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3 years ago

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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.

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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regrine falcons frequently grab prey birds from the air. Sometimes they strike at high enough speeds that the force of the impac

solmaris [256]

Answers:

a) 30 m/s

b) 480 N

Explanation:

The rest of the question is written below:

a. What is the final speed of the falcon and pigeon?

b. What is the average force on the pigeon during the impact?

<h3>a) Final speed</h3>

This part can be solved by the Conservation of linear momentum principle, which establishes the initial momentum $p_{i}$ before the collision must be equal to the final momentum $p_{f}$ after the collision:

$p_{i}=p_{f}$ (1)

Being:

$p_{i}=MV_{i}+mU_{i}$

$p_{f}=(M+m) V$

Where:

$M=480 g \frac{1 kg}{1000 g}=0.48 kg$ the mas of the peregrine falcon

$V_{i}=45 m/s$ the initial speed of the falcon

$m=240 g \frac{1 kg}{1000 g}=0.24 kg$ is the mass of the pigeon

$U_{i}=0 m/s$ the initial speed of the pigeon (at rest)

$V$ the final speed of the system falcon-pigeon

Then:

$MV_{i}+mU_{i}=(M+m) V$ (2)

Finding $V$ :

$V=\frac{MV_{i}}{M+m}$ (3)

$V=\frac{(0.48 kg)(45 m/s)}{0.48 kg+0.24 kg}$ (4)

$V=30 m/s$ (5) This is the final speed

<h3>b) Force on the pigeon</h3>

In this part we will use the following equation:

$F=\frac{\Delta p}{\Delta t}$ (6)

Where:

$F$ is the force exerted on the pigeon

$\Delta t=0.015 s$ is the time

$\Delta p$ is the pigeon's change in momentum

Then:

$\Delta p=p_{f}-p_{i}=mV-mU_{i}$ (7)

$\Delta p=mV$ (8) Since $U_{i}=0$

Substituting (8) in (6):

$F=\frac{mV}{\Delta t}$ (9)

$F=\frac{(0.24 kg)(30 m/s)}{0.015 s}$ (10)

Finally:

$F=480 N$

7 0

3 years ago

A baseball player slides into third base with an initial speed of 4.0 m/s. If the coefficient of

musickatia [10]

The frictional force is given by F = μmg

<span>where μ is the coeficient of friction. </span>

<span>Work done by frictional force = Fd = μmgd </span>

<span>Kinetic energy "lost" = 1/2 mv² </span>

<span>Fd = μmgd = 1/2 mv² </span>

<span>The m's cancel μgd = v² / 2 </span>

<span>d = v² / 2μg </span>

<span>d = 8² / 2(0.41)(9.8) </span>

<span>d = 32 / (0.41)(9.8) </span>

<span>d = 7.96 </span>

<span>Player slides 8 m . </span>

<span>Note. In your other example μ = 0.46 and v = 4 m/s </span>

<span>d = v² / 2μg </span>

<span>= 4² / 2(0.46)(9.8) </span>

<span>= 8 / (0.46)(9.8) </span>

<span>= 1.77 or 1.8 m.
</span>
Hope i Helped :D

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lianna [129]

Answer:

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Explanation:

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dimaraw [331]

You multiply force times friction

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Bob is pushing a box across the floor at a constant speed of 1.5 m/s, applying a horizontal force whose magnitude is 60 n. alice

earnstyle [38]

120n

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