth if the moon did not rotate?
andrey2020 [161]
It would mean that only one side of earth would be light and the other dark all the time also we would only see the sun on one side and on the other we see the moon
6 0
3 years ago
Read 2 more answers
The Moon requires about 1 month (0.08 year) to orbit Earth. Its distance from us is about 400,000 km (0.0027 AU). Use Kepler’s t
dem82 [27]

Answer:

\frac{M_e}{M_s} = 3.07 \times 10^{-6}

Explanation:

As per Kepler's III law we know that time period of revolution of satellite or planet is given by the formula

T = 2\pi \sqrt{\frac{r^3}{GM}}

now for the time period of moon around the earth we can say

T_1 = 2\pi\sqrt{\frac{r_1^3}{GM_e}}

here we know that

T_1 = 0.08 year

r_1 = 0.0027 AU

M_e = mass of earth

Now if the same formula is used for revolution of Earth around the sun

T_2 = 2\pi\sqrt{\frac{r_2^3}{GM_s}}

here we know that

r_2 = 1 AU

T_2 = 1 year

M_s = mass of Sun

now we have

\frac{T_2}{T_1} = \sqrt{\frac{r_2^3 M_e}{r_1^3 M_s}}

\frac{1}{0.08} = \sqrt{\frac{1 M_e}{(0.0027)^3M_s}}

12.5 = \sqrt{(5.08 \times 10^7)\frac{M_e}{M_s}}

\frac{M_e}{M_s} = 3.07 \times 10^{-6}

4 0
3 years ago
Please help!
andrew-mc [135]

Answer:

V=W/Q

107V= W/17C

= We= 107×17 J

= 1819 J

Explanation:

hope it helps

3 0
3 years ago
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