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Svetradugi [14.3K]
3 years ago
13

Which is example of a physical change?

Physics
1 answer:
Xelga [282]3 years ago
8 0
The correct answer would be C) Grinding Pepper.
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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
Marina CMI [18]

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

3 0
3 years ago
Which statement is true about acceleration?
Ilia_Sergeevich [38]
The answer is C is think
8 0
3 years ago
A car travels with an average speed of 22 m/s.what is this speed in km/s
Debora [2.8K]
It is .022 kilometers.
4 0
4 years ago
Read 2 more answers
A volumetric flask made of Pyrex is calibrated at 20.0°C. It is filled to the 285-mL mark with 40.5°C glycerin. After the flask
Charra [1.4K]

Answer:

V_f = 287.04 mL

Explanation:

We are given the initial/original volume of the glycerine as 285 mL.

Now, after it is finally cooled back to 20.0 °C , its volume is given by the formula;

V_f = V_i (1 + βΔT)

Where;

V_f is the final volume

V_i is the original volume = 285 mL

β is the coefficient of expansion of glycerine and from online tables, it has a value of 5.97 × 10^(-4) °C^(−1)

Δt is change in temperature = final temperature - initial temperature = 32 - 20 = 12 °C

Thus, plugging in relevant values;

V_f = 285(1 + (5.97 × 10^(-4) × 12))

V_f = 287.04 mL

7 0
3 years ago
Your friend decides to generate electrical power by rotating a 100,000 turn coil of wire around an axis in the plane of the coil
MakcuM [25]

Answer:

a) I=35mA

b) P=1.73W

Explanation:

a) The max emf obtained in a rotating coil of N turns is given by:

emf_{max}=NBA\omega

where N is the number of turns in the coil, B is the magnitude of the magnetic field, A is the area and w is the angular velocity of the coil.

By calculating A and replacing in the formula (1G=10^{-4}T) we get:

A=\pi r^2 =\pi(0.23m)^2=0.16m^2

emf_{max}=(100000)(0.3*10^{-4}T)(0.166m^2)(140\frac{rev}{s})=69.72V

Finally, the peak current is given by:

I=\frac{emf}{R}=\frac{69.72V}{1400\Omega}=49.8mA

b)

we have that

I_{rms}=\frac{I}{\sqrt{2}}=\frac{0.0498A}{\sqrt{2}}=0.035A

P_{rms}=I^2{rms}R=(0.035A)^2(1400\Omega)=1.73W

hope this helps!!

6 0
3 years ago
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