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3 years ago

Where would you find the lowest density seawater?

Physics

2 answers:

choli [55]3 years ago

8 0

<h2>Answer:</h2>

<u>The lowest density seawater is present in </u><u>Surface Mixed layer</u>

<h2>Explanation:</h2>

The ocean is layered into three zones based on the density of seawater.

The mixed layer,
The pycnoclinen layer
The deep layer.

The surface mixed layer is a layer where this turbulence is generated by winds, surface heat fluxes, or processes such as evaporation or sea ice formation which result in an increase in salinity and the decrease in the density.

VladimirAG [237]3 years ago

7 0

Surface mixed zone. :)

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3 years ago

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If the mass of the sun is 2x, at least one planet will fall into the habitable zone if I place a planet in orbits___, ____, ____

nalin [4]

So, the complete sentence is If the mass of the sun is 2x, at least one planet will fall into the habitable zone if I place a planet in orbits<u> 84, 1, </u>and <u>5</u>, and all planets will orbit the sun successfully for the best conditions.

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2 years ago

023 (part 1 of 2) 10.0 points

Annette [7]

Answer:

Part 1

The angular speed is approximately 1.31947 rad/s

Part 2

The change in kinetic energy due to the movement is approximately 675.65 J

Explanation:

The given parameters are;

The rotation rate of the merry-go-round, n = 0.21 rev/s

The mass of the man on the merry-go-round = 99 kg

The distance of the point the man stands from the axis of rotation = 2.8 m

Part 1

The angular speed, ω = 2·π·n = 2·π × 0.21 rev/s ≈ 1.31947 rad/s

The angular speed is constant through out the axis of rotation

Therefore, when the man walks to a point 0 m from the center, the angular speed ≈ 1.31947 rad/s

Part 2

Given that the kinetic energy of the merry-go-round is constant, the change in kinetic energy, for a change from a radius of of the man from 2.8 m to 0 m, is given as follows;

$\Delta KE_{rotational} = \dfrac{1}{2} \cdot I \cdot \omega ^2 = \dfrac{1}{2} \cdot m \cdot v ^2$

I = m·r²

Where;

m = The mass of the man alone = 99 kg

r = The distance of the point the man stands from the axis, r = 2.8 m

v = The tangential velocity = ω/r

ω ≈ 1.31947 rad/s

Therefore, we have;

I = 99 × 2.8² = 776.16 kg·m²

$\Delta KE_{rotational}$ = 1/2 × 776.16 kg·m² × (1.31947)² ≈ 675.65 J

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3 years ago

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Alekssandra [29.7K]

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4 years ago

11. A bowling ball has a mass of 7 kg and a velocity of 3 m/s.<br> (A) What is the ball's momentum?

DaniilM [7]

Answer: 21 kg x m/s

Momentum can be found using the equation p = mv, or momentum = mass times velocity.

Therefore, to find the momentum of the bowling bowl, you can plug the mass and velocity values into the equation to get p = (7kg)(3m/s)

Solve the equation to get a momentum of 21kg x m/s

I hope I could help :)

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3 years ago