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Ivanshal [37]
3 years ago
7

A propeller is modeled as five identical uniform rods extending radially from its axis. The length and mass of each rod are 0.82

7 m and 2.89 kg, respectively. When the propeller rotates at 507 rpm (revolutions per minute), what is its rotational kinetic energy K
Physics
1 answer:
melisa1 [442]3 years ago
4 0

Answer:

3241.35J

Explanation:

No. Of rods = 5

Mass = 2.89kg

Length (L) = 0.827m

W = 507rpm

Kinetic energy of rotation = ½I*ω²

For each rod, the moment of inertia (I) = ML² / 3

I = ML² / 3

I = [2.89*(0.827)²] / 3

I = 1.367 / 3 = 0.46kgm²

ω = 507 rev/min. Convert rev/min to rev/sec.

507 * 2Πrads/60s = 53.09rad/s

ω = 53.09rad/s

k.e = ½ I * ω²

K.E = ½ * 0.46 * (53.09)²

K.E = 648.27.

But there five (5) rods, so kinetic energy is equal to

K.E = 5 * 648.27 = 3241.35J

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An ideal monatomic gas initially has a temperature of 300 K and a pressure of 5.79 atm. It is to expand from volume 420 cm3 to v
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P_{f} = \frac{P_{i}V_{i}}{V_{f}} = \frac{5.79 atm*420 cm^{3}}{1450 cm^{3}} = 1.68 atm

Hence, the final pressure is 1.68 atm.

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