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ElenaW
2 months ago
5

While dangling a hairdryer by its cord, you observe that the cord is vertical when the hairdryer isoff and, once it is turned on

, the hairdryer moves to the right and comes to rest when the cordmakes an angle of 5​ degrees​ with the vertical, as shown below. In a different experiment, you determine that the same hairdryer is pushing 0.06 m​3​ of air through itself every two seconds. The mass ofthe hairdryer is 420 g. Determine the speed of the air leaving the hairdryer, v ​air​. Assume that themass of 1 m​3​ air is 1.2 kg and that the hairdryer is blowing air perpendicular to the wire.

Physics
1 answer:
Yuliya22 [3.3K]2 months ago
6 0

Answer:

The air exiting from the hairdryer is moving at a speed of 10 m/s.

Explanation:

The thrust generated by the hairdryer enables it to maintain an elevation angled at 5° from vertical; thus, we derive from the force diagram

(1).\: tan (5^o) = \dfrac{F_t}{Mg}

by substituting M =0.420kg, g = 9.8m/s^2 into the equation and resolving for F_t we find:

F_t = Mg\:tan(5^o)

F_t = (0.420kg)(9.8m/s^2)\:tan(5^o)

\boxed{F_t = 0.3601N.}

This thrust is linked to the speed of air ejection v through the equation

(2).\: F_t = v\dfrac{dM}{dt}

where dM/dt signifies the rate of air ejection, which is known to be

0.06m^3/2s  = 0.03m^3/s

and since 1m^3 = 1.2kg,

0.03m^3/s \rightarrow 0.036kg/s

\dfrac{dM}{dt}  = 0.036kg/s,

by inserting these values into equation (2), we obtain the value of F_t as:

0.3601N = 0.036v

resulting in

v= \dfrac{0.3601}{0.036}

\boxed{v =10m/s.}

which indicates the air velocity discharged from the hairdryer.

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