Answer:
your question in not complete.
you need to the high too.
Explanation:
The given value of P is as follows.
P = 1.06F + 22.18, 
or, P' = 1.06
As p' is defined and non-zero. Hence, only critical points are boundary points.
For F = 10, the value of P will be calculated as follows.
P = 
= 32.78
For F = 70, the value of P will be calculated as follows.
P = 
= 96.38
Therefore, the minimum value of P is 32.78 and maximum value of P is 96.38.
Answer:
The maximum permissible propagation delay per flip flop stage is<u> 100 </u>n sec
Explanation:
1024 ripple counter has 10 J-K flip flops(210 = 1024).
So the total delay will be 10×x where x is the delay of each J-K flip flops.
The period of the clock pulse is 1× 10⁻⁶ s.
Now
10x <= 10⁻⁶ s
x <= 100 ns
x= 100 ns for prpoer operation.
pulse train with a frequency of 1 MHz is counted using a modulo-1024 ripple-counter built with J-K flip flops. For proper operation of the counter, the maximum permissible propagation delay per flip flop stage is <u>100 </u>n sec.
a)You throw a stone horizontally at a speed of 5.0 m/s from the top of a cliff that is 78.4 m high.
from above statement we got
height = 78.4 m
since the ball is thrown, so its vertical velocity would be zero
u = 0
taking g = 9.8m/s^2
now, using the equation of motion
h = ut + gt^2/2
now putting all the values in it
we got ,
78.4 = 9.8 * t^2/ 2
by solving we got,
t = 4 sec
b) now, since along the horizontal , no force acting and accelaration is zero so
R = ut , R is RANGE
R = 5 * 4
range = 20 m
c) vertical components of the stone’s velocity just before it hits the ground = v sin θ =
horizontal components of the stone’s velocity just before it hits the ground = v cos θ
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