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mario62 [17]
3 years ago
8

What happens when a candle burns?

Physics
1 answer:
slega [8]3 years ago
5 0

Answer:

oxygen is used up is the answer

Explanation:

These vaporized molecules are drawn up into the flame, where they react with oxygen from the air to create heat, light, water vapor (H2O) and carbon dioxide (CO2).

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Like-charged bodies, when brought closer together, will: repel more strongly be neutralized repel less strongly attract more str
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Answer:

That's almost the true

Explanation:

it does not happen all the time

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Fiberglass insulation trap layers of air to prevent heat loss due to all but one that is
pochemuha

the answer is C. evaporation

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1. Does the validity of the principle of conservation of momentum depend on the validity of Newton's 3rd law of motion?​
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I truly believe so, that’s a definite yes
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4 years ago
Which of the following is most likely not a case of uniform circular motion
Ivan

Answer:

Motion of a racing car on a circular track

Explanation:

Uniform circular motion means the motion of the object is in a circle with a CONSTANT SPEED.

The racing car will accelerate during its motion. Hence, it is not a uniform circular motion.

7 0
3 years ago
A train starts at rest in a station and accelerates at a constant 0.987 m/s2 for 182 seconds. Then the train decelerates at a co
algol [13]

Answer:

\displaystyle X_T=66.6\ km

Explanation:

<u>Accelerated Motion </u>

When a body changes its speed at a constant rate, i.e. same changes take same times, then it has a constant acceleration. The acceleration can be positive or negative. In the first case, the speed increases, and in the second time, the speed lowers until it eventually stops. The equation for the speed vf at any time t is given by

\displaystyle V_f=V_o+a\ t

where a is the acceleration, and vo is the initial speed .

The train has two different types of motion. It first starts from rest and has a constant acceleration of 0.987 m/s^2 for 182 seconds. Then it brakes with a constant acceleration of -0.321 m/s^2 until it comes to a stop. We need to find the total distance traveled.

The equation for the distance is

\displaystyle X=V_o\ t+\frac{a\ t^2}{2}

Our data is

\displaystyle V_o=0,a=0.987m/s^2,\ t=182\ sec

Let's compute the first distance X1

\displaystyle X_1=0+\frac{0.987\times 182^2}{2}

\displaystyle X_1=16,346.7\ m

Now, we find the speed at the end of the first period of time

\displaystyle V_{f1}=0+0.987\times 182

\displaystyle V_{f1}=179.6\ m/s

That is the speed the train is at the moment it starts to brake. We need to compute the time needed to stop the train, that is, to make vf=0

\displaystyle V_o=179.6,a=-0.321\ m/s^2\ ,V_f=0

\displaystyle t=\frac{v_f-v_o}{a}=\frac{0-179.6}{-0.321}

\displaystyle t=559.5\ sec

Computing the second distance

\displaystyle X_2=179.6\times559.5\ \frac{-0.321\times 559.5^2}{2}

\displaystyle X_2=50,243.2\ m

The total distance is

\displaystyle X_t=x_1+x_2=16,346.7+50,243.2

\displaystyle X_t=66,589.9\ m

\displaystyle \boxed{X_T=66.6\ km}

8 0
4 years ago
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