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2 years ago

Thermodynamic Processes

Physics

1 answer:

goblinko [34]2 years ago

5 0

Answer:

A. Part a is the attachment

B. total work = 10.4kj

Explanation:

$workdone=nRT1ln\frac{Vb}{Va}$

T1 = constant temperature

nRT1 = PaVa = PbVb

We write equation as

$workdone =(PaVa)ln\frac{Vb}{Va}$

5ma = Pa, 5L = Va, Vb = 10L(temperature is doubled)

$w1 = workdone =(5mpa*5L)ln\frac{10L}{5L}$

W1 = 25 ln2

W1 = 25 x 0.693

= 17.327kj

The isochoric expansion has no change in volume. So,

W2 = 0

Isothermal compression

$w3=nRT3ln\frac{Vd}{Vc}$

T3 = constant temperature

nRT3 = PcVc = PdVd

$workdone=(PcVc)ln\frac{Vd}{Vc}$

Pc = 1mpa Vc = 10L Vd = 5L

$w3=(1)(10)ln\frac{5L}{10L}$

= 10x-0.693

= -6.93kj

Isochoric compression has no change in volume. Workdone w4 = 0

Total workdone = w1 + w2 + w3 + w4

= 17.33 + 0 + (-6.93) + 0

= 10.4kj

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Answer:

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Explanation:

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mass of the ball
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3 years ago

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Answer:

Check body of the explanation

Explanation:

Ooook, quick theory rushdown. if you're at a depth of h in a tank of a fluid, the pressure is the sum of the atmosferic pressure (if the tank is open on top) plus a term which is the product of acceleration of gravity - about $10 ms^-^2$ , the density of whatever you're sinking in, and the depth at which you are. In formula, $p(h) = p_0 + \rho g h$ , and the pressure is the same for every point of the tank at the same depth.

At this point, we can start answering!

1a. The pressure at A is - not counting atmosferic pressure - $1000 * 10 * 1 = 10^4 Pa$ , while in B is $1000*10*2 = 2*10^4 Pa$ , so it's half of it.

1b. The two points are at the same depth, so the pressure is the same - they would be even if the two cilinders weren't linked!

1c. Ditto. Same depth? same pressure!

1d. Usual equation, this time density is 800. Pressure is $800*10*2 = 1,6*10^4 Pa$ : Since the density is 4/5 of water, the pressure is also 4/5 of the one exerted by water

2a. The volume is simply the product, so $4m*3m*2m = 24m^3$

2b. Density is defined as mass over volume, so you simply multiply the volume you found earlier by the density of paraffine: $800* 24 = 1,92 *10^4kg$

2c. Weight is defined as the mass of something times the acceleration due to gravity, in our case it's $1.92 *10^4 kg * 10 ms^{-2} = 1.92 * 10^5 N$

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2 years ago

A pressure sensor inside of a mixing tank is designed to turn red when the pressure inside the tank exceeds 1.9 kPa. If the sens

spin [16.1K]

Answer:

19 N

Explanation:

From the question given above, the following data were obtained:

Pressure (P) = 1.9 kPa

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Force (F) =?

Next, we shall convert 1.9 KPa to N/m². This can be obtained as follow:

1 KPa = 1000 N/m²

Therefore,

1.9 KPa = 1.9 KPa × 1000 N/m² / 1 KPa

1.9 KPa = 1900 N/m²

Thus, 1.9 KPa is equivalent to 1900 N/m².

Next, we shall convert 10 cm to m. This can be obtained as follow:

100 cm = 1 m

Therefore,

10 cm = 10 cm × 1 m / 100 cm

10 cm = 0.1 m

Thus, 10 cm is equivalent to 0.1 m

Next, we shall determine the area of the square. This can be obtained as follow:

Length (L) = 0.1 m

Area of square (A) =?

A = L²

A = 0.1²

A = 0.01 m²

Thus, the area of the square is 0.01 m².

Finally, we shall determine the force that must be exerted on the sensor in order for it to turn red. This can be obtained as follow:

Pressure (P) = 1900 N/m²

Area (A) = 0.01 m²

Force (F) =?

P = F/A

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Cross multiply

F = 1900 × 0.01

F = 19 N

Therefore, a force of 19 N must be exerted on the sensor in order for it to turn red.

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Answer:

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