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3 years ago

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A piston above liquid in a closed container has an area of 1m2. The piston carries a load of 350 kg. What will be the external p

ressure on the upper surface of the liquid in KPA

2 answers:

grandymaker [24]3 years ago

5 0

<span>Pressure = force / area</span>

I assume that 350kg is the mass
Therefore,

350 x 9.8 (gravity) = 3430N

3430 / 1 = 3430Pa

3.43 KPa

ollegr [7]3 years ago

3 0

Answer:

3.43 kPa

Explanation:

Pressure is defined as the ratio between the force applied and the area of the surface on which the force is applied:

$p=\frac{F}{A}$

in this problem, the force corresponds to the weight of the load, which is given by the product between its mass (350 kg) and the acceleration due to gravity (9.8 m/s^2):

$F=mg=(350 kg)(9.8 m/s^2)=3430 N$

And since the area on which this force is applied is $A=1 m^2$ , the pressure exerted on the upper surface of the liquid is

$p=\frac{F}{A}=\frac{3430 N}{1 m^2}=3430 Pa = 3.43 kPa$

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Answer:

The speed of the package of mass m right before the collision $= 7.668\ ms^-1$

Their common speed after the collision $= 2.56\ ms^-1$

Height achieved by the package of mass m when it rebounds $= 0.33\ m$

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

where $K$ is Kinetic energy and $U$ is Potential energy.

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Considering the fact $K_{initial} = 0\ and U_{final} =0$ we will plug out he values of the given terms.

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Sum of energies and momentum are conserved in all collisions.
Sum of KE and PE is also known as Mechanical energy.
Only KE is conserved for elastic collision.
for elastic collison we have $e=1$ that is co-efficient of restitution.

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b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

$m_1V_1(i) = (m_1+m_2)V_f$ where $V_1{i} =7.668\ ms^-1$ .

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$V_f = 2.56\ ms^-1$

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c.Now the collision is perfectly elastic $e=1$

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