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Cloud [144]
3 years ago
10

A student built a fire using coal. the coal started to burn. which statement best describes the energy changes that occurred.

Physics
2 answers:
BlackZzzverrR [31]3 years ago
5 0
ANSWER: B. The thermal energy stored in the coal changed to light energy and chemical energy.
.
EXPLANATION: thermal energy is already within the coal and because of the light from the fire, that's obviously light energy. And in the end, there is a chemical reaction after the coal burns to ash, and that is chemical energy! I hope this helps!☺️ Please give Brainliest!!
sergiy2304 [10]3 years ago
4 0

Answer:

98% sure the answer is A, as coal, when burning, releases thermal and light energy through the fire.

Explanation:

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A 10-kg disk-shaped flywheel of radius 9.0 cm rotates with a rotational speed of 320 rad/s. Part A Determine the rotational mome
Leya [2.2K]

Answer:

(A). The rotational momentum of the flywheel is 12.96 kg m²/s.

(B). The rotational speed of sphere is 400 rad/s.

Explanation:

Given that,

Mass of disk = 10 kg

Radius = 9.0 cm

Rotational speed = 320 m/s

(A). We need to calculate the rotational momentum of the flywheel.

Using formula of momentum

L=I\omega

L=\dfrac{1}{2}mr^2\omega

Put the value into the formula

L=\dfrac{1}{2}\times10\times(9.0\times10^{-2})^2\times320

L=12.96\ kg m^2/s

(B). Rotation momentum of sphere is same rotational momentum of the  flywheel

We need to calculate the magnitude of the rotational speed of sphere

Using formula of rotational momentum

L_{sphere}=L_{flywheel}

I\omega_{sphere}=I\omega_{flywheel}

\omega_{sphere}=\dfrac{I\omega_{flywheel}}{I_{sphere}}

\omega_{sphere}=\dfrac{I\omega_{flywheel}}{\dfrac{2}{5}mr^2}

Put the value into the formula

\omega_{sphere}=\dfrac{12.96}{\dfrac{2}{5}\times10\times(9.0\times10^{-2})^2}

\omega_{sphere}=400\ rad/s

Hence, (A). The rotational momentum of the flywheel is 12.96 kg m²/s.

(B). The rotational speed of sphere is 400 rad/s.

4 0
3 years ago
Please help me with this question
valkas [14]

Answer: m∠P ≈ 46,42°

because using the law of sines in ΔPQR

=> sin 75°/ 4 = sin P/3

so ur friend is wrong due to confusion between edges

+) we have: sin 75°/4 = sin P/3

=> sin P = sin 75°/4 . 3 = (3√6 + 3√2)/16

=> m∠P ≈ 46,42°

Explanation:

4 0
3 years ago
Yuri built a model of the Solar System on his desk. But, it is not exactly the same as the real Solar System. What limits his mo
agasfer [191]

Answer:

It would be hard to make a model that shows the real sizes and distances between planets ( B )

Explanation:

Yuri building a model of the solar system will face the difficult of replicating the correct distances between the planets and the real sizes of the planets, because in model building the key factors of the Model must be represented properly.

The size of the planets and the distance between the planets are key factors when trying to model the solar system. but the distance between the planets depends on the position of the planets on their orbits which means the distances are not constant ( fixed ) hence that  would be the limitation of his model.

5 0
3 years ago
What is the relationship between electricity and magnetism? A. A moving magnet creates a current, and a still magnet creates a c
motikmotik
I'm trusting my gut and saying it's B.
5 0
3 years ago
A system has 4 independent and identical motors, in order for the system to work, at least one of the motors must operate. Each
grin007 [14]

Answer:

The system's Mean Time to Failure is 1,041.\bar 6 hours

Explanation:

The number of independent identical motors = 4

The failure rate of each motor, λ = 0.002 failures per hour

Given that in order for the system to work, at least one of the motors must operate, therefore, the system are in parallel

The Mean Time to Failure for a parallel system of four identical components is given as follows;

MTTF = \dfrac{1 }{\lambda} \times \left (1 + \dfrac{1}{2} +  \dfrac{1}{3} + \dfrac{1}{4} \right)

Which gives;

MTTF = \dfrac{1 }{0.002} \times \left (1 + \dfrac{1}{2} +  \dfrac{1}{3} + \dfrac{1}{4} \right) = 1,041.\bar 6

The system's Mean Time to Failure = 1,041.\bar 6 hours.

8 0
3 years ago
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