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dlinn [17]
3 years ago
7

( Image is 6 C carbon with the numbers 12.011 under it ) According to the image, the atomic mass of carbon is 12.011. How is the

atomic mass of carbon, or any other element, determined? A. By adding the number of protons and the number of electrons B. By adding the number of protons and the number of neutrons C. By adding the number of protons, the number of neutrons, and the number of electrons D. By identifying the number of protons
Physics
1 answer:
Oduvanchick [21]3 years ago
4 0

Answer:

B. By adding the number of protons and the number of neutrons

Explanation:

The atomic mass is determined by adding the number of protons and neutrons in an atom. An atom is made up of three fundamental particles: Electrons, Protons and Neutrons.

The protons and neutrons occupy a central region in an atom known as the nucleus. The nucleus is positively charged and mass concentrated.

If we compare the relative masses of the subatomic particles, the masses of protons and neutrons would be 1 and that of an electron would be 1/1840. This shows that the mass of electrons are negligible.

In order to ascertain atomic mass, we therefore add the number of protons and neutrons together. This is how we arrive at 12.011 as the value of the atomic mass of C and for other elements.

The atomic mass is also known as the mass number.

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Suppose the ski patrol lowers a rescue sled carrying an injured skier, with a combined mass of 97.5 kg, down a 60.0-degree slope
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a. 1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b.21,835 J work, in joules, is done by the rope on the sled this distance.

c. 23,170 J   the work, in joules done by the gravitational force on the sled d. The net work done on the sled, in joules is 43,670 J.

       

<h3>What is friction work?</h3>

The work done by friction is the force of friction times the distance traveled times the cosine of the angle between the friction force and displacement

a. How much work is done by friction as the sled moves 28m along the hill?

ans. We use the formula:

friction work = -µ.mg.dcosθ

  = -0.100 * 97.5 kg * 9.8 m/s² * 28 m * cos 60

= -1337.3 J

-1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b. How much work is done by the rope on the sled in this distance?

We use the formula:

Rope work = -m.g.d(sinθ - µcosθ)

rope work = - 97.5 kg * 9.8 m/s² * 28 m (sin 60 – 0.100 * cos 60)

                     = 26,754 (0.816)

                     = 21,835 J

21,835 J work, in joules, is done by the rope on the sled this distance.

c.  What is the work done by the gravitational force on the sled?

By using  the formula:

Gravity work = mgdsinθ

                    = 97.5 kg * 9.8 m/s² * 28 m * sin 60

                    = 23,170 J

23,170 J   the work, in joules done by the gravitational force on the sled .

       

D. What is the total work done?

By adding all the values

work done =  -1337.3 + 21,835 + 23,170

                 = 43,670 J

The net work done on the sled, in joules is 43,670 J.

Learn more about friction work here:

brainly.com/question/14619763

#SPJ1

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Compare the mass of block A, the mass of block B is:
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