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2 years ago

4

an ideal gas occupies a volume of 22.4litres at stp while absorbing 2.53KJ of heat from the surrounding the gas expands isobaric

ally to 32.4 litres. What is the change in internal energy

1 answer:

puteri [66]2 years ago

3 0

<span>delta U = Q - W and W in this case = P delta V. Use standard pressure, change in V is given and heat added is given.
</span>

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Assume that segment r exerts a force of magnitude t on segment l. what is the magnitude flr of the force exerted on segment r by

mrs_skeptik [129]

If we are talking on the force being exerted by a segment of a rope of lenght R on the right on a point M which is being also pulled from the Left by a segment of rope R as shown in the figure attached. Then we invoke Newton's Third Law:
"Any force exerted by an object (in this case a segment of the rope) also suffers a equal and opposite force".
If we pick $T_R=T$ whis is the tension exerted by the right segment then the left segment will also exert an equal and opposite force so we have that $T_L=-T$

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2 years ago

When a surface is submerged in a fluid, the resultant pressure force on the body acts in what manner? Assume no shear forces are

Alex17521 [72]

Answer:

Perpendicular to the surface of the body.

Explanation:

When a surface is submerged in a fluid the resultant pressure force on the body acts perpendicular to the surface of the body. This is because fluids cannot withstand nor exert sideways forces. One could obtain this analytically considering that the origin of this force comes from the movement of the fluid molecules.

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2 years ago

What is the net force required to give an automobile with a mass of 1,600 kg an acceleration of 4.5 m/s2

Paladinen [302]

Answer:

Fnet=7200 N

Explanation:

Fnet=mass x acceleration

mass= 1600kg acceleration=4.5m/s^2

8 0

2 years ago

An archer fires an arrow, which produces a muffled "thwok" as it hits a target. If the archer hears the "thwok" exactly 1 s afte

aniked [119]

Answer:

35,79 meters

Explanation:

So, we got an archer, and we got a target. Lets call the distance between this two d.

Now, the archer fires the arrow, that, in a time $t_{arrow}$ travels the distance d with a speed $v_{arrow}$ of 40 m/s and hits the target. We can see that the equation will be:

$v_{arrow} * t_{arrow} = d\\ \\40 \frac{m}{s} * t_{arrow} = d$

Immediately after this, the arrow produces a muffled sound, which will travel the distance d at 340 m/s in a time $t_{sound}$ . Obtaining :

$v_{sound} * t_{sound} = d\\ \\340 \frac{m}{s} * t_{sound} = d$ .

Finally, the sound reaches the archer, exactly 1 second after he fired the bow, so:

$t_{arrow} + t _{sound} = 1 s$ .

This equation allows us to write:

$t _{sound} = 1 s - t_{arrow}$ .

Plugging this relationship in the distance equation for the sound:

$340 \frac{m}{s} * t_{sound} = d \\ \\ 340 \frac{m}{s} * (1 s- t_{arrow}) = d$ .

Now, we can replace d from the first equation, and obtain:

$40 \frac{m}{s} * t_{arrow} = d \\ 40 \frac{m}{s} * t_{arrow} = 340 \frac{m}{s} * (1 s- t_{arrow})$ .

Now, we can just work a little bit:

$40 \frac{m}{s} * t_{arrow} = 340 \frac{m}{s} * 1 s - 340 \frac{m}{s} * t_{arrow} \\ \\ 40 \frac{m}{s} * t_{arrow} + 340 \frac{m}{s} * t_{arrow} = 340 m \\ \\ 380 \frac{m}{s} * t_{arrow} = 340 m \\ \\ t_{arrow} = \frac{340 m}{380 \frac{m}{s}} \\ \\ t_{arrow} = 0.8947 s$ .

Now, we can just plug this value into the first equation:

$40 \frac{m}{s} * t_{arrow} = d$

$40 \frac{m}{s} * 340/380 s = 35,79 s = d$

6 0

2 years ago

A heat engine accepts 200,000 Btu of heat from a source at 1500 R and rejects 100,000 Btu of heat to a sink at 600 R. Calculate

diamong [38]

To solve the problem it is necessary to apply the concepts related to the conservation of energy through the heat transferred and the work done, as well as through the calculation of entropy due to heat and temperatra.

By definition we know that the change in entropy is given by

$\Delta S = \frac{Q}{T}$

Where,

Q = Heat transfer

T = Temperature

On the other hand we know that by conserving energy the work done in a system is equal to the change in heat transferred, that is

$W = Q_{source}-Q_{sink}$

According to the data given we have to,

$Q_{source} = 200000Btu$

$T_{source} = 1500R$

$Q_{sink} = 100000Btu$

$T_{sink} = 600R$

PART A) The total change in entropy, would be given by the changes that exist in the source and sink, that is

$\Delta S_{sink} = \frac{Q_{sink}}{T_{sink}}$

$\Delta S_{sink} = \frac{100000}{600}$

$\Delta S_{sink} = 166.67Btu/R$

On the other hand,

$\Delta S_{source} = \frac{Q_{source}}{T_{source}}$

$\Delta S_{source} = \frac{-200000}{1500}$

$\Delta S_{source} = -133.33Btu/R$

The total change of entropy would be,

$S = \Delta S_{source}+\Delta S_{sink}$

$S = -133.33+166.67$

$S = 33.34Btu/R$

Since $S\neq 0$ the heat engine is not reversible.

PART B)

Work done by heat engine is given by

$W=Q_{source}-Q_{sink}$

$W = 200000-100000$

$W = 100000 Btu$

Therefore the work in the system is 100000Btu

4 0

2 years ago