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
P = 210 kPa (gage)
Pabs = 210 + 101 = 311 KPa
Before the temperature rise
P1 V1 = m1 R1 T1
m1 =
After the temperature rise
P2 =
after bleeding the pressure and the volume returns to its first value
P1 = P2 and V1 = V2
then
m2 =
m2 =
mbleed = m1 - m2 --> mbleed = 0.91 - 0.84 = 0.007 Kg
P2 = 337.43 KPa
mbleed = 0.007 Kg
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.
Buffering with operational amplifiers (op amps) is a crucial application in electronics because it helps to overcome impedance-matching problems that cannot be easily resolved with only resistive circuits. Impedance-matching issues can cause signal distortion, and buffering solves this problem by creating a high-input impedance and low-output impedance circuit that separates the input and output signals, preventing the signal from being affected by the load impedance.
By using op amp buffering, the signal can be efficiently transmitted and received without signal loss or distortion, making it a useful technique in many applications, such as audio amplification and signal conditioning.
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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
delay issues
tolerance issues
temperature issues
loading effects
impedance-matching problems
"
You can learn more about Buffering at
#SPJ11
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.
For example in a car wash management system, where green and red lights are used to indicate if a car is being washed, green light will be on and then red light means that there no car in washing que
Fluorescent lamps
Mercury-containing lamps
All of the above
Answer: D all above
Explanation:
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Answer:
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and exit cross-sectional areas. The inlet cross-sectional area is
6 cm26cm
2. At the duct exit, the pressure of the air is 100 kPa and the velocity is 250 m/s. Neglecting potential energy
effects and modeling air as an ideal gas with constant cp=1.008 kJ/kg⋅Kc
p =1.008kJ/kg⋅K, determine
(a) the velocity of the air at the inlet, in m/s.
(b) the temperature of the air at the exit, in K.
(c) the exit cross-sectional area, in cm2
(a) the velocity of the air at the inlet, in m/s.
(b) the temperature of the air at the exit, in K.
(c) the exit cross-sectional area, in cm
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 / /4 × D²
tensile stress at yield = 80 ×10³ / /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 / /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 / /4 × d²
average stress = 70 ×10³ / /4 × 0.010²
average stress = 892 MPa