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SSSSS [86.1K]
3 years ago
5

Later in this course we will learn how to combine flip-flops to make transitions to desired outputs. The flip-flop is holding bi

t of information, waiting for a signal to change (memory). But in this example YOU are still making the transition by flipping a switch. How can we make the flip-flop change without us providing the input directly
Engineering
1 answer:
expeople1 [14]3 years ago
8 0

Answer:

Clock input( or Timer Oscillator)   in used to change the state of the output.  

Explanation:

Flip-flop is an device that is the state changes with the rising or falling of the clock. It has a control signal (clock) input that stores the input state and outputs the state only in response to the clock signal.

The output changes with the change in the input in synchronization with the clock.  

Whenever the input of the clock changed, it will result in change in the output.

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A metal rod is 0.600 m in length at a temperature of 15.0∘C. When you raise its temperature to 37.0∘C, its length increases by 0
Katyanochek1 [597]

Answer:

The coefficient of linear expansion of the metal is ∝ = 2.91 x 10⁻⁵  °C⁻¹.

Explanation:

We know that Linear thermal expansion is represented by the following equation

Δ L = L x ∝ x Δ T ---- (1)

where Δ L is the change in length, L is for length, ∝ is the coefficient of linear expression and  Δ T is the change in temperature.

Given that:

L = 0.6 m

T₁ = 15° C

T₂ = 37° C

Δ L = 0.28 mm

∝ = ?

Solution:

We know that Δ T = T₂ ₋ T₁

Putting the values of T₁  and T₂ in above equation, we get

Δ T = 37 - 15

Δ T =  22 °C

Also Δ L = 0.28 mm

Converting the mm to m

Δ L = 0.00028 m

Putting the values of Δ T, Δ L, L in equation 1, we get

0.00028 = 0.6 x ∝ x 22

Rearranging the equation, we get

∝ = 0.00028 / (0.6 x 16)

∝ = 0.00028 / 13.2

∝ = 2.12 x 10⁻⁵  °C⁻¹

4 0
4 years ago
How to find Catenary length in sag
Gnom [1K]

Answer:

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6 0
2 years ago
What position would be the lowest paid in a technology company?
murzikaleks [220]
I think the answers is b
3 0
3 years ago
The boost converter of Fig. 6-8 has parameter Vs 20 V, D 0.6, R 12.5 , L 10 H, C 40 F, and the switching frequency is 200 kHz. (
mr Goodwill [35]

Answer:

a) the output voltage is 50 V

b)

- the average inductor current is 10 A

- the maximum inductor current is 13 A

- the maximum inductor current is 7 A

c) the output voltage ripple is 0.006 or 0.6%V₀

d) the average current in the diode under ideal components is 4 A

Explanation:

Given the data in the question;

a) the output voltage

V₀ = V_s/( 1 - D )

given that; V_s = 20 V, D = 0.6

we substitute

V₀ = 20 / ( 1 - 0.6 )

V₀ = 20 / 0.4

V₀ = 50 V

Therefore, the output voltage is 50 V

b)

- the average inductor current

I_L = V_s / ( 1 - D )²R

given that R = 12.5 Ω, V_s = 20 V, D = 0.6

we substitute

I_L = 20 / (( 1 - 0.6 )² × 12.5)

I_L = 20 / (( 0.4)² × 12.5)

I_L = 20 / ( 0.16 × 12.5 )

I_L = 20 / 2

I_L = 10 A

Therefore, the average inductor current is 10 A

- the maximum inductor current

I_{Lmax = [V_s / ( 1 - D )²R] + [ V

given that, R = 12.5 Ω, V_s = 20 V, D = 0.6, L = 10 μH, T = 1/200 kHz = 5 hz

we substitute

I_{Lmax = [20 / (( 1 - 0.6 )² × 12.5)] + [ (20 × 0.6 × 5) / (2 × 10) ]

I_{Lmax = [20 / 2 ] + [ 60 / 20 ]    

I_{Lmax = 10 + 3

I_{Lmax = 13 A

Therefore, the maximum inductor current is 13 A

- The minimum inductor current

I_{Lmax = [V_s / ( 1 - D )²R] - [ V

given that, R = 12.5 Ω, V_s = 20 V, D = 0.6, L = 10 μH, T = 1/200 kHz = 5 hz

we substitute

I_{Lmin = [20 / (( 1 - 0.6 )² × 12.5)] - [ (20 × 0.6 × 5) / (2 × 10) ]

I_{Lmin = [20 / 2 ] -[ 60 / 20 ]    

I_{Lmin = 10 - 3

I_{Lmin  = 7 A

Therefore, the maximum inductor current is 7 A

 

c)  the output voltage ripple

ΔV₀/V₀ = D/RCf

given that; R = 12.5 Ω, C = 40 μF = 40 × 10⁻⁶ F, D = 0.6, f = 200 Khz = 2 × 10⁵ Hz

we substitute

ΔV₀/V₀ = 0.6 / (12.5 × (40 × 10⁻⁶) × (2 × 10⁵) )

ΔV₀/V₀ = 0.6 / 100

ΔV₀/V₀ = 0.006 or 0.6%V₀

Therefore, the output voltage ripple is 0.006 or 0.6%V₀

d) the average current in the diode under ideal components;

under ideal components; diode current = output current

hence the diode current will be;

I_D = V₀/R

as V₀ = 50 V and R = 12.5 Ω

we substitute

I_D = 50 / 12.5

I_D = 4 A

Therefore, the average current in the diode under ideal components is 4 A

7 0
3 years ago
A dead-man system should shut off fuel flow within ____ of the system’s maximum flow rate.
AysviL [449]

Answer:

I think it is 10 percent because

3 0
4 years ago
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