Question

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.

Answer 1

Answer:

The speed of the air leaving the hairdryer is 10m/s.

Explanation:

The thrust that the dryer produces is what keeps it elevated at an angle of 5° from the vertical; therefore, from the force diagram we get

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

putting in $M =0.420kg$, $g = 9.8m/s^2$ and solving for $F_t$ we get:

$F_t = Mg\:tan(5^o)$

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

$\boxed{F_t = 0.3601N.}$

Now, this thrust produced is related to to the air ejection speed $v$ by the relation

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

where $dM/dt$ is the rate of air ejection which we know is

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

and since $1m^3 = 1.2kg$,

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

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

putting this into equation (2) and the value of $F_t$ we get:

$0.3601N = 0.036v$

which gives

$v= \dfrac{0.3601}{0.036}$

$\boxed{v =10m/s.}$

which is the speed of the air ejected.