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damaskus [11]
2 years ago
5

g A CD is spinning on a CD player. You open the CD player to change out the disk and notice that the CD comes to rest after 15 r

evolutions with a constant deceleration of 120 r a d s 2 . What was the initial angular speed of the CD
Physics
1 answer:
Umnica [9.8K]2 years ago
7 0

Answer:

\omega_1=150rads/sec

Explanation:

From the question we are told that:

Number of Revolution N=15=30\pi

Deceleration d= -120 rads/2

Generally the equation for  initial angular speed \omega_1 is mathematically given by

 \omega_2^2=\omega_1^2 +2(d)(N)

 0=\omega_1^2 +2(-120)(20 \pi)

 \omega_1^2=7200 \pi

 \omega_1=150rads/sec

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Is the wavelength comparable to the size of atoms?
Helen [10]

It totally depends on what kind of wave you're talking about.

-- a sound wave from a trumpet or clarinet playing a concert-A pitch is about 78 centimeters long ... about 2 and 1/2 feet.  This is bigger than atoms.

-- a radio wave from an AM station broadcasting on 550 KHz, at the bottom of your radio dial, is about 166 feet long ... maybe comparable to the height of a 10-to-15-story building.  This is bigger than atoms.

-- a radio wave heating the leftover meatloaf inside your "microwave" oven is about 4.8 inches long ... maybe comparable to the length of your middle finger.  this is bigger than atoms.

-- a deep rich cherry red light wave ... the longest one your eye can see ... is around 750 nanometers long.  About 34,000 of them all lined up will cover an inch.  These are pretty small, but still bigger than atoms.

-- the shortest wave that would be called an "X-ray" is 0.01 nanometer long.     You'd have to line up 2.5 billion of <u>those</u> babies to cover an inch.  Hold on to these for a second ... there's one more kind of wave to mention.

-- This brings us to "gamma rays" ... our name for the shortest of all electromagnetic waves.  To be a gamma ray, it has to be shorter than 0.01 nanometer.

Talking very very very very roughly, atoms range in size from about 0.025 nanometers to about 0.26 nanometers.

The short end of the X-rays, and on down through the gamma rays, are in this neighborhood.

5 0
3 years ago
a person with a mass of 75kg jumps on the trampoline the trampoline creates a fprce of 375n on them what is the acceleration of
Anarel [89]

Newton's second law allows calculating the response for the person's acceleration while leaving the trampoline is:

            -4.8 m / s²

Newton's second law says that the net force is proportional to the product of the mass and the acceleration of the body

            F = m a

Where the bold letters indicate vectors, F is the force, m the masses and the acceleration

The free body diagram is a diagram of the forces without the details of the body, in the attached we can see the free body diagram for this system

 

               F_t -W = m a

Whera F_t is the trampoline force

Body weight is

                W = mg

We substitute

              F_t - mg = ma

              a =\frac{F_t - m g}{m}

Let's calculate

              a = \frac{375 - 75 \ 9.8 }{75}

              a = -4.8 m / s²

The negative sign indicates that the acceleration is directed downward.

In conclusion using Newton's second law we can calculate the acceleration of the person while leaving the trampoline is

            -4.8 m / s²

Learn more here:  brainly.com/question/19860811

7 0
3 years ago
A string is stretched by two equal and opposite forces F newton each then the tension in sting is?
Arlecino [84]

Answer: The tension in the string is zero

Explanation:

3 0
3 years ago
g A ping pong ball (thin shell sphere) rolls down an incline at 30° from rest. What is its acceleration
Aleonysh [2.5K]

Answer:

Explanation:

According to newtons second law of motion;

F = ma .... 1

Also the force acting aong the inclines is expressed as;

F = mgsintheta

m is the mass of the object

a is the acceleration

theta is angle of inclination

Equate 1 and 2

ma = mg sin theta

a = gsin(theta)

a = 9.8sin30

a = 9.8(0.5)

a = 4.9m/s²

Hence the acceleration of the ping pong is 4.9m/s²

3 0
2 years ago
Consider an acrylic sheet of thickness L = 5 mm that is used to coat a hot, isothermal metal substrate at Th = 300°C. The proper
Ad libitum [116K]

Answer:

74.52s

Explanation:

The solution is shown in the picture below

7 0
2 years ago
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