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3 years ago

How are electromagnetic waves different from mechanical waves?

Physics

2 answers:

Valentin [98]3 years ago

5 0

Electromagnetic waves can travel through a vacuum; mechanical waves require a medium.

Brilliant_brown [7]3 years ago

5 0

The answer is B) Electromagnetic waves can travel through a vacuum; mechanical waves require a medium.

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An electric motor can drive grinding wheel at two different speeds. When set to high the angular speed is 2000 rpm. The wheel tu

shutvik [7]

a) The initial angular speed is 209.3 m/s

b) The angular acceleration is $-1.74 rad/s^2$

c) The angular speed after 40 s is 139.7 rad/s

d) The wheel makes 1501 revolutions

Explanation:

The initial angular speed of the wheel is

$\omega_i = 2000 rpm$

which means 2000 revolutions per minute.

We have to convert it into rad/s. Keeping in mind that:

$1 rev = 2\pi rad$

$1 min = 60 s$

We find:

$\omega_i = 2000 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=209.3 rad/s$

To find the angular acceleration, we have to convert the final angular speed also from rev/min to rad/s.

Using the same procedure used in part a),

$\omega_f = 1000 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=104.7 rad/s$

Now we can find the angular acceleration, given by

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

where

$\omega_i = 209.3 rad/s$ is the initial angular speed

$\omega_f = 104.7 rad/s$ is the final angular speed

t = 60 s is the time interval

Substituting,

$\alpha = \frac{104.7-209.3}{60}=-1.74 rad/s^2$

To find the angular speed 40 seconds after the initial moment, we use the equivalent of the suvat equations for circular motion:

$\omega' = \omega_i + \alpha t$

where we have

$\omega_i = 209.3 rad/s$

$\alpha = -1.74 rad/s^2$

And substituting t = 40 s, we find

$\omega' = 209.3 + (-1.74)(40)=139.7 rad/s$

The angular displacement of the wheel in a certain time interval t is given by

$\theta=\omega_i t + \frac{1}{2}\alpha t^2$

where

$\omega_i = 209.3 rad/s$

$\alpha = -1.74 rad/s^2$

And substituting t = 60 s, we find:

$\theta=(209.3)(60) + \frac{1}{2}(-1.74)(60)^2=9426 rad$

So, the wheel turns 9426 radians in the 60 seconds of slowing down. Converting this value into revolutions,

$\theta = \frac{9426 rad}{2\pi rad/rev}=1501 rev$

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

8 0

3 years ago

An object of mass 100 kg is accelerated uniformly from a velocity of 5 ms–1

solniwko [45]

Answer:

Initial velocity of the object, u = 5 m/s

Final velocity of the object, v = 8 m/s

Mass of the object, m = 100 kg

Time take by the object to accelerate, t = 6 s

Initial momentum = mu = 100 — 5 = 500 kg m sˆ’1

Final momentum = mv = 100 — 8 = 800 kg m sˆ’1

Force exerted on the object, F = mv – mu / t

= m (v-u) / t

= 800 – 500

= 300 / 6

= 50 N

Initial momentum of the object is 500 kg m sˆ’1.

Final momentum of the object is 800 kg m sˆ’1.

Force exerted on the object is 50 N.

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6 0

4 years ago

Is kane brown a true country singer?<br> A.yes<br> B.no

Kryger [21]

Answer: hes not a country singer if thats what your asking

Explanation:

6 0

3 years ago

Read 2 more answers

If two bodies have the same kinetic energy do they have the same momentum or not and why?

Mnenie [13.5K]

They do because if one thing is moving and the is there both moveing in once

6 0

3 years ago

You have a 160-Ω resistor and a 0.430-H inductor. Suppose you take the resistor and inductor and make a series circuit with a vo

S_A_V [24]

Answer:

A. Z = 185.87Ω

B. I = 0.16A

C. V = 1mV

D. VL = 68.8V

E. Ф = 30.59°

Explanation:

A. The impedance of a RL circuit is given by the following formula:

$Z=\sqrt{R^2+\omega^2L^2}$ (1)

R: resistance of the circuit = 160-Ω

w: angular frequency = 220 rad/s

L: inductance of the circuit = 0.430H

You replace in the equation (1):

$Z=\sqrt{(160\Omega)^2+(220rad/s)^2(0.430H)^2}=185.87\Omega$

The impedance of the circuit is 185.87Ω

B. The current amplitude is:

$I=\frac{V}{Z}$ (2)

V: voltage amplitude = 30.0V

$I=\frac{30.0V}{185.87\Omega}=0.16A$

The current amplitude is 0.16A

C. The current I is the same for each component of the circuit. Then, the voltage in the resistor is:

$V=\frac{I}{R}=\frac{0.16A}{160\Omega}=1*10^{-3}V=1mV$ (3)

D. The voltage across the inductor is:

$V_L=L\frac{dI}{dt}=L\frac{d(Icos(\omega t))}{dt}=-LIsin(\omega t)\\\\V_L=-(0.430H)(160\Omega)sin(220 t)=68.8sin(220t)\\\\V_L_{max}=68.8V$

E. The phase difference is given by:

$\phi=tan^{-1}(\frac{\omega L}{R})=tan^{-1}(\frac{(220rad/s)(0.430H)}{160\Omega})\\\\\phi=30.59\°$

5 0

3 years ago