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Alla [95]
2 years ago
8

State advantages of ultrasonic sound in determining the depth of the ocean ​

Physics
1 answer:
agasfer [191]2 years ago
5 0

Answer:

It has <u>greater accuracy than other nondestructive methods in determining the depth of internal flaws and the thickness of parts with parallel surfaces.</u>

Explanation:

Hope this helps you!

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In a belly-flop diving contest, the winner is the diver who makes the biggest splash upon hitting the water. the size
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The second diver have to leap to make a competitive splash by 4.08 m high.

<h3>What is potential energy?</h3>

The energy by virtue of its position is called the potential energy.

PE = mgh

where, g = 9.81 m/s²

Given is the diver jumps from a 3.00-m platform. one diver has a mass of 136 kg and simply steps off the platform. another diver has a mass of 100 kg and leaps upward from the platform.

The potential energy of the first diver must be equal to the second diver.

P.E₁ = P.E₂

m₁gh₁ = m₂gh₂

Substitute the vales, we have

136 x 3  = 100 x h₂

h₂ = ₂4.08 m

Thus, the second diver need to leap by 4.08 m high.

Learn more about potential energy.

brainly.com/question/24284560

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Calculate the orbital period of a dwarf planet found to have a semimajor axis of a = 4.0x 10^12 meters in seconds and years.
padilas [110]

Explanation:

We have,

Semimajor axis is 4\times 10^{12}\ m

It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :

T^2=\dfrac{4\pi ^2}{GM}a^3

G is universal gravitational constant

M is solar mass

Plugging all the values,

T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.98\times 10^{30}}\times (4\times 10^{12})^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.98\times10^{30}}\times(4\times10^{12})^{3}}\\\\T=4.37\times 10^9\ s

Since,

1\ s=3.17\times 10^{-8}\ \text{years}\\\\4.37\times 10^9\ s=4.37\cdot10^{9}\cdot3.17\cdot10^{-8}\\\\4.37\times 10^9\ s=138.52\ \text{years}

So, the orbital period of a dwarf planet is 138.52 years.

3 0
3 years ago
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