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1 year ago

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Physics

1 answer:

alexdok [17]1 year ago

7 0

The angular velocity of the propeller is 136.1 rad/s; linear velocity is 153.1 m/s; centripetal acceleration 20835.2 m/s² and 2123.8 g.

<h3>What is the angular velocity of the propeller?</h3>

The angular velocity of the propeller in rad/s is given as follows:

1 rev/m = 2π/60 rad/s.

1300 rev/min = 1300 * 2π/60 = 136.1 rad/s.

b. The linear velocity, v = radius * angular velocity

Linear velocity, v = 2.25/2 * 136.1

v = 153.1 m/s

c. Centripetal acceleration, $a = \frac{v^{2}}{r}$

$a = \frac{(153.1)^{2}}{1.125} = 20835.2\:ms^{2}$

Centripetal acceleration in terms of g; $g = \frac{20835.2}{9.81} = 2123.8 g$

Therefore, the angular velocity of the propeller is 136.1 rad/s; linear velocity is 153.1 m/s; centripetal acceleration 20835.2 m/s² and 2123.8 g.

Learn more about angular velocity and centripetal acceleration at: brainly.com/question/10703948

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Explanation:

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2 years ago

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The Ideal Mechanical Advantage (IMA) is given by:

$IMA = \frac{d_i}{d_o}$

where

$d_i$ is the input distance

$d_o$ is the output distance

For the pulley system in this problem, $d_i = 3.90 m$ and $d_o = 0.975 m$ , so the IMA is

$IMA=\frac{3.90 m}{0.975 m}=4$

B. MA: 3.59

The actual mechanical advantage (AMA), or simply the Mechanical Advantage (MA), is given by

$MA=\frac{F_o}{F_i}$

where $F_o$ is the output force and $F_i$ is the input force. For the pulley system in this problem, $F_i = 375 N$ and $F_o = 1345 N$ , so the MA is

$MA=\frac{1345 N}{375 N}=3.59$

C. Efficiency: 89.8 %

The efficiency of a machine is equal to the ratio between the MA and the AMA:

$\eta = \frac{MA}{AMA} \cdot 100$

Therefore, in this case,

$\eta=\frac{3.59}{4}\cdot 100=0.898=89.8 \%$

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3 years ago

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2 years ago

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To calculate the time it takes for the cargo to reach the ground, we just consider the vertical motion of the cargo.

The vertical position at time t is given by

$y(t) = h +u_y t - \frac{1}{2}gt^2$

where

h = 2.5 km = 2500 m is the initial height

$u_y = 0$ is the initial vertical velocity of the cargo

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The cargo reaches the ground when

$y(t) = 0$

So substituting it into the equation and solving for t, we find the time of flight of the cargo:

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So the horizontal distance travelled is

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And if we substitute the time of flight,

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