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3 months ago

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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 m3 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 m3 air is 1.2 kg and that the hairdryer is blowing air perpendicular to the wire.

1 answer:

Yuliya22 [3.3K]3 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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