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strojnjashka [21]

3 years ago

10

Each blade of a fan has a radius of 11 inches. If the fan’s rate of turn is 1440o /sec, find the following. (a) The angular spee

d in units of radians/sec. (b) The linear speed in units of inches/sec of a point on the outer edge of the blade.

1 answer:

notka56 [123]3 years ago

3 0

Answer:

a) 24.43 radians per second

b) 268.73 inches per second

Explanation:

a) The angular speed of the fan on Celsius degrees/second is 1400, so we should convert that value to radians using the fact that 2π rad = 360 °C:

$\omega = 1400\frac{C}{s}=1400\frac{C}{s}*\frac{2\pi\,rad}{360\,C}$

$\omega = 1400\frac{C}{s}=24.43\frac{rad}{s}$

b) Linear speed on a point of the blade is related with angular speed of the fan by the equation

$v=\omega r$

with v linear speed, ω angular speed and r the radius of the blades. So:

$v=(24.43\frac{rad}{s})(11 in)$

Radians isn't really a unity; it is dimensionless so we can put it or not. So:

$v=268.73\frac{in}{s}$

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A vector → A has a magnitude of 56.0 m and points in a direction 30.0° below the negative x axis. A second vector, → B , has a m

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Answer:

The magnitude of the vector $\vec{C}$ is 107.76 m

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To find the components of the vectors we can use:

$\vec{A} = | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )$

where $| \vec{A} |$ is the magnitude of the vector, and θ is the angle over the positive x axis.

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$\vec{A} = 56.0 \ m \ ( \ cos( 180 \° + 30 \°) \ , \ sin (180 \° + 30 \°) \ )$

$\vec{A} = 56.0 \ m \ ( \ cos( 210 \°) \ , \ sin (210 \°) \ )$

$\vec{A} = ( \ -48.497 \ m \ , \ - 28 \ m \ )$

$\vec{B} = 82.0 \ m \ ( \ cos( 180 \° - 49 \°) \ , \ sin (180 \° - 49 \°) \ )$

$\vec{B} = 82.0 \ m \ ( \ cos( 131 \°) \ , \ sin (131 \°) \ )$

$\vec{B} = ( \ -53.797 \ m \ , \ 61.886\ m \ )$

Now, we can perform vector addition. Taking two vectors, the vector addition is performed:

$(a_x,a_y) + (b_x,b_y) = (a_x+b_x,a_y+b_y)$

So, for our vectors:

$\vec{C} = ( \ -48.497 \ m \ , \ - 28 \ m \ ) + ( \ -53.797 \ m \ , ) = ( \ -48.497 \ m \ -53.797 \ m , \ - 28 \ m \ + \ 61.886\ m \ )$

$\vec{C} = ( \ - 102.294 \ m , \ 33.886 m \ )$

To find the magnitude of this vector, we can use the Pythagorean Theorem

$|\vec{C}| = \sqrt{C_x^2 + C_y^2}$

$|\vec{C}| = \sqrt{(- 102.294 \ m)^2 + (\ 33.886 m \)^2}$

$|\vec{C}| =107.76 m$

And this is the magnitude we are looking for.

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