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
Answer:
Mechanical average of a wheel = 3
Explanation:
Given:
Radius of wheel = 1.5 ft = 1.5 x 12 = 18 inches
Radius of axle = 6 inches
Find:
Mechanical average of a wheel
Computation:
Mechanical average of a wheel = Radius of wheel / Radius of axle
Mechanical average of a wheel = 18 / 6
Mechanical average of a wheel = 3
Answer:
The coefficient of thermal expansion tells us how much a material can expand due to heat.
Explanation:
Thermal expansion occurs when a material is subjected to heat and changes it's shape, area and volume as a result of that heat. How much that material changes is dependent on it's coefficient of thermal expansion.
Different materials have different coefficients of thermal expansion (i.e. It is a material property and differs from one material to the next). It is important to understand how materials behave when heated, especially for engineering applications when a change in dimension might pose a problem or risk (eg. building large structures).
Answer:
dislocations play an important role in controlling as
Explanation:
As dislocations plays an important role in the ductility, elasticity and plurality of materials
- The elastic and elastic deflections play a large role in their properties as the metallic materials, because the dislocation of a glass material does not play a major role in their properties.
