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3 years ago

Somebody please help me with this question..!

Physics

1 answer:

Rudik [331]3 years ago

3 0

8.1) Here, Pressure increases with depth in the ocean at higher rate then that of increase in altitude in atmosphere. So, the rate of change of pressure is different in these journeys.So, your most correct answer would be option D.

8.2) Difference in the shape of lines is due to different density of air & water. So, that physical property and your answer would be option A

So, in Short Answers of your questions are:
8.1) - Option D
8.2) - Option A

Hope this helps!

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An object is placed on a surface. A student tries to apply various combinations of forces on the object. Which pair of forces wi

AVprozaik [17]

Answer:

See Explanation

Explanation:

The question is incomplete, as there are no diagrams or options to provide more information to the question.

The general explanation is as follows:

For the object not to move

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(2) The two forces must be equal.

So, for instance:

If the pair of forces are 5N and 5N in opposite directions, the object wil not move.

However, if one of the forces is greater, the object will move towards the direction of the greater force.

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2 years ago

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Answer:

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Explanation:

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3 years ago

Why a small car can stop easier than an 18 wheeler while traveling at 70 mph down the interstate?

almond37 [142]

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2 years ago

Find the density of seawater at a depth where the pressure is 500 atm if the density at the surface is 1100 kg/m^3 . Seawater ha

mixer [17]

The density of seawater at a depth where the pressure is 500 atm is $1124kg/m^3$

Explanation:

The relationship between bulk modulus and pressure is the following:

$B=\rho_0 \frac{\Delta p}{\Delta \rho}$

where

B is the bulk modulus

$\rho_0$ is the density at surface

$\Delta p$ is the variation of pressure

$\Delta \rho$ is the variation of density

In this problem, we have:

$B=2.3\cdot 10^9 N/m^2$ is the bulk modulus

$\rho_0 =1100 kg/m^3$

$\Delta p = p-p_0 = 500 atm - 1 atm = 499 atm = 5.05\cdot 10^7 Pa$ is the change in pressure with respect to the surface (the pressure at the surface is 1 atm)

Therefore, we can find the density of the water where the pressure is 500 atm as follows:

$\rho = \rho_0 + \Delta \rho = \rho_0+\frac{\rho_0 \Delta p}{B}=\rho_0 (1+\frac{\Delta p}{B})=(1100)(1+\frac{5.05\cdot 10^7}{2.3\cdot 10^9})=1124kg/m^3$