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fomenos
3 years ago
7

Determine the boundary work done by a gas during an expansion process if the pressure and volume values at various states are me

asured to be 300 kPa, 1 L; 290 kPa, 1.1 L; 270 kPa, 1.2 L; 250 kPa, 1.4 L; 220 kPa, 1.7 L; and 200 kPa, 2 L.
Physics
1 answer:
astraxan [27]3 years ago
7 0

Answer:

W=0.217kJ

Explanation:

#Boundary Work is energy expended when a force acts through a displacement

The boundary of work can be calculated as the area under the P-V curve for all the 4 stages measured:

W= \sum_{n=1}^{5} \frac{P_{n+1}+P_{n+2}+P_{n+3}+P_{n+4}+P_{n+5}}{2}\\\\=0.5((290+300)(1.1-1)+(270+290)(1.2-1.1)+(250+270)(1.4-1.3)+(220+250)(1.7-1.4)+(200+220)(2-1.7))\times 10^-^3kJ\\\\=0.217kJ

Hence the boundary of work done by a gas during this 4-stage process is 0.217kJ

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A helicopter is ascending vertically with a speed of 5.10m/s. At a height of 105m above the Earth, a package is dropped from a w
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(b) The maximum height of the projectile is 4,591.84 m

(c) The speed with which the projectile hits the ground is 670.82 m/s.

Explanation:

Given;

initial speed of the projectile, u = 600 m/s

angle of projection, θ = 30⁰

acceleration due to gravity, g = 9.8 m/s²

(a) The range of the projectile in meters;

R = \frac{u^2sin \ 2\theta}{g} \\\\R = \frac{600^2 sin(2\times 30^0)}{9.8} \\\\R = \frac{600^2 sin (60^0)}{9.8} \\\\R = 31,813.18 \ m

(b) The maximum height of the projectile in meters;

H = \frac{u^2 sin^2\theta}{2g} \\\\H = \frac{600^2 (sin \ 30)^2}{2\times 9.8} \\\\H = \frac{600^2 (0.5)^2}{19.6} \\\\H = 4,591.84 \ m

(c) The speed with which the projectile hits the ground is;

v^2 = u^2 + 2gh\\\\v^2 = 600^2 + (2 \times 9.8)(4,591.84)\\\\v^2 = 360,000 + 90,000.064\\\\v = \sqrt{450,000.064} \\\\v = 670.82 \ m/s

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