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mihalych1998 [28]
4 years ago
6

The stored energy in a piece of coal can be _________ into heat energy.

Physics
2 answers:
Leviafan [203]4 years ago
7 0
The answer is transformed
because you changed the kind of energy
gavmur [86]4 years ago
4 0
The Law of Conservation of Energy states that energy can either be transferred from one object to another, or transformed from one type to another, but never destroyed. The stored energy in coal is chemical energy because it is ready to react, and when the coal is burned, the chemical energy is transformed by the reaction into heat energy.

Answer: a. transformed

Hope this helps!
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Zarrin [17]

To solve this problem we will use the linear motion description kinematic equations. We will proceed to analyze the general case by which the analysis is taken for the second car and the tenth. So we have to:

x = v_0 t \frac{1}{2} at^2

Where,

x= Displacement

v_0 = Initial velocity

a = Acceleration

t = time

Since there is no initial velocity, the same equation can be transformed in terms of length and time as:

L = \frac{1}{2} a t_1 ^2

For the second cart

2L \frac{1}{2} at_2^2

When the tenth car is aligned the length will be 9 times the initial therefore:

9L = \frac{1}{2} at_3^2

When the tenth car has passed the length will be 10 times the initial therefore:

10L = \frac{1}{2}at_4^2

The difference in time taken from the second car to pass it is 5 seconds, therefore:

t_2-t_1 = 5s

From the first equation replacing it in the second one we will have that the relationship of the two times is equivalent to:

\frac{1}{2} = (\frac{t_1}{t_2})^2

t_1 = \frac{t_2}{\sqrt{2}}

From the relationship when the car has passed and the time difference we will have to:

(t_2-\frac{t_2}{\sqrt{2}}) = 5

t_2 (\sqrt{2}-1) = 3\sqrt{2}

t_2= (\frac{5\sqrt{2}}{\sqrt{2}-1})^2

Replacing the value found in the equation given for the second car equation we have to:

\frac{L}{a} = \frac{1}{4} (\frac{5\sqrt{2}}{\sqrt{2}-1})^2

Finally we will have the time when the cars are aligned is

18 \frac{1}{4} (\frac{5\sqrt{2}}{\sqrt{2}-1})^2 = t_3^2

t_3 = 36.213s

The time when you have passed it would be:

20\frac{1}{4} (\frac{5\sqrt{2}}{\sqrt{2}-1})^2 = t_4^2

t_4 = 38.172

The difference between the two times would be:

t_4-t_3 = 38.172-36.213 \approx 2s

Therefore the correct answer is C.

4 0
3 years ago
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