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mario62 [17]
3 years ago
13

When the play button is pressed, a CD accelerates uniformly from rest to 450 rev/min in 3.0 revolutions. If the CD has a radius

of 7.0 cm and a mass of 17 g , what is the torque exerted on it? Express your answer using two significant figures.
Physics
1 answer:
Marina CMI [18]3 years ago
3 0

To solve this problem it is necessary to apply the kinematic equations of angular motion.

Torque from the rotational movement is defined as

\tau = I\alpha

where

I = Moment of inertia \rightarrow \frac{1}{2}mr^2 For a disk

\alpha = Angular acceleration

The angular acceleration at the same time can be defined as function of angular velocity and angular displacement (Without considering time) through the expression:

2 \alpha \theta = \omega_f^2-\omega_i^2

Where

\omega_{f,i} = Final and Initial Angular velocity

\alpha = Angular acceleration

\theta = Angular displacement

Our values are given as

\omega_i = 0 rad/s

\omega_f = 450rev/min (\frac{1min}{60s})(\frac{2\pi rad}{1rev})

\omega_f = 47.12rad/s

\theta = 3 rev (\frac{2\pi rad}{1rev}) \rightarrow 6\pi rad

r = 7cm = 7*10^{-2}m

m = 17g = 17*10^{-3}kg

Using the expression of angular acceleration we can find the to then find the torque, that is,

2\alpha\theta=\omega_f^2-\omega_i^2

\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}

\alpha = \frac{47.12^2-0^2}{2*6\pi}

\alpha = 58.89rad/s^2

With the expression of the acceleration found it is now necessary to replace it on the torque equation and the respective moment of inertia for the disk, so

\tau = I\alpha

\tau = (\frac{1}{2}mr^2)\alpha

\tau = (\frac{1}{2}(17*10^{-3})(7*10^{-2})^2)(58.89)

\tau = 0.00245N\cdot m \approx 2.45*10^{-3}N\cdot m

Therefore the torque exerted on it is 2.45*10^{-3}N\cdot m

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Answer: Please find the answer in the explanation

Explanation:

Under what circumstances does distance traveled equal magnitude of displacement?

When a body's motion is linear in one direction. Or a body moving in a straight line without turning back.

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When the body is moving in a straight line with without changing direction or without turning back.

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When you are on a roller coaster, you are constantly transforming from Potential to Kinetic energy and back. Explain how these e
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Answer:

The two types of energy possessed by the roller coaster are:

- Potential energy: it is the energy possessed by the roller coaster due to its position. It is calculated as

PE=mgh

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m is the mass of the roller coaster

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h is the height of the roller coaster relative to the ground

- KInetic energy: it is the energy possessed by the roller coaster due to its motion. It is calculated as

KE=\frac{1}{2}mv^2

where

v is the speed of the roller coaster

Moreover, according to the law of conservation of energy, the total mechanical energy of the roller coaster (the sum of potential+kinetic energy) is constant during the motion:

E=PE+KE=const.

This implies that:

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- When PE decreases (because h decreases), KE increases (because v increases)

Now we can apply these conclusions to the motion of the roller coaster:

- When it moves from A to B, potential energy is converted into kinetic energy, so PE decreases and KE increases

- When it moves from B to C, kinetic energy is converted into potential energy, so PE increases and KE decreases

- When it moves from C to D, potential energy is converted into kinetic energy, so PE decreases and KE increases

- When it moves from D to E,  kinetic energy is converted into potential energy, so PE increases and KE decreases

8 0
3 years ago
Please help 25 points and brainliest
GREYUIT [131]
No. Terrance cannot move the sled. This is because, the force that he produces is 80*3 Newton’s, which is only 240 Newton’s, but since the sled needs a force of 250 Newton’s, he cannot move the sled (even though the difference is only by 10 Newton’s).
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Why is ammeter connected in series in an electric circuit?
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In order to measure the size of the electrical current flowing in the circuit,
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3 years ago
A 2. 0 μf and a 4. 0 μf capacitor are connected in series across an 8. 0-v dc source. what is the charge on the 2. 0 μf capacito
Nezavi [6.7K]

voltage across 2.0μf capacitor is 5.32v

Given:

C1=2.0μf

C2=4.0μf

since two capacitors are in series there equivalent capacitance will be

[tex] \frac{1}{c} = \frac{1}{c1} + \frac{1}{c2} [/tex]

c =  \frac{c1 \times c2}{c1 + c2}

=  \frac{2 \times 4}{2 + 4}

=1.33μf

As the capacitance of a capacitor is equal to the ratio of the stored charge to the potential difference across its plates, giving: C = Q/V, thus V = Q/C as Q is constant across all series connected capacitors, therefore the individual voltage drops across each capacitor is determined by its its capacitance value.

Q=CV

given,V=8v

= 1.33 \times 10 {}^{ - 6}  \times 8

= 10.64 \times 10 {}^{ - 6}

charge on 2.0μf capacitor is

\frac{Qeq}{2 \times 10 {}^{ - 6} }

=  \frac{10.64 \times 10 {}^{ - 6} }{2 \times 10 {}^{ - 6} }

=5.32v

learn more about series capacitance from here: brainly.com/question/28166078

#SPJ4

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