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otez555 [7]
2 years ago
15

Write a Nios II assembly program that reads binary data from the Slider Switches, SW11-0, on the DE2-115 Simulator, and display

the decimal equivalent on the seven-segment displays HEX3-0.​
Engineering
1 answer:
Citrus2011 [14]2 years ago
6 0

Algorithm of the Nios II assembly program.

  • Attain data for simulation from the  SW11-0, on the DE2-115 Simulator
  • The data will be  read from the switches in loop.
  • The decimal output is displayed using the seven-segment displays and done using the loop.
  • The program is ended by the user operating the SW1 switch

and

The decimal equivalent on the seven-segment displays HEX3-0 is

  • DE2-115
  • DE2-115_SW11
  • DE2-115_HEX3
  • DE2-115_HEX4
  • DE2-115_HEX5
  • DE2-115_HEX6
  • DE2-115_HEX7

<h3>The  Algorithm and decimal equivalent on the seven-segment displays HEX3-0</h3>

Generally,  the program will be written using a  cpulator simulator in order to attain best result.

We are to

  • Attain data for simulation from the  SW11-0, on the DE2-115 Simulator
  • The data will be  read from the switches in loop.
  • The decimal output is displayed using the seven-segment displays and done using the loop.
  • The program is ended by the user operating the SW1 switch

This will be the Algorithm of the Nios II assembly program .

Hence, the decimal equivalent on the seven-segment displays HEX3-0 is

  • DE2-115
  • DE2-115_SW11
  • DE2-115_HEX3
  • DE2-115_HEX4
  • DE2-115_HEX5
  • DE2-115_HEX6
  • DE2-115_HEX7

For more information on Algorithm

brainly.com/question/11623795

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A freezer is maintained at 20°F by removing heat from it at a rate of 75 Btu/min. The power input to the freezer is 0.70 hp, and
Igoryamba

Answer:

Explanation:

Cop of reversible refrigerator = TL / ( TH - TL)

TL = low temperature of freezer = 20 °F

TH = temperature of air around = 75 °F

Heat removal rate QL = 75 Btu/min

W actual, power input = 0.7 hp

conversion on F to kelvin = (T (°F) + 460 ) × 5 / 9

COP ( coefficient of performance) reversible = (20 + 460) × 5/9 / (5/9 ( ( 75 +460) - (20 + 460) ))

COP reversible = 480 / 55 = 8.73

irreversibility expression, I = W actual - W rev

COP r = QL / Wrev

W rev = QL /  COP r  where 75 Btu/min = 1.76856651 hp  where W actual = 0.70 hp

a) W rev =  1.76856651 hp  /  8.73  = 0.20258 hp is reversible power

I = W actual - W rev

b) I = 0.7 hp - 0.20258 hp = 0.4974 hp

c) the second-law efficiency of this freezer = W rev / W actual =  0.20258 hp / 0.7 hp = 0.2894 × 100 = 28.94 %

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Which option identifies the next step Kristin should take to test the design in the following scenario?
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If an airplane travels at a speed of 1120 km/hr at an altitude of 15 km, what is the required speed at an altitude of 8 km to sa
jek_recluse [69]

Answer:

1170 km/hr

Explanation:

Let's first state the formula to be used

c = √(KRT)

The temperature at an altitude of 15km is -56.5° C

Let's not convert this to °K, we have

-56.5° + 273.15 = 216.65° K

Also, the temperature at 8km is -36.9° C, on converting to °K we have

-36.9° + 273.15 = 236.25° K

Then again, we look for the speed at both 15 km and 8 km both of which are 295 m/s and 308 km

Finally, we use the mach similarity formula

(V/c) of 15km = (V/c) of 8km

V of 8km = c of 8km * (V/c) of 15km

V of 8km = 1170 km/hr

6 0
3 years ago
A steel wire is suspended vertically from its upper end. The wire is 400 ft long and has a diameter of 3/16 in. The unit weight
jek_recluse [69]

Answer:

a) the maximum tensile stress due to the weight of the wire is 1361.23 psi

b) the maximum load P that could be supported at the lower end of the wire is 624.83 lb

Explanation:

Given the data in the question;

Length of wire L = 400 ft = ( 400 × 12 )in = 4800 in

Diameter d = 3/16 in

Unit weight w = 490 pcf

First we determine the area of the wire;

A = π/4 × d²

we substitute

A = π/4 × (3/16)²

A = 0.0276 in²

Next we get the Volume

V = Area × Length of wire

we substitute

V = 0.0276 × 4800

V = 132.48 in³

Weight of the steel wire will be;

W = Unit weight × Volume

we substitute

W = 490 × ( 132.48 / 12³ )

W = 490 × 0.076666

W = 37.57 lb

a) the maximum tensile stress due to the weight of the wire;

σ_w = W / A

we substitute

σ_w = 37.57 / 0.0276

= 1361.23 psi

Therefore, the maximum tensile stress due to the weight of the wire is 1361.23 psi

b) the maximum load P that could be supported at the lower end of the wire. Allowable tensile stress is 24,000 psi

Maximum load P that the wire can safely support its lower end will be;

P = ( σ_{all - σ_w )A

we substitute

P = ( 24000 - 1361.23  )0.0276

P = 22638.77 × 0.0276

P = 624.83 lb

Therefore, the maximum load P that could be supported at the lower end of the wire is 624.83 lb

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