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:
(i) 12 V in series with 18 Ω.
(ii) 0.4 A; 1.92 W
(iii) 1,152 J
(iv) 18Ω — maximum power transfer theorem
Explanation:
<h3>(i)</h3>
As seen by the load, the equivalent source impedance is ...
10 Ω + (24 Ω || 12 Ω) = (10 +(24·12)/(24+12)) Ω = 18 Ω
The open-circuit voltage seen by the load is ...
(36 V)(12/(24 +12)) = 12 V
The Thevenin's equivalent source seen by the load is 12 V in series with 18 Ω.
__
<h3>(ii)</h3>
The load current is ...
(12 V)/(18 Ω +12 Ω) = 12/30 A = 0.4 A . . . . load current
The load power is ...
P = I^2·R = (0.4 A)^2·(12 Ω) = 1.92 W . . . . load power
__
<h3>(iii)</h3>
10 minutes is 600 seconds. At the rate of 1.92 J/s, the electrical energy delivered is ...
(600 s)(1.92 J/s) = 1,152 J
__
<h3>(iv)</h3>
The load resistance that will draw maximum power is equal to the source resistance: 18 Ω. This is the conclusion of the Maximum Power Transfer theorem.
The power transferred to 18 Ω is ...
((12 V)/(18 Ω +18 Ω))^2·(18 Ω) = 144/72 W = 2 W
Answer:
t = 1.06 sec
Explanation:
Once disconnected from the battery, the capacitor discharges through the internal resistance of the dielectric, which can be expressed as follows:
R = (1/σ)*d/A, where d is is the separation between plates, and A is the area of one of the plates.
The capacitance C , for a parallel plates capacitor filled with a dielectric of a relative permittivity ε, can be expressed in this way:
C = ε₀*ε*A/d = 8.85*10⁻¹² *12*A/d
The voltage in the capacitor (which is proportional to the residual charge as it discharges through the resistance of the dielectric) follows an exponential decay, as follows:
V = V₀*e(-t/RC)
The product RC (which is called the time constant of the circuit) can be calculated as follows:
R*C = (1/10⁻¹⁰)*d/A*8.85*10⁻¹² *12*A/d
Simplifying common terms, we finally have:
R*C = 8.85*10⁻¹² *12 / (1/10⁻¹⁰) sec = 1.06 sec
If we want to know the time at which the voltage will decay to 3.67 V, we can write the following expression:
V= V₀*e(-t/RC) ⇒ e(-t/RC) = 3.67/10 ⇒ -t/RC = ln(3.67/10)= -1
⇒ t = RC = 1.06 sec.
Answer:
#Initialise a tuple
team_names = ('Rockets','Raptors','Warriors','Celtics')
print(team_names[0])
print(team_names[1])
print(team_names[2])
print(team_names[3])
Explanation:
The Python code illustrates or printed out the tuple team names at the end of a season.
The code displayed is a function that will display these teams as an output from the program.
Answer:
battery life in year = 9 years and 48 days
Explanation:
given data
Battery Ampere-hours = 1.5
Pulse voltage = 2 V
Pulse width = 1.5 m sec
Pulse time period = 1 sec
Electrode heart resistance = 150 Ω
Current drain on the battery = 1.25 µA
to find out
battery life in years
solution
we get first here duty cycle that is express as
duty cycle =
...............1
duty cycle = 1.5 × 
and applied voltage will be
applied voltage = duty energy × voltage ...........2
applied voltage = 1.5 ×
× 2
applied voltage = 3 mV
so current will be
current =
................3
current = 
current = 20 µA
so net current will be
net current = 20 - 1.25
net current = 18.75 µA
so battery life will be
battery life = 
battery life = 80000 hours
battery life in year = 
battery life in year = 9.13 years
battery life in year = 9 years and 48 days