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My name is Ann [436]
3 years ago
5

Which sentence about particles in matter is true?

Physics
2 answers:
AlexFokin [52]3 years ago
7 0

Answer:

B. The particles of matter are in constant motion.

Explanation:

The particles of matter are always in constant motion at different speeds when exposed to certain temperatures.

finlep [7]3 years ago
4 0
B. The particles of matter are in constant motion.
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In which state of matter are the particles closest together?
vovikov84 [41]

Answer: The particles of a solid are [generally] closest together while those of a gas are farthest apart

Explanation:

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3 years ago
How much kinetic energy does a 55 kg object have that is traveling 20m/s?
shutvik [7]

Answer:

11000 J

Explanation:

Apply the formula: K = 1/2.m.v²

K = 1/2.55.20²

K = 1/2.55.400

K = 55.200

K = 11000 Joules

4 0
3 years ago
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Give main forms before and after for the following energy transformations: A gasoline engine.
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➷ It would be chemical energy to kinetic energy. The chemical energy from the gasoline is being transferred to kinetic energy.

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4 0
4 years ago
Potential energy is the energy an object has because of its
Naya [18.7K]
Radiation compared to other potential
3 0
2 years ago
The mean distance of an asteroid from the Sun is 2.98 times that of Earth from the Sun. From Kepler's law of periods, calculate
lutik1710 [3]

Answer:

The asteroid requires 5.14 years to make one revolution around the Sun.

Explanation:

Kepler's third law establishes that the square of the period of a planet will be proportional to the cube of the semi-major axis of its orbit:

T^{2} = a^{3} (1)

Where T is the period of revolution and a is the semi-major axis.

In the other hand, the distance between the Earth and the Sun has a value of 1.50x10^{8} Km. That value can be known as well as an astronomical unit (1AU).

But 1 year is equivalent to 1 AU according with Kepler's third law, since 1 year is the orbital period of the Earth.

For the special case of the asteroid the distance will be:

a = 2.98(1.50x10^{8}Km)

a = 4.47x10^{8}Km

That distance will be expressed in terms of astronomical units:

4.47x10^{8}Km.\frac{1AU}{1.50x10^{8}Km} ⇒ 2.98AU

Finally, from equation 1 the period T can be isolated:

T = \sqrt{a^{3}}

T = \sqrt{(2.98)^{3}}  

T = \sqrt{26.463592}

T = 5.14AU

Then, the period can be expressed in years:

5.14AU.\frac{1yr}{1AU} ⇒ 5.14 yr

T = 5.14 yr

Hence, the asteroid requires 5.14 years to make one revolution around the Sun.

8 0
3 years ago
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