Answer:
my_mul:
.globl my_mul
my_mul:
//Multiply X0 and X1
// Does not handle negative X1!
// Note : This is an in efficient way to multipy!
SUB SP, SP, 16 //make room for X19 on the stack
STUR X19, [SP, 0] //push X19
ADD X19, X1, XZR //set X19 equal to X1
ADD X9 , XZR , XZR //set X9 to 0
mult_loop:
CBZ X19, mult_eol
ADD X9, X9, X0
SUB X19, X19, 1
B mult_loop
mult_eol:
LDUR X19, [SP, 0]
ADD X0, X9, XZR // Move X9 to X0 to return
ADD SP, SP, 16 // reset the stack
BR X30
Explanation:
Answer:
0.5°c
Explanation:
Humidity ratio by mass can be expressed as
the ratio between the actual mass of water vapor present in moist air - to the mass of the dry air
Humidity ratio is normally expressed in kilograms (or pounds) of water vapor per kilogram (or pound) of dry air.
Humidity ratio expressed by mass:
x = mw / ma (1)
where
x = humidity ratio (kgwater/kgdry_air, lbwater/lbdry_air)
mw = mass of water vapor (kg, lb)
ma = mass of dry air (kg, lb)
It can be as:
x = 0.005 (100) / [(100 - 100)]
x = 0.005 x 100 / (100 - 100)
x = 0.005 x 100 / 0
x = 0.5°c
So the temperature to which atmospheric air must be cooled in order to have humidity ratio of 0.005 lb/lb is 0.5°c
Answer:
The required pumping head is 1344.55 m and the pumping power is 236.96 kW
Explanation:
The energy equation is equal to:
For the pipe 1, the flow velocity is:
Q = 18 L/s = 0.018 m³/s
D = 6 cm = 0.06 m
The Reynold´s number is:
Using the graph of Moody, I will select the f value at 0.0043 and 335339.4, as 0.02941
The head of pipe 1 is:
For the pipe 2, the flow velocity is:
The Reynold´s number is:
The head of pipe 1 is:
The total head is:
hi = 1326.18 + 21.3 = 1347.48 m
The required pump head is:
The required pumping power is:
The answer is number 2) Increase the resistance of the concrete to freeze-thaw damage.
Answer:
V = 0.5 m/s
Explanation:
given data:
width of channel = 4 m
depth of channel = 2 m
mass flow rate = 4000 kg/s = 4 m3/s
we know that mass flow rate is given as
Putting all the value to get the velocity of the flow
V = 0.5 m/s
