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:
a) Ef = 0.755
b) length of specimen( Lf )= 72.26mm
diameter at fracture = 9.598 mm
c) max load ( Fmax ) = 52223.24 N
d) Ft = 51874.67 N
Explanation:
a) Determine the true strain at maximum load and true strain at fracture
True strain at maximum load
Df = 9.598 mm
True strain at fracture
Ef = 0.755
b) determine the length of specimen at maximum load and diameter at fracture
Length of specimen at max load
Lf = 72.26 mm
Diameter at fracture
= 9.598 mm
c) Determine max load force
Fmax = 52223.24 N
d) Determine Load ( F ) on the specimen when a true strain et = 0.25 is applied during tension test
F = 51874.67 N
attached below is a detailed solution of the question above
Answer:
a. Rotational speed of the drill = 375.96 rev/min
b. Feed rate = 75 mm/min
c. Approach allowance = 3.815 mm
d. Cutting time = 0.67 minutes
e. Metal removal rate after the drill bit reaches full diameter. = 9525 mm³/min
Explanation:
Here we have
a. N = v/(πD) = 15/(0.0127·π) = 375.96 rev/min
b. Feed rate = fr = Nf = 375.96 × 0.2 = 75 mm/min
c. Approach allowance = tan 118/2 = (12.7/2)/tan 118/2 = 3.815 mm
d. Approach allowance T∞ =L/fr = 50/75 = 0.67 minutes
e. R = 0.25πD²fr = 9525 mm³/min.
Answer:
Mechanical dissection
Explanation:
Taking apart old things and innovating to improve upon them is mechanical dissection.
Volume=Hh(b1+b2)/2. b1 and b2 are the base of the trapezoid
