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SVEN [57.7K]
3 years ago
10

​A piston–cylinder assembly contains 5.0 kg of air, initially at 2.0 bar, 30 oC. The air undergoes a process to a state where th

e pressure is 1.0 bar, during which the pressure–volume relationship is pV = constant. Assume ideal gas behavior for the air.
Determine the work and heat transfer, in kJ.

For an ideal gas, the specific volume is equal to:

A: the gas constant R times the gage temperature divided by the gage pressure.

B: the universal gas constant times the absolute temperature divided by the gage pressure.

C: the gas constant R times the absolute temperature divided by the absolute pressure.

D: the universal gas constant times the absolute temperature divided by the absolute pressure.
Physics
1 answer:
vlada-n [284]3 years ago
5 0

Answer:

Explanation:

The process is isothermic,  as P V = constant .

work done = 2.303 n RT log P₁ / P₂

= 2.303 x 5 / 29 x 8.3 x 303  log 2 / 1 kJ

= 300.5k J

This energy in work done by the gas will come fro heat supplied as internal energy is constant due to constant temperature.

heat supplied  = 300.5k J

specific volume is volume per unit mass

v / m

pv = n RT

pv  = m / M  RT

v / m = RT / p M

specific volume = RT / p M

option B is correct.

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A biker is pedaling at a constant speed of 36 km/h. During the last 10 s of the race, he increases his speed with a constant acc
adell [148]

Answer:

54 km/h

Explanation:

given,

speed of the biker = 36 Km/h

time = 10 s

acceleration = 0.5 m/s²

speed at which it crosses the finish line  = ?

v = 36 x 0.278 = 10 m/s

using equation of motion

v = u + a t

v = 10 + 0.5 x 10

v = 15 m/s

v = 15 x 3.6 = 54 km/hr

speed at which the biker crosses the finish line is equal to 54 km/h

4 0
3 years ago
Calculate the internal axial load at a point D if length L=7 ft. The part is subjected to loads P1=632 lbs, P2=888 lbs (applied
liraira [26]

Answer:

- 256 lbs

Explanation:

The internal axial load at point D can be calculated as the change in the subjected loads. if the magnitude of the horizontal direction = zero

EF_x = 0; Then:

internal axial load at point D = Δ P

= -(P₂ - P₁)

= - ( 888 lbs - 632 lbs)

= - 256 lbs

5 0
3 years ago
When was the first airplane made?
expeople1 [14]
The first ariplanr was made December 17, 1903
3 0
3 years ago
Read 2 more answers
A 10 cm diameter pulley is used to lift a bucket of cement weighing 400 N. How much force must be applied to the rope to lift th
Alina [70]

hi brainly user! ૮₍ ˃ ⤙ ˂ ₎ა

⊱┈────────────────────────┈⊰

\large \bold {ANSWER}

Considering that the pulley is fixed, the force applied should be equal to the weight of the object - of 400N.

\large \bold {EXPLANATION}

Pulleys or pulleys are mechanical tools used to assist in the movement of objects and bodies. There are two types of pulleys: fixed and movable. While the fixed pulley changes the direction of force, the moving pulley helps to decrease the force needed to move the object or body in question.

As the statement only tells us a pulley, we must consider that it is fixed, <u>because generally when it is mobile, this information is highlighted in the question</u>.

In this way, a fixed pulley only changes the direction of the applied force. Thus, the force must have the same magnitude as the weight of the object to be moved. If the bucket weighs 400N, the force applied to the pulley must be 400N.

<u>Therefore, having a fixed pulley, the force applied must be equal to the weight of the object, and will be 400N.</u>

3 0
2 years ago
A person slaps her leg with her hand, which results in her hand coming to rest in a time interval of 2.65 ms from an initial spe
dezoksy [38]

Answer:

the magnitude of the average contact force exerted on the leg is  3466.98 N

Explanation:

Given the data in the question;

Initial velocity of hand v₀ = 5.25 m/s

final velocity of hand v = 0 m/s

time interval t = 2.65 ms = 0.00265 s

mass of hand m = 1.75 kg

We calculate force on the hand F_{hand

using equation for impulse in momentum

F_{hand × t = m( v - v₀ )

we substitute

F_{hand × 0.00265  = 1.75( 0 - 5.25 )

F_{hand × 0.00265  = 1.75( - 5.25 )

F_{hand × 0.00265  = -9.1875

F_{hand = -9.1875 / 0.00265  

F_{hand = -3466.98 N

Next we determine force on the leg F_{leg

Using Newton's third law of motion

for every action, there is an equal opposite reaction;

so, F_{leg = - F_{hand

we substitute

F_{leg = - ( -3466.98 N )

F_{leg = 3466.98 N

Therefore, the magnitude of the average contact force exerted on the leg is  3466.98 N

5 0
3 years ago
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