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shtirl [24]
3 years ago
6

A large bottle contains 150 L of water, and is open to the atmosphere. If the bottle has a flat bottom with an area of 2 ft, cal

culate the absolute pressure at the bottom of the bottle. Give your answer in Pa. Assume the water has a density of 1000 kg/m, and assume the system is at sea level on Earth. Answer: over 100,000 Pa
Physics
1 answer:
Yuri [45]3 years ago
6 0

Answer:

Total pressure exerted at bottom =  119785.71 N/m^2

Explanation:

given data:

volume of water in bottle = 150 L = 0.35 m^3

Area of bottle = 2 ft^2

density of water = 1000 kg/m

Absolute pressure on bottom of bottle will be sum of atmospheric pressure and pressure due to water

Pressure due to water P = F/A

F, force exerted by water = mg

m, mass of water = density * volume

                             =  1000*0.350 = 350 kg

F  = 350*9.8 = 3430 N

A = 2 ft^2 = 0.1858 m^2  

so, pressure P = 3430/ 0.1858 = 18460.71 N/m^2

Atmospheric pressure

At sea level atmospheric pressure is 101325 Pa

Total pressure exerted at bottom  = 18460.71 + 101325 = 119785.71 N/m^2

Total pressure exerted at bottom =  119785.71 N/m^2

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3 years ago
A person is trying to lift a crate that has a mass of 30 kg. The normal force of the floor is currently supplying 150N of force.
alexdok [17]

Here when an object is placed on the level floor then in that case there are two forces on the object

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2). Normal force due to floor which will counterbalance the weight (N)

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Now here it is given that A person tried to lift the box upwards

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So now these two forces will counter balance the weight of the crate

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So force applied by the person must be 150 N

7 0
3 years ago
An experimenter finds that standing waves on a string fixed at both ends occur at 24 Hz and 32 Hz , but at no frequencies in bet
Vera_Pavlovna [14]

Answer:

8 Hz

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Standing wave at one end is 24 Hz

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f(m+1) - f(m) = (m + 1).f(1) - m.f(1) = f(1)

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4 0
3 years ago
A 0.5-kg ball moving at 5 m/s strikes a wall and rebounds in the opposite direction with a speed of 2 m/s. If the impulse occurs
mart [117]

<em>Given that:</em>

                       mass of the ball (m) = 0.5 Kg ,

                    ball strikes the wall (v₁) = 5 m/s ,

rebounds in opposite direction (v₂) = 2 m/s,

                                time duration (t) = 0.01 s,

        <em> Determine the force (F) = ?</em>

We know that from Newton's II law,

                                <em>F = m. a</em>  Newtons  

                                  (velocity acting in opposite direction, so <em>a = ( (v₁ + v₂)/t</em>

                                   = m × (v₁ + v₂)/t

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<em>The force acting up on the ball is 350 N</em>

                                     

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