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Sergio [31]
3 years ago
11

Calculate the pressure on the bottom of a swimming pool 3. 5 m deep. How does the pressure compare with atmospheric pressure, 10

5 Pa? (density of water=1000kg/m3 ).
Physics
1 answer:
Nostrana [21]3 years ago
5 0

Answer:

  1.343 atm

Explanation:

The mass of water above 1 square meter of swimming pool bottom is ...

  M = (3.5 m)·(1000 kg/m^3) = 3500 kg/m^2

Then the force exerted by the water on the pool bottom is ...

  F = Mg = (3500 kg/m^2)(9.8 m/s^2) = 34300 N/m^2 = 34300 Pa

Compared with atmospheric pressure, this is ...

  34,300/10^5 = 0.343 . . . . atmospheres

Added to the atmospheric pressure on the water's surface, the total pressure on the pool bottom is 1.343 atmospheres.

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joja [24]

Answer:

Force on the object is 20 N

Explanation:

As we know that work done to raise the book from initial position to final position is known as potential energy stored in it

So here we know that

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here we know that

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so we have

30 = F(1.5)

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6 0
3 years ago
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BabaBlast [244]

Answer:

yo they deleted my answer. The answer is 0N

Explanation:

so when two forces pull on an object from opposite sides with the same force (in this case its 20N), then the object is in equilibrium at 0N.

So its clear that there is one person on the the opposite side.

SOOO  generally<u>: (left or down) would be considered </u><u>negative</u><u> in an equation. And the other person  (right or up) would be considered </u><u>positive</u><u>.</u> So if both forces are the same numbers on opposite sides then the answer is 0 (if you add both of them).

<em>0 is the number of equilibrium.</em>

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thanks for coming to my TED talk. I hope they don't delete this answer.

5 0
3 years ago
When a certain gas under a pressure of 4.90 106 pa at 20.0°c is allowed to expand to 3.00 times its original volume, its final p
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As per question the initial states of the gases are given as

INITIAL STATE:                                          FINAL STATE:

p_{1} =4.90106 pa                       p_{2} =1.06106 pa

v_{1} =v[say]                                 v_{2} =3v[say]

T_{1} =20 degree celcius  =293 K              T_{2} =?

AS  per combined gas equation obtained from the combination of Boyle's law and Charles law [Basic ideal gas laws]

              \frac{p_{1} v_{1} }{T_{1} } =\frac{p_{2}v_{2}  }{T_{2} }

Hence T_{2} =\frac{p_{2} v_{2}T_{1}  }{p_{1} v_{1} }

                    =\frac{1.06106*3v*293}{4.90106*v}

                    =190.3 K [ANS]

                 

   

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