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Olegator [25]
3 years ago
11

Helicopter blades withstand tremendous stresses. In addition to supporting the weight of a helicopter, they are spun at rapid ra

tes and experience large centripetal accelerations, especially at the tip. (a) Calculate the magnitude (in m/s2) of the centripetal acceleration at the tip of a 3.10 m long helicopter blade that rotates at 280 rev/min. m/s2 (b) Compare the linear speed of the tip with the speed of sound (taken to be 340 m/s). vtip vsound =
Physics
1 answer:
BabaBlast [244]3 years ago
4 0

Explanation:

It is given that,

Length of the helicopter, l = 3.1 m

The helicopter rotates, the length of helicopter will become the radius of circular path, r = 3.1 m

Angular speed of the helicopter, \omega=280\ rev/min=29.32\ rad/s

(a) The centripetal acceleration in terms of angular velocity is given by :

a_c=r\times \omega^2

a_c=3.1\times (29.32)^2

a_c=2664.95\ m/s^2

(b) Let v is the linear speed of the tip. The relation between the linear and angular speed is given by :

v=r\times \omega

v=3.1\times 29.32

v = 90.89 m/s

\dfrac{v_{tip}}{v_{sound}}=\dfrac{90.89}{340}=0.267

Hence, this is the required solution.

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Fill in the blanks for the following:
storchak [24]

Answer:

<em>a. 4.21 moles</em>

<em>b. 478.6 m/s</em>

<em>c. 1.5 times the root mean square velocity of the nitrogen gas outside the tank</em>

Explanation:

Volume of container = 100.0 L

Temperature = 293 K

pressure = 1 atm = 1.01325 bar

number of moles n = ?

using the gas equation PV = nRT

n = PV/RT

R = 0.08206 L-atm-mol^{-1}K^{-1}

Therefore,

n = (1.01325 x 100)/(0.08206 x 293)

n = 101.325/24.04 = <em>4.21 moles</em>

The equation for root mean square velocity is

Vrms = \sqrt{\frac{3RT}{M} }

R = 8.314 J/mol-K

where M is the molar mass of oxygen gas = 31.9 g/mol = 0.0319 kg/mol

Vrms = \sqrt{\frac{3*8.314*293}{0.0319} }= <em>478.6 m/s</em>

<em>For Nitrogen in thermal equilibrium with the oxygen, the root mean square velocity of the nitrogen will be proportional to the root mean square velocity of the oxygen by the relationship</em>

\frac{Voxy}{Vnit} = \sqrt{\frac{Mnit}{Moxy} }

where

Voxy = root mean square velocity of oxygen = 478.6 m/s

Vnit = root mean square velocity of nitrogen = ?

Moxy = Molar mass of oxygen = 31.9 g/mol

Mnit = Molar mass of nitrogen = 14.00 g/mol

\frac{478.6}{Vnit} = \sqrt{\frac{14.0}{31.9} }

\frac{478.6}{Vnit} = 0.66

Vnit = 0.66 x 478.6 = <em>315.876 m/s</em>

<em>the root mean square velocity of the oxygen gas is </em>

<em>478.6/315.876 = 1.5 times the root mean square velocity of the nitrogen gas outside the tank</em>

6 0
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