Answer:
Explanation:
1. With the operands R0, R1, the program would compute AND operation and ADD operation .
2. The operands could truly be signed 2's complement encoded (i.e Yes) .
3. The overflow truly occurs when two numbers that are unsigned were added and the result is larger than the capacity of the register, in that situation, overflow would occur and it could corrupt the data.
When the result of an operation is smaller in magnitude than the smallest value represented by the data type, then arithmetic underflow will occur.
Answer:
d
Explanation:
is the because that's the amount of work in making machine can do producing heat
Answer:
26.7 min
Explanation:
First, we will find the <u>time required to drill each hole</u>:
- N = 300 x 12/0.75
= 1527.7 rev/min
- fr = 1527.7 (0.015) = 22.916 in/min
Formula for <u>distance per hole</u>: 0.5 + A + 1.75
- A = 0.5 (0.75) tan (90-100 / 2) = 0.315 in
- Tm = (0.5 + 0.315 + 1.75) / 22.916 = 0.112 min
Now, we will calculate the <u>time required to draw back the drill form hole</u>:
= 0.112 / 2 = 0.056 min
Time to move between holes = 1.5 / 15 = 0.1 min
For 100 holes, the number of moves between holes = 99
Total time required to drill 100 holes (t):
t = 100 (0.112 + 0.056) + 99 (0.1) = 26.7 min
Answer:
The exit temperature is 293.74 K.
Explanation:
Given that
At inlet condition(1)
P =80 KPa
V=150 m/s
T=10 C
Exit area is 5 times the inlet area
Now

If consider that density of air is not changing from inlet to exit then by using continuity equation

So 
m/s
Now from first law for open system

Here Q=0 and w=0

When air is treating as ideal gas

Noe by putting the values



So the exit temperature is 293.74 K.