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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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A bicycle pump contains 200 cm3 of air and is connected to a bicycle tyre. The volume of the tyre is 800 cm3. The pressure of th
Aleksandr [31]

Answer:

The total volume of the air is 1000 cubic centimeters.

Explanation:

Since the bicycle pump and the bicycle tyre have the same pressure, then the total volume of the air is the sum of the volume of each element, then we translate this into the following artihmetical expression:

V = 200\,cm^{3}+800\,cm^{3}

V = 1000\,cm^{3}

The total volume of the air is 1000 cubic centimeters.

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3 years ago
Which of the following is true about the easy walk harness
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A flea jumps straight up to a maximum height of 0.550 m . what is its initial velocity v0 as it leaves the ground?
Alexxx [7]
Since my givens are x = .550m [Vsub0] = unknown
 [Asubx] = =9.80
 
 [Vsubx]^2 = [Vsub0x]^2 + 2[Asubx] * (X-[Xsub0]

[Vsubx]^2 = [Vsub0x]^2 + 2[Asubx] * (X-[Xsub0]) 

Vsubx is the final velocity, which at the max height is 0, and Xsub0 is just 0 as that's where it starts so I just plug the rest in

0^2 = [Vsub0x]^2 + 2[-9.80]*(.550)

0 = [Vsub0x]^2 -10.78

10.78 = [Vsub0x]^2

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3 0
3 years ago
How much work must be done to bring three electrons from a great distance apart to 5.0×10^−10 m from one another (at the corners
Inessa05 [86]

Answer:

1.38 x 10^-18 J

Explanation:

q = - 1.6 x 10^-19 C

d = 5 x 10^-10 m

the potential energy of the system gives the value of work done

The formula for the potential energy is given by

U =\frac{Kq_{1}q_{2}}{d}

So, the total potential energy of teh system is

U =\frac{Kq_{1}q_{2}}{d}+\frac{Kq_{2}q_{3}}{d}+\frac{Kq_{1}q_{3}}{d}

As all the charges are same and the distance between the two charges is same so the total potential energy becomes

U =3\times \frac{Kq^{2}}{d}

K = 9 x 10^9 Nm^2/C^2

By substituting the values

U =3\times \frac{9\times 10^{9}\times \ 1.6 \times 1.6 \times 10^{-38}}{5\times 10^{-10}}

U = 1.38 x 10^-18 J

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