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2 years ago

El aspa de un ventilador de 0.30 m de diametro gira 1200 rev/min. calcule la rapidez tangencial.

Physics

1 answer:

Keith_Richards [23]2 years ago

4 0

Answer:

v = 18.84 m/s

Explanation:

Let the question says,"The blade of a 0.30 m diameter fan rotates 1200 rev / min. calculate the tangential speed."

It is given that,

Diameter of a blade of fan is 0.3 m

Angular velocity of the fan is 1200 rev/min or 125.66 rad/s

We need to find the tangential speed of the fan. The relation is given by :

$v=r\omega$

r is radius of fan

$v=0.15\times 125.66 \\\\v=18.84\ m/s$

So, the tangential velocity is 18.84 m/s.

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A wheel rotates about a fixed axis with a constant angular acceleration of 3.3 rad/s2. The diameter of the wheel is 21 cm. What

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Answer:

The the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

Explanation:

We are given that

Angular acceleration, $\alpha=3.3 rad/s^2$

Diameter of the wheel, d=21 cm

Radius of wheel, $r=\frac{d}{2}=\frac{21}{2}$ cm

Radius of wheel, $r=\frac{21\times 10^{-2}}{2} m$

1m=100 cm

Magnitude of total linear acceleration, a= $1.7 m/s^2$

We have to find the linear speed of a at an instant when that point has a total linear acceleration with a magnitude of 1.7 m/s2.

Tangential acceleration, $a_t=\alpha r$

$a_t=3.3\times \frac{21\times 10^{-2}}{2}$

$a_t=34.65\times 10^{-2}m/s^2$

Radial acceleration, $a_r=\frac{v^2}{r}$

We know that

$a=\sqrt{a^2_t+a^2_r}$

Using the formula

$1.7=\sqrt{(34.65\times 10^{-2})^2+(\frac{v^2}{r})^2}$

Squaring on both sides

we get

$2.89=1200.6225\times 10^{-4}+\frac{v^4}{r^2}$

$\frac{v^4}{r^2}=2.89-1200.6225\times 10^{-4}$

$v^4=r^2\times 2.7699$

$v^4=(10.5\times 10^{-2})^2\times 2.7699$

$v=((10.5\times 10^{-2})^2\times 2.7699)^{\frac{1}{4}}$

$v=0.418 m/s$

Hence, the the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

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2 years ago