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Vladimir [108]
3 years ago
13

The pressure in a compressed air storage tank is 1200 kPa. What is the tank’s pressure in (a) kN and m units; (b) kg, m, and s u

nits; and (c) kg, km, and s units?
Physics
1 answer:
elixir [45]3 years ago
8 0

Explanation:

Given data

Pressure P=1200 kPa

To find

Pressure in

(a) kN/m²

(b) kg/m.s²

(c) kg/km.s²

Solution

For Part (a)

P=(1200kPa)(\frac{1kN/m^{2} }{1kPa} )\\P=1200kN/m^{2}

For Part (b)

P=(1200kPa)(\frac{1kN/m^{2} }{1kPa} )(\frac{1000kg.m/s^{2} }{1kN} )\\P=1,200,000kg/m.s^{2}

For Part (c)

P=(1200kPa)(\frac{1kN/m^{2} }{1kPa} )(\frac{1000kg.m/s^{2} }{1kN} )(\frac{100m}{1km} )\\P=1,200,000,000kg/km.s^{2}

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sergiy2304 [10]

Answer:

Because Kinetic Energy(KE) is not the same as Momentum(P)

Kinetic Energy is a scalar(has magnitude only). For a body of mass M, velocity V:

KE = 0.5MV^2

The units of KE: Joules.

Energy is the ability to do work.

Momentum is not a form of energy.

Momentum is a vector(has magnitude and direction).

P = MV

Units of momentum: kg m/s

If you have rifles of mass 2, 4, 8, 16 kg, using the same cartridge, with the same load, barrel length(remember momentum of projectile is proportional to velocity), they all have the same recoil momentum.

But the kinetic energy of recoil would be inversely proportional to the mass of the gun.

Thus the 2kg gun(possible even in large powerful calibers due to modern materials like titanium etc), would have 8 times the recoil ENERGY of the 16kg gun.

A lot of confusion exists in America because of retention of old units, namely Foot Pounds(force) for KE, and Pounds(mass) Feet Per Second for Momentum(P). Because of the more awkward momentum units, a lot of old books had a bad habit of calling the momentum units Pounds Feet, leaving out the rest. Naturally this created confusion with Foot Pounds. Multiplication being commutative and all that:).

Remember that the momentum of the rifles is the same. But the ones with the highest recoil energy hurt the most.

Speaking of hurt:

If momentum killed, then consider two dinosaur killer asteroids with the same masses and velocities, striking vertically at the same time antipodal points on the Earth’s surface. Total momentum delivered would be Zero. That would not make us safe at all:)

Similarly, being shot simultaneously at close range from opposite sides with a 5 round burst from each from two M4 assault rifles(by definition must be able to fire full auto) delivered in 0.3 seconds, would deliver zero momentum. But not zero harm.

Also, the recoil momentum of any firearm is equal to the mass of projectile x velocity + mass of propellant x exit velocity of propellant. This is obviously greater, often much greater, depending on range, than the striking momentum of the projectile at the target.

The recoil kinetic energy is vastly less than the kinetic energy of the bullet/projectile. Neglecting propellant contribution:

recoil Momentum = bullet momentum

BUT:

recoil KE/bullet KE = projectile mass/gun mass

This is a very small fraction.

If we consider the M4 carried by American military:

M855(SS109 equivalent) 5.56 bullet of mass 0.004kg(62 grains)is fired from M4 assault rifle of mass, with optic and full mag 4kg, a thousand times as much!

Even allowing for the 0.0015kg powder charge, and the higher velocity of the powder(approx 1400=1500 m/s vs approx 900 m/s muzzle velocity of the bullet), the recoil energy is hundreds of times less than the muzzle energy of the bullet.

That’s why you want to be behind the gun, and not in front.

Explanation:

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