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zhuklara [117]
2 years ago
7

Imagine you transfer heat into a gas. This heat splits up in two different ways: Additional thermal energy making the gas hotter

and “fuel” for the gas to do work. Let’s suppose, 64% of the transferred heat transforms into thermal energy. How many percent of the initial heat can be used to do work?
Physics
1 answer:
navik [9.2K]2 years ago
7 0

Answer:

  • <u>36% of the initial heat can be used to do work.</u>

Explanation:

The initial heat is the 100%, i.e. it is all the heat that will be transferred.

It wil be transferred into two different kinds of energy:

  • additional thermal energy that will increase the thermal energy of the gas, making it hotter. Call it U. U = 64%. And,

  • "fuel" for the gas to do work. Call it W.

Then,

  • 100% = U + W
  • W = 100% - U
  • W = 100% - 64% = 36%

That is, 36% of the initial heat is available to do work.

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Read 2 more answers
An automobile traveling 95 km/h overtakes a 1.30-km-long train traveling in the same direction on a track parallel to the road.
SOVA2 [1]

Answer:

Same direction: t=234s; d=6.175Km

Opposite direction: t=27.53s; d=0.73Km

Explanation:

If the automobile and the train are traveling in the same direction, then the automobile speed relative to the train will be v_{AT}=v_A-v_T (<em>the train must see the car advancing at a lower speed</em>), where v_A is the speed of the automobile and v_T the speed of the train.

So we have v_{AT}=(95km/h)-(75Km/h)=20Km/h.

So the train (<em>anyone in fact</em>) will watch the automobile trying to cover the lenght of the train L at that relative speed. The time required to do this will be:

t = \frac{L}{v_{AT}} = \frac{1.3Km}{20Km/h} = 0.065h=234s

And in that time the car would have traveled (<em>relative to the ground</em>):

d=v_At=(95Km/h)(0.065h)=6.175Km

If they are traveling in opposite directions, <u>we have to do all the same</u> but using v_{AT}=v_A+v_T (<em>the train must see the car advancing at a faster speed</em>), so repeating the process:

v_{AT}=(95km/h)+(75Km/h)=170Km/h

t = \frac{L}{v_{AT}} = \frac{1.3Km}{170Km/h} = 0.00765h=27.53s

d=v_At=(95Km/h)(0.00765h)=0.73Km

5 0
3 years ago
PLEASE HELP FAST The object distance for a convex lens is 15.0 cm, and the image distance is 5.0 cm. The height of the object is
IgorLugansk [536]

Answer:

The image height is 3.0 cm

Explanation:

Given;

object distance, d_o = 15.0 cm

image distance, d_i = 5.0 cm

height of the object, h_o = 9.0 cm

height of the image, h_i = ?

Apply lens equation;

\frac{h_i}{h_o} = -\frac{d_i}{d_o}\\\\ h_i = h_o(-\frac{d_i}{d_o})\\\\h_i = -9(\frac{5}{15} )\\\\h_i = -3 \ cm

Therefore, the image height is 3.0 cm. The negative values for image height indicate that the image is an inverted image.

4 0
3 years ago
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