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KiRa [710]
3 years ago
9

Is the wavelength comparable to the size of atoms?

Physics
1 answer:
Helen [10]3 years ago
5 0

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.

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A ball has a mass of 1.5kg and is thrown straight up with a speed of 60m/s, what is the ball’s momentum:
madam [21]

Answer:

Assumption: the air resistance on this ball is negligible. Take g = 10\; \rm m \cdot s^{-2}.

a. The momentum of the ball would be approximately 60\;\rm kg \cdot m \cdot s^{-1} two seconds after it is tossed into the air.

b. The momentum of the ball would be approximately \rm \left(-45\; \rm kg \cdot m \cdot s^{-1}\right) three seconds after it reaches the highest point (assuming that it didn't hit the ground.) This momentum is smaller than zero because it points downwards.

Explanation:

The momentum p of an object is equal its mass m times its velocity v. That is: \vec{p} = m \cdot \vec{v}.

Assume that the air resistance on this ball is negligible. If that's the case, then the ball would accelerate downwards towards the ground at a constant g \approx -10\; \rm m \cdot s^{-2}. In other words, its velocity would become approximately 10\; \rm m \cdot s^{-1} more negative every second.

The initial velocity of the ball is 60\; \rm m \cdot s^{-1}. After two seconds, its velocity would have become 60\;\rm m \cdot s^{-1} + 2\; \rm s \times \left(-10\;\rm m \cdot s^{-1}\right) = 40\; \rm m \cdot s^{-1}. The momentum of the ball at that time would be around p = m \cdot v \approx 60\; \rm kg \cdot m \cdot s^{-1}.

When the ball is at the highest point of its trajectory, the velocity of the ball would be zero. However, the ball would continue to accelerate downwards towards the ground at a constant g \approx -10\; \rm m \cdot s^{-2}. That's how the ball's velocity becomes negative.

After three more seconds, the velocity of the ball would be 0\; \rm m \cdot s^{-1} + 3\; \rm s \times \left(-10\; \rm m \cdot s^{-2}\right) = -30 \; \rm m \cdot s^{-1}. Accordingly, the ball's momentum at that moment would be p = m \cdot v \approx \left(-45\; \rm kg \cdot m \cdot s^{-1}\right).

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3 years ago
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In this exercise we have to know the definition of energy to understand what is transferred to a body, like this:

Work

<h2>What is energy?</h2>

Despite being used in many different contexts, the scientific use of the word energy has a well-defined and precise meaning: Innate potential to perform work or perform an action. Anything that is working, moving another object, or heating it up, for example, is expending (transferring) energy.

With this definition we can say that the only alternative that responds to this is work.

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Incomplete question.The complete question is here

Determine the torque applied to the shaft of a car that transmits 225 hp and rotates at a rate of 3000 rpm.

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Torque=0.51 Btu

Explanation:

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Revolutions =3000 rpm

To find

T( torque )=?

Solution

As

T(Torque)=\frac{W(Work)}{2\pi n(Revolutions) }

As force moves an object through a distance, work is done on the object. Likewise, when a torque rotates an object through an angle, work is done.

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