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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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5 0
3 years ago
5.
iris [78.8K]

Answer:

t = 1.659s

Explanation:

We can use the kinematics equations to solve this questions:

v = u + at

v^{2} = u^{2} +2as

where v = Final Velocity, u = initial velocity, a = acceleration, t = time, s = displacement

a) Given information from the question,

u = \frac{70km}{h} =\frac{(70*1000)m}{(1*3600)s} = 19.444m/s (Convert km/h to m/s first)

a = 2m/s^{2}

s = 35m

Now we can substitute these values into the 2nd kinematics equation to find v, final velocity.

v^{2} =(19.444)^{2} +2(2)(35)\\v=\sqrt{(19.444)^{2} +2(2)(35)} \\v= 22.761m/s (5.sf)\\

b) Now we have the final velocity, we can substitute the values into the first kinematics equation to find t , the time taken.

v = u + at

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t =\frac{22.761-19.444}{2}

t = 1.659s

7 0
2 years ago
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