Query 5

course Phy 202

005. `query 5

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Question: query introset plug of water from cylinder given gauge pressure, c-s hole area, length of plug.

Explain how we can determine the velocity of exiting water, given the pressure difference between inside and outside, by considering a plug of known cross-sectional area and length?

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Your Solution:

Force= Pressure/Area acting with the distance the tube covers

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Given Solution:

** The net force on the plug is P * A, where A is its cross-sectional area and P the pressure difference.

• If L is the length of the plug then the net force F_net = P * A acts thru distance L doing work `dW = F_net * L = P * A * L.

If the initial velocity of the plug is 0 and there are no dissipative forces, this is the kinetic energy attained by the plug.

The volume of the plug is A * L so its mass is rho * A * L.

• Thus we have mass rho * A * L with KE equal to P * A * L.

Setting .5 m v^2 = KE we have

.5 rho A L v^2 = P A L so that

v = sqrt( 2 P / rho).

Your Self-Critique:

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Question: prin phy and gen phy 10.25 spherical balloon rad 7.35 m total mass 930 kg, Helium => what buoyant force

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Your Solution:

Buoyant force is equal to the amount of space that the object is displaced.

Vol sphere= (4/3)pi r^3

Vol= (4/3)pi (7.35m)^3= 529.4 pi m^3

m air displaced = 1.3 kg / m^3 * 1660 m^3 = 2160 kg* 9.8 m/s^2= 21168 N

Weight= Consider the force gravity combined with the mass of the balloon= 930kg* 9.8 m/s^2= 9114 N

The net force is the buoyant force minus the weight (mass +gravity)= 21168N-9114N= 12054 N

The Helium must be considered in the weight.

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Given Solution:

** The volume of the balloon is about 4/3 pi r^3 = 1660 cubic meters and mass of air displaced is about 1.3 kg / m^3 * 1660 m^3 = 2160 kg.

The buoyant force is equal in magnitude to the force of gravity on the displaced air, or about 2160 kg * 9.8 m/s^2 = 20500 Newtons, approx.. If the total mass of the balloon, including helium, is 930 kg then the weight is 930 kg * 9.8 m/s^2 = 9100 N approx., and the net force is

• Net force = buoyant force - weight = 20,500 N - 9100 N = 11,400 N, approximately.

If the 930 kg doesn't include the helium we have to account also for the force of gravity on its mass. At about .18 kg/m^3 the 1660 m^3 of helium will have mass about 300 kg on which gravity exerts an approximate force of 2900 N, so the net force on the balloon would be around 11,400 N - 2900 N = 8500 N approx.

The mass that can be supported by this force is m = F / g = 8500 N / (9.8 m/s^2) = 870 kg, approx.. **

Your Self-Critique:

I’m not exactly sure what is going on during the last part of the given solution. I’m just confused about where the numbers come from.

Can you be more specific? I'm not sure where 'last part' starts or what numbers you don't understand.

The only new number introduced there is the density of the helium, which can be looked up from various sources or calculated from P V = n R T and the fact that a mole of helium has mass 4 grams (mass of a mole in grams is equal atomic weight).