The pressure in an automobile tire depends on thetemperature of the air in the tire. When the air temperature is25°C, the pressure gage reads 210 kPa. If the volume of the tire is 0.025 m3, determine the pressure rise in the tire whenthe air temperature in the tire rises to 50°C. Also, determinethe amount of air that must be bled off to restore pressure toits original value at this temperature. Assume the atmosphericpressure to be 100 kPa.

Answers

Answer 1
Answer:

Answer:

The pressure rise in the tire when the air temperature in the tire rises to 50°C is 337.43 KPa.

The amount of air that must be bled off to restore pressure 0.007 Kg

Explanation:

Knowing

T1 = 25°C = 298 K

T2 = 50°C = 323 K

volume of the tire = V = 0.025 m^(3)

P = 210 kPa (gage)

Pabs = 210 + 101 = 311 KPa

Before the temperature rise

P1 V1 = m1 R1 T1

m1 = (P1 V1)/(R1 T1) = (310 * 10^(3) * 0.25 )/(287 - 298) = 0.091 Kg\n \n

After the temperature rise

P2 = (m2 * R * T2)/(V2) = (0.091 *287*323 )/(0.025) = 337.43 KPa

after bleeding the pressure and the volume returns  to its first value

P1 = P2 and V1 = V2

then

(m2 * R * T2)/(V) = (m1 * R * T1)/(V)

m2 = (m1*T1)/(T2)

m2 = (0.091*298)/(332) = 0.084 Kg\n\n

mbleed = m1 - m2 --> mbleed = 0.91 - 0.84 = 0.007 Kg

P2 = 337.43 KPa

mbleed = 0.007 Kg


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Buffering is an important op amp application because it solves impedance-matching problems that can't easily be solved with purely resistive circuits. This allows for the transfer of maximum power between two circuits without any loss of signal strength.

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Complete question

Buffering is an important op amp application because it solves _____ that can't easily be solved with purely resistive circuits.

Group of answer choices

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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.

A case where debouncing can be compromised where a system is run partially through human intervention or that has different indications for one operation.

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Answers

Answer:

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At steady state, air at 200 kPa, 325 K, and mass flow rateof 0.5 kg/s enters an insulated duct having differing inlet
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Answers

Letra A

A letra

A.
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A tensile test is carried out on a bar of mild steel of diameter 20 mm. The bar yields under a load of 80 kN. It reaches a maximum load of 150 kN, and breaks finally at a load of 70 kN. Find (i) the tensile stress at the yield point (1i) the ultimate tensile stress; (iii) the average stress at the breaking point, if the diameter of the fractured neck is 10mm

Answers

Answer:

tensile stress at yield = 254 MPa

ultimate stress = 477 MPa

average stress = 892 MPa

Explanation:

Given data in question

bar yields load = 80 kN

load maximum = 150 kN

load fail = 70 kN

diameter of steel (D) = 20 mm i.e. = 0.020 m

diameter of breaking point (d) = 10 mm i.e. 0.010 m

to find out

tensile stress at the yield point , ultimate tensile stress and average stress at the breaking point

solution

in 1st part we calculate tensile stress at the yield point by this formula

tensile stress at yield =  yield load / area

tensile stress at yield =  80 ×10³ / \pi /4 × D²

tensile stress at yield =  80 ×10³ / \pi /4 × 0.020²

tensile stress at yield = 254 MPa

in 2nd part we calculate ultimate stress by given formula

ultimate stress = maximum load / area

ultimate stress = 150 ×10³   / \pi /4 × 0.020²

ultimate stress = 477 MPa

In 3rd part we calculate average stress at breaking point by given formula

average stress = load fail / area

average stress = 70 ×10³  / \pi /4 × d²

average stress = 70 ×10³  / \pi /4 × 0.010²

average stress = 892 MPa