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baherus [9]
3 years ago
14

If the cd rotates clockwise at 500 rpm (revolutions per minute) while the last song is playing, and then spins down to zero angu

lar speed in 2.60 s with constant angular acceleration, what is α, the magnitude of the angular acceleration of the cd, as it spins to a stop? express your answer in radians per second squared
Physics
1 answer:
Katen [24]3 years ago
3 0

The solution for this problem is:

500 revolution per minute = 8.33rev /s = 2π*8.33 rad /s = 52.36 rad /s 

Angular velocity ω = 2π N

Angular acceleration α= (ω2 - ω1) /t

ω2 = 0

α = - ω1/t = -2π N /t

N = 500 rpm = 8.33 r p s.

α = -2π 8.33 /2.6 =- 20 rad/s^2

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a fan acquires a speed of 180 rpm in 4s, starting from rest. calculate the speed of the fan at the end of the 5th second startin
KengaRu [80]

Answer:

225 rpm

Explanation:

The angular acceleration of the fan is given by:

\alpha = \frac{\omega_f - \omega_i}{\Delta t}

where

\omega_f is the final angular speed

\omega_i is the initial angular speed

\Delta t is the time interval

For the fan in this problem,

\omega_i = 0\\\omega_f = 180 rpm\\\Delta t=4 s

Substituting,

\alpha = \frac{180-0}{4}=45 rpm/s

Now we can find the angular speed of the fan at the end of the 5th second, so after t = 5 s. It is given by:

\omega' = \omega_i + \alpha t

where

\omega_i = 0\\\alpha = 45 rpm/s\\t = 5 s

Substituting,

\omega' = 0 + (45)(5)=225 rpm

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2 years ago
In a closed system, as kinetic energy increases, what happens to potential energy?
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Students in an introductory physics lab are performing an experiment with a parallel-plate capacitor made of two circular alumin
Alex777 [14]

Answer:

0.92 μC

Explanation:

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E = \frac{o}{e_0} = \frac{Q}{Ae_0} \\ Q = E*A*e_0 = 3*10^6 N/C * (0.25*\pi *(0.21m)^2)*8.85*10^{-12}C^2/Nm^2 = 9.196 *10^{-7} C

or 0.92 μC

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Answer:

Well first for criteria think what would the rover need in order to sustain itself on Venus. And for constraints think of anything that could possibly affect the rover( ex: gasses, active volcanoes)

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