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Pepsi [2]
3 years ago
12

The crest of one wave with an amplitude of 4 m meets up with the crest of a second

Physics
1 answer:
trapecia [35]3 years ago
3 0

Answer:

Constructive

Explanation:

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Kobotan [32]

Answer:

No

Explanation:

Cause a monster truck don

3 0
2 years ago
The sun rotates once in 25.05 days. That is 601.2 hours. so the sun's rotational speed is 0.0017 rotations per hour. what is the
Margaret [11]

Answer:

v=2019.09\ m.s^{-1}

Explanation:

Given:

  • time taken by the sun to complete one revolution, t=601.2\ hr
  • radial distance of the sunspot, r=6.955\times 10^8\ m.s^{-1}

<u>Therefore, angular speed of rotation of sun:</u>

\omega=\frac{2\pi}{601.2\times 3600} \ rad.s^{-1}

<u>Now the tangential velocity of the sunspot can be given by:</u>

v=r.\omega

v=6.955\times 10^8\times \frac{2\pi}{601.2\times 3600}

v=2019.09\ m.s^{-1}

4 0
3 years ago
19. The particles in an object move very quickly and are spread apart. Then they move slower and get slower and get closer toget
svlad2 [7]

Answer:

The transformation went from a gas to a solid which is a Deposition

Explanation:

Deposition is the phase transition in which gas transforms into solid without passing through the liquid phase. Deposition is a thermodynamic process. The reverse of deposition is sublimation and hence sometimes deposition is called desublimation.

4 0
3 years ago
Apa saja manfaat sumber daya hewan?
wlad13 [49]

Hewan peliharaan, seperti ternak, memberi kita makanan, serat, dan kulit. Hewan liar, termasuk burung, ikan, serangga, dan penyerbuk, penting untuk mendukung jaring aktivitas dalam ekosistem yang berfungsi. Populasi tumbuhan dan hewan yang sehat sangat penting bagi kehidupan.

8 0
3 years ago
The volume of water in the Pacific Ocean is about 7.0 × 10 8 km 3 . The density of seawater is about 1030 kg/m3. (a) Determine t
Novay_Z [31]

To solve the problem it is necessary to consider the concepts related to Potential Energy and Kinetic Energy.

Potential Energy because of a planet would be given by the equation,

PE=\frac{GMm}{r}

Where,

G = Gravitational Universal Constant

M = Mass of Ocean

M = Mass of Moon

r = Radius

From the data given we can calculate the mass of the ocean water through the relationship of density and volume, then,

m = \rho V

m = (1030Kg/m^3)(7*10^8m^3)

m = 7.210*10^{11}Kg

It is necessary to define the two radii, when the ocean is far from the moon and when it is facing.

When it is far away, it will be the total diameter from the center of the earth to the center of the moon.

r_1 = 3.84*10^8 + 6.4*10^6 = 3.904*10^8m

When it's near, it will be the distance from the center of the earth to the center of the moon minus the radius,

r_2 = 3.84*10^8-6.4*10^6 - 3.776*10^8m

PART A) Potential energy when the ocean is at its furthest point to the moon,

PE_1 = \frac{GMm}{r_1}

PE_1 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.904*10^8}

PE_1 = 9.05*10^{15}J

PART B) Potential energy when the ocean is at its closest point to the moon

PE_2 = \frac{GMm}{r_2}

PE_2 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.776*10^8}

PE_2 = 9.361*10^{15}J

PART C) The maximum speed. This can be calculated through the conservation of energy, where,

\Delta KE = \Delta PE

\frac{1}{2}mv^2 = PE_2-PE_1

v=\sqrt{2(PE_2-PE_1)/m}

v = \sqrt{\frac{2*(9.361*10^{15}-9.05*10^{15})}{7.210*10^{11}}}

v = 29.4m/s

8 0
3 years ago
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