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3 years ago

A heat engine operates with 70.0 kcal of heat supplied and exhausts 30.0 kcal of heat. How much work did the engine do?

Physics

1 answer:

gladu [14]3 years ago

4 0

The work done by the heat engine is 40 kCal.

The given parameters;

input heat of the engine, Q₁ = 70 kCal

output heat of the engine, Q₂ = 30 kCal

To find:

the work done by the heat engine

The work done by the heat engine is the change in the heat energy of the engine;

W = Q₂ - Q₁

Substitute the given parameters and solve work done (W)

W = 70 kCal - 30 kal

W = 40 kCal

Thus, the work done by the heat engine is 40 kCal.

Learn more here: brainly.com/question/4280097

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A cylinder with radius R spins around its axis with an angular speed ω. On its inner surface, there lies a small block; the coef

ValentinkaMS [17]

Answer:

A. Mrŵ² = ųMg

Ŵ = (ųg/r)^½

Ŵ =[ (g /r)* tan á]^½

Explanation:

T.v.= centrepetal force = mrŵ²

Where m = mass of block,

r = radius

Ŵ = angular momentum

On a horizontal axial banking frictional force supplies the Pentecostal force is numerically equal.

So there for

Mrŵ² = ųMg

Ŵ = (ųg/r)^½

g = Gravitational pull

ų = coefficient of friction.

B. The net external force equals the horizontal centerepital force if the angle à is ideal for the speed and radius then friction becomes negligible

So therefore

N *(sin á) = mrŵ² .....equ 1

Since the car does not slide the net vertical forces must be equal and opposite so therefore

N*(cos á) = mg.....equ 2

Where N is the reaction force of the car on the surface.

Equ 2 becomes N = mg/cos á

Substituting N into equation 1

mg*(sin á /cos á) =mrŵ²

Tan á = rŵ²/g

Ŵ =[ (g /r)* tan á]^½

8 0

3 years ago

How does the graph of velocity vs. time look for something with a constant positive acceleration?

Pie

Answer:

The correct answer is option b.

Explanation:

In velocity vs time graph, vertical axis represent velocity vector and horizontal axis represents time. The slope of the line obtained in this graph is equal to the acceleration.

$Slope= \frac{Velocity}{Time}=Acceleration$

If the line is straight makes an acute angle with horizontal axis then it is considered as positive acceleration.
If the line is straight makes an obtuse angle with horizontal axis then it is considered as negative acceleration or deceleration.