Answer:
a. true
Explanation:
Firstly, we need to understand what takes places during the compression process in a quasi-equilibrium process. A quasi-equilibrium process is a process in during which the system remains very close to a state of equilibrium at all times. When a compression process is quasi-equilibrium, the work done during the compression is returned to the surroundings during expansion, no exchange of heat, and then the system and the surroundings return to their initial states. Thus a reversible process.
While for a non-quasi equilibrium process, it takes more work to move the piston against this high-pressure region.
Answer:
a scale model each size is a certain amount smaller
Explanation:
Answer:
A centrifugal clutch works, as the name suggests, through centrifugal force.
One of the most common methods is by mounting the clutch onto the parallel or taper crank shaft of the engine.
When the crank shaft rotates the shaft of the clutch rotates at the same speed as the engine
Answer:
Timing Diagrams 15 pts. A 10 MHz clock that generates a 0 to 5V pulse train with a 30% duty cycle is connected to input X of a two input OR gate that has a 20nS propagation delay. The clock also goes to an inverter with a 10 ns propagation delay. The output of the inverter goes to the Y input of the OR gate. a) Draw the circuit. 2 pts. b) Plot the output of the clock for two cycles. Show times and voltages. 5 pts. c) On the same page as part (b) plot the output of the inverter. Show times and voltages. 3 pts. d) On the same page as parts (b & c) plot the output of the OR gate. Show times and voltages. 5 pts.
Answer:
V = 6.33 m/s
Explanation:
Given:
- The length of the wire L = 0.02 m
- The diameter of the wire D = 0.0005 m
- The calibration expression V = 0.0000625*h^2
- Environment temperature T_inf = 298 K
- Surface temperature T_s = 348 K
- The voltage drop dV = 5 V
- The electric current I = 0.1 A
Find:
- the velocity of Air
Solution:
- Calculate the surface area of the wire:
A = pi*D*L
A = pi*(0.0005)*(0.02) = 0.00003142 m^2
- The rate of energy in the wire P:
P = I*dV = 0.1*5 = 0.5 W
- Apply Newton's Law of Cooling:
P = h*A*(T_s - T_inf)
h = P /A*(T_s - T_inf)
Plug in the values:
h= 0.5/ 0.00003142*(348 - 298)
h = 318.27 W /m^2K
- Using the calibration relationship given, compute the velocity of air:
V = 6.25*10^-5 * h^2
V = 6.25*10^-5 * (318.27)^2
V = 6.33 m/s
