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ValentinkaMS [17]
3 years ago
11

To produce work a gas is expanded adiabatically from 3 MPa and 300oC to 80 kPa in a piston-cylinder device. Which of these two c

hoices –air with an isentropic expansion efficiency of 90% or neon with an isentropic expansion efficiency of 80% – will produce the most work?

Physics
1 answer:
ANEK [815]3 years ago
5 0

Answer:

Explanation:

Check attachment for solution

Given that, .

The initial gas pressure is

P1 = 3 MPa

The initial gas temperature

T1 = 300°C

The final gas pressure is

P2 = 80 KPa

Which are of two choices

1. Air with an isentropic expansion of 90% efficiency

2. Neon with an isentropic expansion of 80% efficiency

We want to know which will produce more work

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The volume electric charge density of a solid sphere is given by the following equation: The variable r denotes the distance fro
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Answer:

62.8 μC

Explanation:

Here is the complete question

The volume electric charge density of a solid sphere is given by the following equation: ρ = (0.2 mC/m⁵)r²The variable r denotes the distance from the center of the sphere, in spherical coordinates. What is the net electric charge (in μC) of the sphere if the radius of the sphere is 0.5 m?

Solution

The total charge on the sphere Q = ∫∫∫ρdV where ρ = volume charge density = 0.2r² and dV = volume element in spherical coordinates = r²sinθdθdrdΦ

So,  Q =  ∫∫∫ρdV

Q =  ∫∫∫ρr²sinθdθdrdΦ

Q =  ∫∫∫(0.2r²)r²sinθdθdrdΦ

Q =  ∫∫∫0.2r⁴sinθdθdrdΦ

We integrate from r = 0 to r = 0.5 m, θ = 0 to π and Φ = 0 to 2π

So, Q =  ∫∫∫0.2r⁴sinθdθdrdΦ

Q =  ∫∫∫0.2r⁴[∫sinθdθ]drdΦ

Q =  ∫∫0.2r⁴[-cosθ]drdΦ

Q =  ∫∫0.2r⁴-[cosπ - cos0]drdΦ

Q =  ∫∫∫0.2r⁴-[-1 - 1]drdΦ

Q =  ∫∫0.2r⁴-[- 2]drdΦ

Q =  ∫∫0.2r⁴(2)drdΦ

Q =  ∫∫0.4r⁴drdΦ

Q =  ∫0.4r⁴dr∫dΦ

Q =  ∫0.4r⁴dr[Φ]

Q =  ∫0.4r⁴dr[2π - 0]

Q =  ∫0.4r⁴dr[2π]

Q =  ∫0.8πr⁴dr

Q =  0.8π∫r⁴dr

Q =  0.8π[r⁵/5]

Q = 0.8π[(0.5 m)⁵/5 - (0 m)⁵/5]

Q = 0.8π[0.125 m⁵/5 - 0 m⁵/5]

Q = 0.8π[0.025 m⁵ - 0 m⁵]

Q = 0.8π[0.025 m⁵]

Q = (0.02π mC/m⁵) m⁵

Q = 0.0628 mC

Q = 0.0628 × 10⁻³ C

Q = 62.8 × 10⁻³ × 10⁻³ C

Q = 62.8 × 10⁻⁶ C

Q = 62.8 μC

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Answer:

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A 0.500-V potential difference is maintained across a 1.50-m length of tungsten wire that has a cross-sectional area of 0.800 mm
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Answer:

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Explanation:

From the question

R = ρL/A..................... Equation 1

Where R = resistance of the tungsten wire, ρ = Resistivity of the tungsten wire, L = length, A = cross sectional area.

Given: L = 1.5 m, A = 0.8 mm² = 0.8×10⁻⁶ m, ρ = 5.60×10⁻⁸ Ω.m

Substitute these values into equation 1

R = 1.5(5.60×10⁻⁸)/0.8×10⁻⁶

R = 0.084 Ω.

Finally, using Ohm law,

V = IR

Where V = Voltage, I = current

Make I the subject of the equation

I = V/R............... Equation 2

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Answer: 6.12 kg

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Z = 90/14.7

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Thus, the bowling ball weigh 6.12 kilograms

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