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rjkz [21]
4 years ago
8

Pls i need the diffrences between sound waves and light waves​

Physics
1 answer:
shepuryov [24]4 years ago
3 0

Sound waves need a medium to go through and are created through vibrations, whereas light waves need no medium, meaning it can move through space. Light also doesn’t loose energy when reflected, but sound waves get absorbed by the medium they go through causing it to loose energy.

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What can you say about the speed of light in diamond?
Inessa05 [86]
So n=c/v, n= index, c=speed of light and v= speed of light in diamond. 2.42=c/v so v=c/2.42, c≈<span>3x108 m/sec</span><span> so v=</span><span>1.24x108 m/sec</span>.
<span>Hope this helps.</span>
7 0
3 years ago
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
A train on a straight track goes in the positive direction for 5.9 km, and then backs up for 3.8 km
kenny6666 [7]

Answer:

2.1km

Explanation:

Ill take it as u are talking about the displacement

Since displacement has negatives and positves

5.9 - 3.8 = 2.1km

6 0
3 years ago
Challenge! A marshmallow is dropped from a 5-meter high pedestrian bridge and 0.83 seconds later, it lands right on the head of
WINSTONCH [101]

a₀).  You know ...
         -- the object is dropped from 5 meters
             above the pavement;
         --  it falls for 0.83 second.

a₁).  Without being told, you assume ...
         -- there is no air anyplace where the marshmallow travels,
             so it free-falls, with no air resistance;
         -- the event is happening on Earth,
            where the acceleration of gravity is  9.81 m/s² .

b).  You need to find how much LESS than 5 meters
       the marshmallow falls in 0.83 second.
    
c).  You can use whatever equations you like.
       I'm going to use the equation for the distance an object falls in
       ' T ' seconds, in a place where the acceleration of gravity is ' G '.

d).  To see how this all goes together for the solution, keep reading:


The distance that an object falls in ' T ' seconds
when it's dropped from rest is

                                 (1/2 G) x (T²) .

On Earth, ' G ' is roughly  9.81 m/s², so in 0.83 seconds,
such an object would fall

                               (9.81 / 2) x (0.83)² = 3.38 meters .

It dropped from 5 meters above the pavement, but it
only fell 3.38 meters before something stopped it.
So it must have hit something that was

                         (5.00 - 3.38)  =  1.62 meters

above the pavement.  That's where the head of the unsuspecting
person was as he innocently walked by and got clobbered.

7 0
3 years ago
What is the Unit used to measure voltage
muminat
The derived unit for voltage is named volt.
3 0
3 years ago
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