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RUDIKE [14]
2 years ago
10

2. A 3.45-kg centrifuge takes 100 s to spin up from rest to its final angular speed with constant angular acceleration. A point

located 8.00 cm from the axis of rotation of the centrifuge moves with a speed of 150 m/s when the centrifuge is at full speed. (a) What is the angular acceleration (in rad/s2) of the centrifuge as it spins up
Physics
2 answers:
irina1246 [14]2 years ago
4 0

Answer:

18.75 rad/s²

Explanation:

Given that

Mass of the centrifuge, m = 3.45 kg

Time taken to spin, t = 100 s

Distance from axis of rotation, r = 8 cm = 0.08 m

Speed of the centrifuge, v = 150 m/s

First, we find the angular velocity

Angular velocity, w = v / r

w = 150 / 0.08

w = 1875 rad/s

And from the angular velocity, we get our angular acceleration.

Angular acceleration = angular velocity / time taken

Angular acceleration = 1875 rad/s / 100 s

Angular acceleration = 18.75 rad/s²

Therefore, the angular acceleration is 18.75 rad/s²

Yuki888 [10]2 years ago
3 0

Answer:

The angular acceleration of the centrifuge as it spins up is 18.75 rad/s²

Explanation:

Given;

mass of centrifuge, m = 3.45 kg

time taken to spin, t = 100 s

distance from the axis of rotation, r = 8.00 cm = 0.08 m

final velocity of the centrifuge, v = 150 m/s

initial velocity of the centrifuge, u = 0

Determine the linear acceleration of the centrifuge at the given time;

a = \frac{v-u}{t} \\\\a = \frac{150-0}{100} \\\\a = 1.5 \ m/s^2

Finally, determine the angular acceleration of the centrifuge as it spins up;

α = a/r

where;

α is the angular acceleration

α = 1.5 / 0.08

α = 18.75 rad/s²

Therefore, the angular acceleration (in rad/s2) of the centrifuge as it spins up is 18.75 rad/s²

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Five Characteristics of a Mineral

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A soccer player takes a corner kick, lofting a stationary ball 30.0° above the horizon at 18.0 m/s. If the soccer ball has a mas
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Answer:

change in momentum, \Delta p=7.65 \,kg.m.s^{-1}

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Average Force, F=144.3396\,N

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  • F_y=72.1698\,N

Explanation:

Given:

angle of kicking from the horizon, \theta= 30^{\circ}

velocity of the ball after being kicked, v=18 m.s^{-1}

mass of the ball, m=0.425\, kg

time of application of force, t=5.3\times 10^{-2}\,s

We know, since body is starting from the rest

\Delta p=m.v.....................(1)

\Delta p=0.425\times 18

\Delta p=7.65 \,kg.m.s^{-1}

Now the components:

\Delta p_x= 7.65\times cos 30^{\circ}

\Delta p_x= 6.6251 \,kg.m.s^{-1}

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\Delta p_y= 7.65\times sin 30^{\circ}

\Delta p_y= 3.825 \,kg.m.s^{-1}

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Then from eq. (1) & (2)

F\times t=m.v

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F=144.3396\,N

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F_x=144.3396\times cos 30^{\circ}

F_x=125.0018\,N

&

F_y=144.3396\times sin 30^{\circ}

F_y=72.1698\,N

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3 years ago
When a 58g tennis ball is served, it accelerates from rest to a constant speed of 36 m/s. The impact with the racket gives the b
inysia [295]
We first calculate the acceleration on the ball using:
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a = (36)²/(2 x  0.35)
a = 1850 m/s²
F = ma
F = 0.058 x 1850
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