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____ [38]
3 years ago
7

How is energy conserved into transformation

Physics
1 answer:
Zina [86]3 years ago
8 0
When one form of energy is transformed to another, no energy is destroyed in the process. So, the total amount of energy is the same before and after any transformation. 
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The answer is C, an educated guess
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Which of the following would likely be the best reflector of heat energy
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Kiran drove from City A to City B, a distance of 228 mi. She increased her speed by 12 mi/h for the 400-mi trip from City B to C
Degger [83]

Answer:

From city A to city B her speed was 38 mi/h

Explanation:

The traveled distance can be calculated using this equation:

From city A to city B

228 mi = v · t₁

Where:

v = velocity

t₁ = time it took Kiran to travel the 228 mi from city A to city B

From city B to city C

400 mi = (v + 12 mi/h) · t₂

We also know that the entire trip took 14 h, then:

t₁ + t₂ = 14 h

So, we have a system of three equations with three unknwons:

228 mi = v · t₁

400 mi = (v + 12 mi/h) · t₂

t₁ + t₂ = 14 h

Let´s solve the third equation for t₁:

t₁ = 14 h - t₂

Now let´s replace t₁ in the first equation and solve it for t₂

228 mi = v · t₁

228 mi = v · (14 h - t₂)

228 mi/v - 14 h =  - t₂

t₂ = 14 h - 228 mi/v

Now let´s replace t₂ in the second equation:

400 mi = (v + 12 mi/h) · t₂

400 mi = (v + 12 mi/h) · (14 h - 228 mi/v)

400 mi = 14 h · v - 228 mi + 168 mi - 2736 mi²/(v · h)

400 mi = 14 h · v - 60 mi - 2736 mi²/(v · h)

460 mi = 14 h · v - 2736 mi²/(v · h)

Multiplicate by v both sides of the equation:

460 mi · v = 14 h · v² - 2736 mi²/h

0 = 14 h · v² - 460 mi · v - 2737 mi²/h

Solving the quadratic equation:

v = 38 mi/h

(The other solution of the equation is negative, and therefore discarded)

From city A to city B her speed was 38 mi/h

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