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A container filled with a sample of an ideal gas at the pressure of 150 Kpa. The gas is compressed isothermally to one-third of its original volume. What is the new pressure of the gas a)-900 kpa b)- 300 kpa c)- 450 kpa d)- 600 kpa

Answers

Answer 1
Answer:

Answer: c) 450 kPa

Explanation:

Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto (1)/(V)     (At constant temperature and number of moles)

P_1V_1=P_2V_2  

where,

P_1 = initial pressure of gas  = 150 kPa

P_2 = final pressure of gas  = ?

V_1 = initial volume of gas   = v L

V_2 = final volume of gas  = (v)/(3)L

150* v=P_2* (v)/(3)  

P_2=450kPa

Therefore, the new pressure of the gas will be 450 kPa.

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Answers

Answer:

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In same Fig. A-3, a force F, having a slope of 2 vertical 3 horizontal, produces a clockwise moment of 330 ft-lb about A and a counterclockwise moment of 420 ft-lb about B. Compute the moment of F about C.​

Answers

The moment of force at the given slope about point C is 210 ft-lb.

The given parameters:

  • Vertical slope, = 2
  • Horizontal slope = 3
  • Clockwise moment = 330 ft-lb
  • Counterclockwise moment = 420 ft-lb

The magnitude of the two moments are in the following simple ratio;

330:420 = 11:14 (divide through by 30)

  • the A coordinate = (0, 5)
  • the B-coordinate = (5,0)

The line of action of the force passes line AB at the final following coordinates;

total ratio of 11:14 = 11 + 14 = 25

= (11 )/(25) * 5, \ \ (14)/(25) * 5\n\n= (2.2, \ 2.8)

The position of C = (3, 1)

The resultant position of point C = (3 - 2.2,  2.8-1) = (0.8, 1.8)

The moment of force at the given slope about point C is calculated as;

3(1.8) + 2(0.8) = 7

Recall that this is the simplest form of the moment produced by the force.

Moment about C = 7 x 30 = 210 ft-lb

Thus, the moment of force at the given slope about point C is 210 ft-lb.

Learn more about moment of force here: brainly.com/question/6278006

Given that the debouncing circuit is somewhat expensive in terms of hardware (2 NAND gates, 2 resistors, and a double-pole, single throw switch), describe applications where you would require switch-debouncing circuits as well as applications where you would not need to include the additional hardware for switch debouncing (in other words, applications where you can tolerate switch bouncing). Note, you cannot use the clock and clear inputs of our lab as example applications; instead you need to think of other examples.

Answers

Explanation:

One of the common application of debouncing g circuit is in microprocessors or microcontrollers or FPGA's where fast processing is required. In such cases, it is extremely important that during the limited processing cycle, the signals remains valid without debouncinng. Because debouncing can complete impact the output of the controller.

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Although the viscoelastic response of a polymer can be very complex (time-dependent stress cycling for instance), two special loading scenarios are fairly simple to describe mathematically. _____________refers to scenarios for which the stress applied to a polymer must decay over time in order to maintain a constant strain. Otherwise, over time, the polymer chains will slip and slide past one another in response to a constant applied load and the strain will increase (in magnitude).
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Answers

Answer:

Viscoelastic stress relaxation

Viscoelastic creep

Explanation:

  • Viscoelastic stress refers to the reduction of tensile stress with the relaxation time that occurs when the body is kept at a certain length under tensile stress. The purpose of this study was to demonstrate viscoelastic stress relaxation
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  • When under constant pressure, the viscoelastic material experiences a time-dependent increase in pressure. This phenomenon is called viscoelastic creep.
  • Therefore the phase constant pressure decreases when there is relaxation in the viscoelastic stress

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Answers

Answer:

I want to believe the program is to be written in java and i hope your question is complete. The code is in the explanation section below

Explanation:

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A student takes 60 voltages readings across a resistor and finds a mean voltage of 2.501V with a sample standard deviation of 0.113V. Assuming that errors are due to random processes, how many of the readings are expected to be greater than 2.70V?

Answers

Answer:

There are 2 expected readings greater than 2.70 V

Solution:

As per the question:

Total no. of readings, n = 60 V

Mean of the voltage, \mu = 2.501 V

standard deviation, \sigma = 0.113 V

Now, to find the no. of readings greater than 2.70 V, we find:

The probability of the readings less than 2.70 V, P(X\leq 2.70):

z = (x - \mu)/(\sigma) = (2.70 - 2.501)/(0.113) = 1.761

Now, from the Probability table of standard normal distribution:

P(z\leq 1.761) = 0.9608

Now,

P(X\geq 2.70) = 1 - P(X\leq 2.70) = 1 - 0.9608 = 0.0392 = 3.92%

Now, for the expected no. of readings greater than 2.70 V:

P(X\geq 2.70) = (No.\ of\ readings\ expected\ to\ be\ greater\ than\ 2.70\ V)/(Total\ no.\ of\ readings)

No. of readings expected to be greater than 2.70 V = P(X\geq 2.70)* Total\ no.\ of\ readings

No. of readings expected to be greater than 2.70 V = 0.0392* 60 = 2.352 ≈ 2
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