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2 years ago

Please answer i dont understand and dont know the answer

Engineering

1 answer:

dsp732 years ago

3 0

I need pointskjjjjjjhhhhhhhhhhhhhhhbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb

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Five kilograms of air at 427°C and 600 kPa are contained in a piston–cylinder device. The air expands adiabatically until the pr

son4ous [18]

Answer:

The entropy change of the air is $0.240kJ/kgK$

Explanation:

$T_{1} =427+273K,T_{1} =700K\\P_{1} =600kPa\\P_{2} =100kPa$

$T_{2}$ is unknown

we can apply the following expression to find $T_{2}$

$-w_{out} =mc_{v} (T_{2} -T_{1} )$

$T_{2} =T_{1} -\frac{w_{out } }{mc_{v} }$

now substitute

$T_{2} =700K-\frac{600kJ}{5kg*0.718kJ/kgK} \\T_{2}=533K$

To find entropy change of the air we can apply the ideal gas relationship

Δ $s_{air}$ $=c_{p} ln\frac{T_{2} }{T_{1} } -Rln\frac{P_{2} }{P_{1} }$

Δ $s_{air} =$ $1.005*ln(\frac{533}{700})-0.287* in(\frac{100}{600} )$

Δ $s_{air} =0.240kJ/kgK$

4 0

2 years ago

I WILL GIVE BRAINLIEST IF ANSWER FAST What is the measurement on this Dial Caliper?

garik1379 [7]

Answer:

b i think i dont see any dial caliper

Explanation:

8 0

2 years ago

Can i have answer of this question please?

cestrela7 [59]

uh its a tough one mate

3 0

2 years ago

Astronauts who landed on the moon during the Apollo 15, 16, and 17 missions brought back a large collection of rocks to the eart

stiks02 [169]

Answer: a) W(earth) = 935.62 lbs

b) Mass of rocks in slugs = 29.06 slugs

Explanation:

a) From Newton's law, W = mg. Whether on the moon or on earth. Although, the mass of the rocks everywhere is the same, that is, mass of rocks on the moon = mass of rocks on earth.

W(moon) = mg(moon)

W(moon) = 154 lbs

g(moon) = 5.30 ft/s2

m = W(moon)/g(moon) = 154/5.3 = 29.06 lb.s2/ft

W(earth) = m g(earth)

g(earth) = 32.2 ft/s2

W(earth) = 29.06 × 32.2 = 935.62 lbs.

b) A slug = 1 lb.s2/ft, therefore the mass of the rocks in slugs is 29.06 slugs.

QED!

8 0

2 years ago

A spacecraft at rest has moment of inertia of 100 kg-m^2 about an axis of interest. If a 1 newton thruster with a 1 meter moment

Kazeer [188]

To solve this problem we will apply the concepts related to translational torque, angular torque and the kinematic equations of angular movement with which we will find the angular displacement of the system.

Translational torque can be defined as,

$\tau = Fd$

Here,

F = Force

d = Distance which the force is applied

$\tau = (1N)(1m)$

$\tau = 1N\cdot m$

At the same time the angular torque is defined as the product between the moment of inertia and the angular acceleration, so using the previous value of the found torque, and with the moment of inertia given by the statement, we would have that the angular acceleration is

$\tau = I\alpha$