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lions [1.4K]
3 years ago
6

The change in internal energy during one complete cycle of a heat engine A. equals the net heat flow into the engine. B. equals

zero. C. equals the heat energy exhausted from the engine. D. equals the heat energy entering the engine.
Physics
1 answer:
Stels [109]3 years ago
7 0

Answer:

B. equals zero

Explanation:

Given data

one complete cycle = heat flow

solution

we have given that when heat engine complete 1 cycle change in energy = net heat flow

that is always equal to zero

from first law of thermodynamics that

ΔU = Q + W

we know ΔU is the change internal energy in system and Q is net heat transfer in system and W is  net work done in system

therefore change of internal energy during one cycle

ΔU = Ufinal -  Uinitial

ΔU  = Uinitial  -  Uinitial  = 0

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You do 120 j of work while pulling your sister back on a swing, whose chain is 5.10 m long. you start with the swing hanging ver
Goryan [66]
The work done to pull the sister back on the swing is equal to the increase in potential energy of the sister:
W= \Delta U = mg \Delta h (1)

where m is the sister's mass, g is the gravitational acceleration and \Delta h is the increase in altitude of the sister with respect to its initial position.

By calling \theta the angle of the chain with respect to the vertical, the increase in altitude is given by
\Delta h = L - L \cos \theta = L(1 - \cos \theta) (2)
where L is the length of the chain.

Putting (2) inside (1), we find
W= m g L (1 - \cos \theta)
from which we can find the mass of the sister:
m =  \frac{W}{g L (1 - \cos \theta)} =  \frac{120 J}{(9.81 m/s^2)(5.10 m)(1- \cos 32.0^{\circ})} =15.8 kg
5 0
3 years ago
Circuit diagram showing three resistors in series with an ammeter,a cell and a switch​
docker41 [41]

Answer:

the mayflower

Explanation:

4 0
3 years ago
someone help pls. Two students, Mia and Peter, leave school to meet at the local coffee shop. Peter decides to jog to the coffee
cluponka [151]

Answer:

1) The distance further it takes Peter to arrive at the Coffee shop than Mia is 1.24 km

2) Mia's average speed is 6.00 km/hour

Peter's average speed is 8.48 km/hour

4) Mia's average velocity = Peter's average velocity = 6.00 km/hour

Explanation:

The given information from the diagram are;

The distance Peter jogs from school to the flower shop = 2.00 km

The distance Peter jogs from the Flower shop to the Coffee shop = 2.24 km.

The distance Mia walks from school directly to the Coffee shop = 3.00 km

The time it takes both Peter and Mia to arrive at the coffee shop = 30 minutes = 0.5 hour

1) The total distance Peter travels from school to the Coffee shop = 2.00 km + 2.24 km = 4.24 km

The distance Mia travels from school to the Coffee shop = 3.00 km

The distance further it takes Peter to arrive at the Coffee shop than Mia = 4.24 km - 3.00 km = 1.24 km

The distance further it takes Peter to arrive at the Coffee shop than Mia = 1.24 km

2) Average \ speed = \dfrac{Total \ distance \ traveled}{Total \ time \ taken \  in \ the \ journey}

Therefore, \ Mia's \ average \ speed = \dfrac{3.00 \ km}{0.5 \ hour}= 6.00 \ km/hour

Mia's average speed = 6.00 km/hour

Peter's \ average \ speed = \dfrac{4.24 \ km}{0.5 \ hour}= 8.48 \ km/hour

Peter's average speed = 8.48 km/hour

4) Average \ velocicty = \dfrac{Displacement }{Time  \ taken}

The displacement from the School to the Coffee shop is 3.00 km for both Mia and Peter

The time it takes both Peter and Mia to arrive at the Coffee shop from the school is 30 minutes = 0.5 hour

Therefore, \ Mia's \ average \ velocity = \dfrac{3.00 \ km}{0.5 \ hour}= 6.00 \ km/hour

Mia's average velocity = 6.00 km/hour

Peter's \ average \ velocity = \dfrac{3.00 \ km}{0.5 \ hour}= 6.00 \ km/hour

Therefore, Peter's average velocity is also = 6.00 km/hour

6 0
3 years ago
block of mass 5kgriding on a horizontal frictionlessxy-plane surface is subjected tothree applied forces:→F1= 12√2N[ 45◦]→F2= (8
dsp73

Answer:

(i) See attached image for the drawing

(ii) net force given in component form: (20, 20)N with magnitude: \sqrt{800} \,\,\,N

Explanation:

First try to write all forces in  vector component form:

The force F1 acting at 45 degrees would have multiplication factors of \frac{\sqrt{2} }{2} on both axes, to take care of the sine and cosine projections. Therefore, the:

x-component of F1 is    F1_x=12\,\sqrt{2} \frac{\sqrt{2} }{2} =12\,\,N

y-component of F1 is    F1_y=12\,\sqrt{2} \frac{\sqrt{2} }{2} =12\,\,N

As far as force F2, it is given already in x and y components, then:

x-component of F2 = 8 N

y-component of F2 = -6 N (negative meaning pointing down the y-axis)

Force F3 has only component (upwards) in the y-direction

x-component of F3 = 0 N

y-component of F3 =14 N

The additions of all these component by component, gives the resultant force (R) acting on the 5 kg mass:

x-component of R = 12 + 8 = 20 N

y-component of R = 12 + 14 - 6 = 20 N

Therefore, the acceleration that the mass receives due to this force is given in component form as:

x-component of acceleration: 20 N / 5 kg = 4\,\,\,m/s^2

y-component of acceleration: 20 N / 5 kg = 4\,\,\,m/s^2

Now we can calculate the components of the velocity of this mass after 2 seconds of being accelerated by this force, using the formula of acceleration times time:

x-component of the velocity is:     v_x=4\,*\,2=8\,m/s

y-component of the velocity is:     v_y=4\,*\,2=8\,m/s

7 0
3 years ago
9.58 A spring of equilibrium length L1 and spring constant k1 hangs from the ceiling. Mass m1 is suspended from its lower end. T
andrey2020 [161]

Answer:

The distance of m2 from the ceiling is L1 +L2 + m1g/k1 + m2g/k1 + m2g/k2.

See attachment below for full solution

Explanation:

This is so because the the attached mass m1 on the spring causes the first spring to stretch by a distance of m1g/k1 (hookes law). This plus the equilibrium lengtb of the spring gives the position of the mass m1 from the ceiling. The second mass mass m2 causes both springs 1 and 2 to stretch by an amout proportional to its weight just like above. The respective stretchings are m2g/k1 for spring 1 and m2g/k2 for spring 2. These plus the position of m1 and the equilibrium length of spring 2 L2 gives the distance of L2 from the ceiling.

4 0
4 years ago
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