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Varvara68 [4.7K]
2 years ago
5

A heat engine receives 7000 j of heat and loses 3000 j in each cycle. what is the efficiency of this engine?

Physics
1 answer:
Nonamiya [84]2 years ago
4 0

Answer: 57.1 %

Explanation:

The efficiency of the engine is given by:

\eta=\frac{Q_{used}}{Q_{in}}

where the numerator corresponds to the heat used by the engine, while the denominator corresponds to the heat in input to the engine.

In this problem, the heat in input is

Q_{in}=7000 J

while the heat used is

Q_{used}=7000 J-3000 J=4000 J

So, the efficiency is

\eta=\frac{4000 J}{7000 J}=0.571=57.1\%


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A friend throws a heavy ball toward you while you are standing on smooth ice. You can either catch the ball or deflect it back t
baherus [9]

Answer:

Explanation:

My speed after the interaction will depend upon the impulse the ball will make on me . Now impulse can be expressed as follows

Impulse = change in momentum

change in momentum in the ball will be maximum when the ball bounces back with the same velocity which can be shown as follows

change in momentum = mv - ( - mv ) = 2mv

So when ball is bounced back with same velocity , it suffers greatest impulse from my hand . In return ,  it reacts with the same impulse on my hand pushing me with greatest impulse according to third law of motion. this maximizes my speed after the interaction.

6 0
2 years ago
Wile E. Coyote wants to launch Roadrunner into the air using a long lever asshown below. The lever starts at rest before the Coy
algol13

Answer:

Explanation:

Find attached the solution

4 0
2 years ago
A 1500 kg car traveling at 20 m/s suddenly runs out of gas while approaching the valley shown in the figure. The alert driver im
geniusboy [140]

Answer:

v_f = 17.4 m / s

Explanation:

For this exercise we can use conservation of energy

starting point. On the hill when running out of gas

          Em₀ = K + U = ½ m v₀² + m g y₁

final point. Arriving at the gas station

         Em_f = K + U = ½ m v_f ² + m g y₂

energy is conserved

         Em₀ = Em_f

         ½ m v₀ ² + m g y₁ = ½ m v_f ² + m g y₂

        v_f ² = v₀² + 2g (y₁ -y₂)

         

we calculate

        v_f ² = 20² + 2 9.8  (10 -15)

        v_f = √302

         v_f = 17.4 m / s

8 0
2 years ago
Which of the following statements is FALSE?
Levart [38]

A thrust fault is a reverse fault with an extremely high dip (close to 90°). This is the false statement.

Answer: Option D

<u>Explanation:</u>

Faults are the fracture or fracture zone occurring on the rocks. These fractures can travel through the rocks leading to massive destruction. So, depending upon the direction of their travel, the faults can be classified as normal, reverse and strike slip fault. Also, the angle of dip along the fault is one of the important criteria for determining the type of faults.

There is dip-slip fault which has its movement along the vertical fault plane while the strike slip fault will be in horizontal direction. Similarly, an oblique fault will be acting in both vertical and the horizontal direction. So, the fourth statement related to thrust fault is false as in reverse fault or thrust fault the dip will be shallow and not high.

5 0
2 years ago
Two speakers, A and B, produce identical sound waves. A listener is 3.2 m away from speaker A. The listener finds the lowest fre
earnstyle [38]

Answer:

  0.83 m or 5.57 m

Explanation:

Destructive interference will occur when the distances from the speakers differ by 1/2 wavelength.

The length of 1 cycle of 72.4 Hz is ...

  λ = v/f = (343 m/s)/(72.4 Hz) ≈ 4.738 m

So, the distance of the listener from speaker B is ...

  3.2 m ± (4.738 m)/2 = {0.83 m, 5.57 m} . . . either of these distances

_____

The location could be at additional multiples of 4.738 m, but we think not. The sound intensity drops off with the square of the distance from the speaker, so identical sound waves from the speakers will sound quite different at different distances from the speakers. For best interference, the distances need to be as close to the same as possible. That will be at 3.2 m and 5.57 m.

_____

<em>Comment on the speed of sound</em>

We don't know what speed you are to use for the speed of sound. We have used 343 m/s. Some sources use 340 m/s, which will give a result different by 2 or 3 cm.

8 0
2 years ago
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