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mr_godi [17]
3 years ago
14

A gas at a pressure of 2.10 atm undergoes a quasi static isobaric expansion from 3.70 to 5.40 L. How much work is done by the ga

s (in J)
Physics
1 answer:
BartSMP [9]3 years ago
5 0

Answer:

Total work done in expansion will be 3.60\times 10^5J

Explanation:

We have given pressure P = 2.10 atm

We know that 1 atm =1.01\times 10^5Pa

So 2.10 atm =2.10\times 1.01\times 10^5=2.121\times 10^5Pa

Volume is increases from 3370 liter to 5.40 liter

So initial volume V_1=3.70liter

And final volume V_2=5.40liter

So change in volume dV=5.40-3.70=1.70liter

For isobaric process work done is equal to W=PdV=2.121\times 10^5\times 1.70=3.60\times 10^5J

So total work done in expansion will be 3.60\times 10^5J

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Alfredo makes a diagram to organize his notes about charging objects.
konstantin123 [22]

The region marked X in the diagram shows that the objects have the same charge.

<h3>What is conduction?</h3>

The term conduction has to do with the manner of charging in which charge is passed from one object to another. Induction involves charging objects without the objects touching each other.

The region marked X in the diagram shows that the objects have the same charge.

Learn ore about conduction:brainly.com/question/15306642

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8 0
1 year ago
A moving particle encounters an external electric field that decreases its kinetic energy from 9520 eV to 7060 eV as the particl
Sati [7]

Given Information:

KEa = 9520 eV

KEb = 7060 eV

Electric potential = Va = -55 V

Electric potential = Vb = +27 V

Required Information:

Charge of the particle = q = ?

Answer:

Charge of the particle = +4.8x10⁻¹⁸ C

Explanation:

From the law of conservation of energy, we have

ΔKE = -qΔV

KEb - KEa = -q(Vb - Va)

-q = KEb - KEa/Vb - Va

-q = 7060 - 9520/27 - (-55)

-q = 7060 - 9520/27 + 55

-q = -2460/82

minus sign cancels out

q = 2460/82

Convert eV into Joules by multiplying it with 1.60x10⁻¹⁹

q = 2460(1.60x10⁻¹⁹)/82

q = +4.8x10⁻¹⁸ C

6 0
3 years ago
Fronts are termed by the temperature of the air mass that overtakes another air mass. A. True B. False
Vikki [24]
A. True
if cold air is replacing warm air it is a cold front and vice versa.
6 0
3 years ago
Read 2 more answers
1.)A tank travels at a rate of 10.0 km/hr for 12.00 minutes, then at 15.0 km/hr for 8.00
rosijanka [135]

12.00 min = 0.2 hr

8.00 min = 0.15 hr

Total distance:

(10.0 km/hr) (0.2 hr) + (15.0 km/hr) (0.15 hr) + (20.0 km/hr) (0.2 hr)

= 8.25 km

Average speed:

(10.0 km/hr + 15.0 km/hr + 20.0 km/hr) / 3

= 15 km/hr

Change in position:

(10.0 km/hr) (0.2 hr) + (15.0 km/hr) (0.15 hr) - (20.0 km/hr) (0.2 hr)

= 0.25 km

Average velocity:

(10.0 km/hr + 15.0 km/hr - 20.0 km/hr) / 3

≈ 1.67 m/s

8 0
3 years ago
A cylinder fitted with a piston exists in a high-pressure chamber (3 atm) with an initial volume of 1 L. If a sufficient quantit
Bess [88]

Answer:

C. 85%

Explanation:

A cylinder fitted with a piston exists in a high-pressure chamber (3 atm) with an initial volume of 1 L. If a sufficient quantity of a hydrocarbon material is combusted inside the cylinder to produce 1 kJ of energy, and if the volume of the chamber then increases to 1.5 L, what percent of the fuel's energy was lost to friction and heat?

A. 15%

B. 30%

C. 85%

D. 100%

work done by the system will be

W=PdV

p=pressure

dV=change in volume

3tam will be changed to N/m^2

3*1.01*10^5

W=3.03*10^5*(1.5-1)

convert 0.5L to m^3

5*10^-4

W=3.03*10^5*5*10^-4

W=152J

therefore

to find the percentage used

152/1000*100

15%

100%-15%

85% uf the fuel's energy was lost to friction and heat

6 0
3 years ago
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