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:
Explanation:
Given
Temperature of solid
Einstein Temperature
Heat Capacity in the Einstein model is given by
Substitute the values
Answer:
hello some part of your question is missing below is the complete question
answer :
A) 162750 Ib.ft
B) - 64950 Ib.ft
Explanation:
Applying Muller-Breslau's law
we will make assumptions which include assuming an imaginary hinge at G
therefore the height of I.LD for B.M at G = ( 12 * 8 ) / 20 = 4.8
height of I.L.D at C = 2.4 ( calculated )
height of I.L.D at F = 1.5 ( calculated )
A) Determine Maximum positive moment produced at G
= [ (1/2 * 20 * 4.8 ) ( 600 + 300 ) ] + [ ( 25 * 4.8 * 10^3 ) ] - [ ( 1/2 *2.4*20 ) * 300 ] + [ (1/2 * 1.5 * 10 ) ( 600 + 300 ) ]
= 162750 Ib.ft
B) Determine the maximum negative moment produced at G
= [ ( 1/2 * 20 * 4.8 ) * 300 ] - [ ( 1/2 * 2.4 * 20 ) ( 600 + 300 ) ] - [ (2.5 * 10^3 * 2.4 ) ] + [ ( 1/2 * 1.5 * 10) * 300 ]
= - 64950 Ib.ft