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frosja888 [35]
3 years ago
12

Fluid pressure changes with depth are assumed to be linear. Which statement best explains why this does not hold true for atmosp

heric pressure changes?
Physics
1 answer:
ankoles [38]3 years ago
4 0

Answer:

Explanation:

Pressure due to fluid is directly proportional to the depth of fluid, density of the fluid and the value of acceleration due to gravity.

P = h d g

Where, h is the depth, d be the density and g be the acceleration due to gravity.

If we talk about teh atmospheric pressure, the density of air goes on decreasing as we go up and up. o we cannot say that it is directly depends only on the depth of air, it also depends on the changing density of air.

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Why might you want to use a single fixed pulley?
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A single fixed pulley can be used to raise or lower lightweight objects.

Option b

<u>Explanation:</u>

A pulley is a simple machine tool which is used to make lifting or lowering tasks easy.  A single fixed pulley is a system involving only one pulley fixed on a constant rigid support with a rope wrapped around the wheel. Such a system can be used only to change the direction of  applied force in raising or lowering small, lightweight objects which need minimal work force.

A single fixed pulley system helps only in redirecting the applied force direction by using a rope and wheel assembly. The work done in such a case remains the same and hence it is not preferred to use it in lifting heavy objects. Neither is the required force reduced in case of a single fixed pulley system. A movable pulley helps in achieving (A) and (C).

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A person experiencing heat
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Answer:

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The 2004 Indian Ocean tsunami was caused by a shift in two plates that pushed one section of the sea floor under another section
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DUE TONIGHT HURRY PLS I give 30 pts
Paha777 [63]

The answer is both.

For kinetic energy:

KE = 1/2*m*v^2 = 0.5*20,000 grams*5 = 50,000 J

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Two current-carrying wires are parallel to each other. The current in one is increased by a factor of 2 , the current in the oth
ladessa [460]

Answer:

The new force F_N will be \frac{6}{13} times the old force F. The change then will be \Delta F=-\frac{7}{13}F

Explanation:

The force between two current-carrying parallel wires is calculated with the formula:

F=\frac{\mu_0I_1I_2\Delta L}{2\pi r}

where r is the distance between them, \Delta L a portion of length of the wires we consider, I_1 and I_2 their current intensity and \mu_0=4\pi\times10^{-7}N/A^2 the vacuum permeability.

If the current in one wire is increased by a factor of 2, the current in the other is increased by a factor of 3 , and the distance between the wires is decreased by a factor of 13, then we would have a new situation N where (considering the previous variables as an initial situation):

I_{N1}=2I_1\\I_{N2}=3I_2\\r_N=13r\\

And the force then will be:

F_N=\frac{\mu_0I_{N1}I_{N2}\Delta L}{2\pi r_N}=\frac{\mu_02I_13I_2\Delta L}{2\pi 13r}=\frac{6(\mu_0I_1I_2\Delta L)}{13(2\pi r)}=\frac{6}{13}F

So the change will be:

\Delta F=F_N-F=\frac{6}{13} F-F=(\frac{6}{13} -1)F=-\frac{7}{13}F

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3 years ago
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