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kvasek [131]
3 years ago
9

A single-engine helicopter has two rotors; a main rotor and a tail rotor. The main rotor has a diameter of 14.4 m and rotates at

the rate of 450 rev/min while the tail rotor with a diameter of 1.6 m rotates at 4100 rev/min. What are the speeds, in m/s, of the tips of each rotor? main rotor speed m/s tail rotor speed m/s
Physics
1 answer:
Ber [7]3 years ago
6 0

Answer:

339.12 m/s, 343.31 m/s

Explanation:

For main rotor

diameter, d = 14.4 m

radius, r = 7.2 m

f = 450 rev/min = 450 / 60 = 7.5 rps

Angular speed, ω = 2 x π x f = 2 x 3.14 x 7.5 = 47.1 rad/s

linear speed, v = r x ω = 7.2 x 47.1 = 339.12 m/s

For tail rotor

diameter, d = 1.6 m

radius, r = 0.8 m

f = 4100 rev/min = 4100 / 60 = 68.33 rps

Angular speed, ω = 2 x π x f = 2 x 3.14 x 68.33 = 429.13 rad/s

linear speed, v = r x ω = 0.8 x 429.13 = 343.31 m/s

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8090 [49]

Answer:

t = 3.516 s

Explanation:

The most useful kinematic formula would be the velocity of the motorcylce as a function of time, which is:

v(t) = v_0 +at

Where v_0 is the initial velocity and a is the acceleration. However the problem states that the motorcyle start at rest therefore v_0 = 0

If we want to know the time it takes to achieve that speed, we first need to convert units from km/h to m/s.

This can be done knowing that

1 km = 1000 m

1 h = 3600 s

Therefore

1 km/h = (1000/3600) m/s = 0.2777... m/s

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6 0
3 years ago
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Paul [167]

Answer:

λ=hc/E

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3 years ago
How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothe
Rus_ich [418]

Explanation:

(a)  For an isothermal process, work done is represented as follows.

             W = -nRT ln(\frac{V_{2}}{V_{1}})

Putting the given values into the above formula as follows.

        W = -nRT ln(\frac{V_{2}}{V_{1}})

             = - 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})

             = -12280.82 \times ln (0.09)

             = -12280.82 \times -2.41

             = 29596.78 J

or,         = 29.596 kJ       (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

         W = \frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}

              = \frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}

              = \frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}

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               P_{1}V_{1} = P_{2}V_{2}

or,       P_{2} = \frac{P_{1}V_{1}}{V_{2}}

                    = 1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})

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Hence, the required pressure is 11 atm.

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          P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}

or,       P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}

                    = 1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}

                    = 28.7 atm

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