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Mama L [17]
4 years ago
6

You live on a planet far from ours. "Based on extensive communication with a physicist on earth", you have determined that all l

aws of physics on your planet are the same as ours and you have adopted the same units of seconds and meters as on earth. But you suspect that the value of g, the acceleration of an object in free fall near the surface of your planet, is different from what it is on earth. To test this, you take a solid uniform cylinder and let it roll down an incline. The vertical height h of the top of the incline above the lower end of the incline can be varied. You measure the speed vcm of the center of mass of the cylinder when it reaches the bottom for various values of h. You plot v2cm (in m2/s2) versus h (in m) and find that your data lie close to a straight line with a slope of 6.42 m/s2
Physics
1 answer:
Ugo [173]4 years ago
3 0

Answer:

8.56 m/s2

Explanation:

Using law of energy conservation while taking into account of the rotational and translation kinetic energy, when the solid cylinder rolls down the incline we have the potential energy converted to kinetic energy:

E_p = E_{kv} + E_{k\omega}

mgh = mv^2/2 + I\omega^2/2

where m is the mass, I = mr^2/2 is the moments of inertia of the solid cylinder \omega = v / r is the angular speed of the cylinder

mgh = mv^2/2 + \frac{mr^2}{2}\frac{(v/r)^2}{2}

mgh = mv^2/2 + mv^2/4 = 3mv^2/4

h = 3v^2/(4g)

So if you plot a liner chart of h vs v^2 and get a slope of 6.42 then that means 3/(4g) = 6.42 so g = 6.42*4/3 = 8.56 m/s^2

The gravitational acceleration on this planet is 8.56 m/s2

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C. Slippery feel

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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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3 years ago
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Keith_Richards [23]

Answer:

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7 0
1 year ago
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Answer: reaction force = -558N

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reaction force = -(f)

reaction force = -558N

if that helps.

8 0
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