If a clock frequency is applied to a cascaded counter, The lowest output frequency available will be
- The lowest output frequency will be =

<h3>
Cascade Counter</h3>
For a cascade counter,
Overall frequency = 
Overall frequency = 
<h3>Lowest F
requency</h3>
Therefore,
the lowest frequency

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Answer:
35.7 kg lid we put
Explanation:
given data
temperature = 105 celcius
diameter = 15 cm
Patm = 101 kPa
to find out
How heavy a lid should you put
solution
we know Psaturated from table for temperature is 105 celcius is
Psat = 120.8 kPa
so
area will be here
area =
..................1
here d is diameter
put the value in equation 1
area =
area = 0.01767 m²
so net force is
Fnet = ( Psat - Patm ) × area
Fnet = ( 120.8 - 101 ) × 0.01767
Fnet = 0.3498 KN = 350 N
we know
Fnet = mg
mass = 
mass = 
mass = 35.7 kg
so 35.7 kg lid we put
Answer:
1. True
2. False
Explanation:
given data
EAX contains = ff ff ff 51
doubleword referenced = ff ff ff f1
conditional jump add = eax
solution
1st statement is true
but 2nd statement is false
as here
- js or jne instruction is the conditional jump that is follow a test
- It jump to the specified location when previous instructions are set the SF (Sign Flag) .
Answer:
<em>Heat rejected to cold body = 3.81 kJ</em>
Explanation:
Temperature of hot thermal reservoir Th = 1600 K
Temperature of cold thermal reservoir Tc = 400 K
<em>efficiency of the Carnot's engine = 1 - </em>
<em> </em>
eff. of the Carnot's engine = 1 -
eff = 1 - 0.25 = 0.75
<em>efficiency of the heat engine = 70% of 0.75 = 0.525</em>
work done by heat engine = 2 kJ
<em>eff. of heat engine is gotten as = W/Q</em>
where W = work done by heat engine
Q = heat rejected by heat engine to lower temperature reservoir
from the equation, we can derive that
heat rejected Q = W/eff = 2/0.525 = <em>3.81 kJ</em>