The author writes that each part of the ecosystem is as important as another. Based

You have gas in a container with a movable piston. The walls of the container are thin enough so that its temperature stays the

bearhunter [10]

Answer:

New pressure of the gas increases by 26.5% with respect to initial pressure, new volume decreases 27% with respect to initial volume and new temperature decreases 8% with respect to initial volume.

Explanation:

If we assume the gas is a perfect gas we can use the perfect gas equation:

$PV=nRT$

For Isothermal process:

$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$ (1)

Where subscripts 1 shows before the isothermal process and 2 after it, because isothermal means constant temperature T1=T2, and pressure increases by 10% means P2=1,1*P1, using these facts on (1) we have:

$V_{2}=\frac{V_{1}}{1.1}$ (2)

For Isobaric process:

$\frac{P_{2}V_{2}}{T_{2}}=\frac{P_{3}V_{3}}{T_{3}}$ (3)

Where subscripts 2 shows before the isobaric process and 3 after it, because isobaric means constant pressure P2=P3, and volume decreases by 20% means V3=0.8*V2, using these facts on (3) we have:

$T_{3}=0.8T_{2}$ (4)

For Isochoric process:

$\frac{P_{3}V_{3}}{T_{3}}=\frac{P_{4}V_{4}}{T_{4}}$ (5)

Where subscripts 3 shows before the isochoric process and 4 after it, because isochoric means constant volume V3=V4, and temperature increases by 15% means T4=1.15*T3, using these facts on (5) we have:

$P_{4}=1.15P_{3}$ (6)

So now because P4=1.15*P3, P2=P3 and P2=1.1*P1:

$P_{4}=1.15*1.1P_{1}=1.265P1$

This is, the new pressure of the gas increases by 26.5% with respect to initial pressure.

Similarly, we have V3=V4, V3=0.8*V2 and V1=1,1*V2:

$V_{4}=\frac{0.8}{1.1}V_{1}=0.72V1$

so the final volume decreases 27% with respect to initial volume.

T4=1,15*T3, T3=0.8*T2 and T1=T2:

$T_{4}=1.15*0.8T_{1}=0.92T1$

The new temperature decreases 8% with respect to initial volume.

3 0

3 years ago

Someone got this paper?

Snezhnost [94]

First question (upper left):
1/Req = 1/12 + 1/24 = 1/8
Req = 8 ohms
Voltage is equal through different resistors, and V1 = V2 = 24 V.
Current varies through parallel resistors: I1 = V1/R1 = 24/12 = 2 A. I2 = 24/24 = 1 A.

Second question (middle left):
V1 = V2 = 6 V (parallel circuits)
I1 = 2 A, I2 = 1 A, IT = 2+1 = 3 A.
R1 = V1/I1 = 6/2 = 3 ohms, R2 = 6/1 = 6 ohms, 1/Req = 1/2 + 1/1, Req = 2/3 ohms

Third question (bottom left):
V1 = V2 = 12 V
IT = 3 A, meaning Req = V/It = 12 V/3 A = 4 ohms
1/Req = 1/R1 + 1/R2, 1/4 = 1/12 + 1/R2, R2 = 6 ohms
I1 = V/R1 = 1 A, I2 = V/R2 = 2 A

Fourth question (top right):
1/Req = 1/20 + 1/20, Req = 10 ohms
IT = 4 A, so VT = IT(Req) = 4*10 = 40 V
Parallel circuits, so V1 = V2 = VT = 40 V
Since the resistors are identical, the current is split evenly between both: I1 = I2 = IT/2 = 2 A.

Fifth question (middle right):
1/Req = 1/5 + 1/20 + 1/4, Req = 2 ohms
IT = VT/Req = 40 V/2 ohms = 20 A
V1 = V2 = V3 = 40 V
The current of 20 A will be divided proportionally according to the resistances of 5, 20, and 4, the factors will be 5/(5+20+4), 20/(5+20+4), and 4/(5+20+4), which are 5/29, 20/29, and 4/29.
I1 = 20(5/29) = 100/29 A
I2 = 20(20/29) = 400/29 A
I3 = 20(4/29) = 80/29 A

Sixth question (bottom right):
V2 = 30V is given, but since these are parallel circuits, V1 = VT = 30 V.
Then I1 = V1/R1 = 30 V/10 ohms = 3 A.
I2 = 30 V/15 ohms = 2 A.
IT = 3 + 2 = 5 A
1/Req = 1/10 + 1/15, Req = 6 ohms

6 0

3 years ago

Why are Watson’s experiments considered controversial?

iris [78.8K]

D, it is considered unethical today

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