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.
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Answer:
Part a)
Part B)
Explanation:
As we know that when both the forces are acting on the object in same direction then we will have
as we know that
m = 10.6 kg
now we will have
Now two forces are in opposite direction then we have
Part A)
Now we will have from above two equation
Part B)
Similarly for other force we have
11. aboard
12. marsh
13. swore
14. starch
15. barge
16.swear
17. coarse
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Answer:
vₓ = xg/2y
Explanation:
In this question, let us find the time it takes for the ball on the right that has zero initial velocity to reach the ground.
By newton equation of motion we know that
y = v₀ t - ½ g t²
t = 2y / g
This is the time it takes for the ball on the right to reach the ground; at this time the ball on the left travels a distance
vₓ = x/t
vₓ = xg/2y
vₓ = xg/2y
Where we assume that x and y are known.
Very high melting<span> points - Substances with giant covalent structures have very high</span>melting<span> points, because a lot of strong covalent bonds must be broken. Graphite, for example, has a </span>melting point<span> of more than 3,600ºC. Variable conductivity - Diamond does not conduct electricity.</span>
