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inysia [295]
3 years ago
11

In a heat engine if 1000 j of heat enters the system the piston does 500 j of work, what is the final internal energy of the sys

tem if the initial energy was 2000 j? 1. write the equation 2.list out your known variables 3.plug the numbers into the equations 4.Solve 5.write your solution statement that includes initial energy and final
Physics
1 answer:
nydimaria [60]3 years ago
5 0

Answer : The final energy of the system if the initial energy was 2000 J is, 3500 J

Solution :

(1) The equation used is,

\Delta U=q+w\\\\U_{final}-U_{initial}=q+w

where,

U_{final} = final internal energy

U_{initial} = initial internal energy

q = heat energy

w = work done

(2) The known variables are, q, w and U_{initial}

initial internal energy = U_{initial} = 2000 J

heat energy = q = 1000 J

work done = w = 500 J

(3) Now plug the numbers into the equation, we get

U_{final}-(2000J)=(1000J)+(500J)

(4) By solving the terms, we get

U_{final}-(2000J)=(1000J)+(500J)

U_{final}-(2000J)=1500J

U_{final}=2000J+1500J

U_{final}=3500J

(5) Therefore, the final energy of the system if the initial energy was 2000 J is, 3500 J

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By applying the wave equation we know that the maximum speed of the element's oscillatory motion is 1716 micrometer / s.

We need to know about wave equations to solve this problem. The displacement of the wave on the y-axis can be explained by the wave equation

y = A cos (kx - ωt)

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From the question above, we know that:

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astra-53 [7]

Answer:

a)

Y0 = 0 m

Vy0 = 15 m/s

ay = -9.81 m/s^2

b) 7.71 m

c) 3.06 s

Explanation:

The knowns are that the initial vertical speed (at t = 0 s) is 15 m/s upwards. Also at that time the dolphin is coming out of the water, so its initial position is 0 m. And since we can safely assume this happens in Earth, the acceleration is the acceleration of gravity, which is 9.81 m/s^2 pointing downwards

Y(0) = 0 m

Vy(0) = 15 m/s

ay = -9.81 m/s^2 (negative because it points down)

Since acceleration is constant we can use the equation for uniformly accelerated movement:

Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2

To find the highest point we do the first time derivative (this is the speed:

V(t) = Vy0 + a * t

We equate this to zero

0 = Vy0 + a * t

0 = 15 - 9.81 * t

15 = 9.81 * t

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At this time it will have a height of:

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The doplhin jumps and falls back into the water, when it falls again it position will be 0 again. So we can equate the position to zero to find how long it was in the air knowing that it started the jump at t = 0s.

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0 = 0 + 15 * t - 1/2 * 9.81 t^2

0 = 15 * t - 4.9 * t^2

0 = t * (15 - 4.9 * t)

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0 = 15 - 4.9 * t2

15 = 4.9 * t2

t2 = 3.06 s This is the moment when it falls again.

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