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Amanda [17]
3 years ago
12

A 5 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 10ns propagation delay. The output of the inverter goes to the Y input of the OR gate.
a. Draw the circuit
b. Plot the output of the clock for two cycles. Show times and voltages.
c. Plot the output of the inverter in the same plot. Show times and voltages.
d. Plot the output of the OR gate in the same plot. Show times and voltages.
Engineering
1 answer:
Reptile [31]3 years ago
4 0

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.

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For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of t
Strike441 [17]

Answer:

the elongation of the metal alloy is 21.998 mm

Explanation:

Given the data in the question;

K = σT/ (εT)ⁿ

given that metal alloy true stress σT = 345 Mpa, plastic true strain εT = 0.02,

strain-hardening exponent n = 0.22

we substitute

K = 345 / 0.02^{0.22

K = 815.8165 Mpa

next, we determine the true strain

(εT) = (σT/ K)^1/n

given that σT = 412 MPa

we substitute

(εT) = (412 / 815.8165 )^(1/0.22)

(εT) = 0.04481 mm

Now, we calculate the instantaneous length

l_i = l_0e^{ET

given that l_0 = 480 mm

we substitute

l_i =480mm × e^{0.04481

l_i =  501.998 mm

Now we find the elongation;

Elongation = l_i - l_0

we substitute

Elongation = 501.998 mm - 480 mm

Elongation = 21.998 mm

Therefore, the elongation of the metal alloy is 21.998 mm

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Answer:

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