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serg [7]
3 years ago
7

A burst of sound takes 0.4 seconds to reach the seabed and return, if speed of sound in water is 1400m/s. what is the depth of w

ater?
Physics
2 answers:
Rama09 [41]3 years ago
6 0

Explanation:

<u>Given:</u>

Speed of sound = 1400 m/s

Total time taken = 0.4 seconds

<u>To find</u><u>:</u>

The depth of water

<u>Solution</u><u>:</u>

As it take 0.4 seconds for the burst to reach the seabed and return (both ways), we can deduce that it takes 0.2 seconds to reach the seabed (one way).

This is because the speed of sound is a constant.

Now, to find the depth, we use the equation

Speed = Distance/Time

Here, we know the values of speed (1400 m/s) and time (0.2 seconds). Distance is the depth of the water, which we need to find. Let the distance be x.

Therefore

1400 = x/0.2

x = 1400 * 0.2

x = 280 m

We get the depth as 280 metres.

Feel free to ask me if you didn't understand any part.

Hope this helps! :D

spayn [35]3 years ago
6 0
0.4 seconds bae hope this helps thanks your welcome
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Answer: 3.7 \times 10^{-4} N

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F_{net] = G\frac{M_eM_v}{r_v^2}+G\frac{M_eM_j}{r_j^2}+G\frac{M_eM_s}{r_s^2}

Mass of Earth, Me = 5.98 × 10²⁴ kg

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Mass of Jupiter, Mj = 318 Me

Mass of Saturn, Ms = 95.1 Me

closest distance between Earth and Venus, rv = 38 × 10⁶ km = 0.25 AU

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Inserting the values:

F_{net} = G\frac{M_e\times 0.815 M_e}{(0.25AU)^2}+G\frac{M_e\times 318 M_e}{(3.93AU)^2}+G\frac{M_e\times 95.1 M_e}{(8.0AU)^2}\\ \Rightarrow F_{net} = \frac{(GM_e^2)}{(1AU)^2}(\frac{0.815}{0.25^2}+\frac{318}{3.93^2}+\frac{95.1}{8.0^2})=\frac{6.67\times 10^{-11} \times (5.98\times 10^{24})^2}{(1.5\times 10^{11})^2}(35.1) = 3.7 \times 10^{-4} N

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During a full time period T, the mass on the spring oscillates back and forth, returning to its original position. This means that the total distance covered by the mass during a period T is 4 times the amplitude (4A), because the amplitude is just half the distance between the maximum and the minimum position, and during a time period the mass goes from the maximum to the minimum, and then back to the maximum.

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posledela

Answer:

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