Answer:
a. 10bits
b. 7 bits
c. 6 bits
Explanation:
a. for 0 to 512
# of numbers = 512 - 0 + 1 =513
[log ₂513] = 9 bits
we actually need 10 bits
b.  for 0 to 75
# of numbers = 75 - 0 + 1 =76
[log ₂76] = 6 bits
we actually need 7 bits
c. for -20 to 13
# of numbers = 13 - (-20) + 1 =34
[log ₂34] = 5 bits
we actually need 6 bits
 
        
             
        
        
        
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
 
        
             
        
        
        
Answer:
0.0432 m^3/s
Explanation:
Internal diameter of smaller pipes = 2.5 cm = 0.025 m 
pipa wall thickness = 3 mm = 0.003 m
internal diameter of larger pipes = 8 cm = 0.8 m
velocity of region between smaller and larger pipe = 10 m/s 
Calculate discharge in m^3/s
First we calculate the area of the smaller pipe
A =  =
 =  = 0.00023571 m^2
  = 0.00023571 m^2
next we calculate area of fluid between the smaller pipes and larger pipe
A = ![[\frac{\pi }{4} D^{2} _{L}  ] - 3(A_{s})](https://tex.z-dn.net/?f=%5B%5Cfrac%7B%5Cpi%20%7D%7B4%7D%20D%5E%7B2%7D%20_%7BL%7D%20%20%5D%20-%203%28A_%7Bs%7D%29)
    = ![[ \frac{\pi }{4} (0.08 )^2 - 3 ( 0.00023571 )]](https://tex.z-dn.net/?f=%5B%20%5Cfrac%7B%5Cpi%20%7D%7B4%7D%20%280.08%20%29%5E2%20-%203%20%28%200.00023571%20%29%5D)
    = [ 0.00502857 - 0.00070713 ]
    = 0.00432144 m^2
hence the discharge in m^3/s
Q = AV 
    = 0.00432144 * 10 
    = 0.0432 m^3/s
 
        
             
        
        
        
Answer:
N = 278.5 rpm 
F = 348.15 mm/min 
machine time = 10.34 seconds
material removal rate = 33420000 mm³/min 
Explanation:
given data 
length L = 400 mm
width W = 60 mm 
diameter D = 80 mm
no of cutting n = 5 
velocity V  = 70 m/min  = 70000 mm/min 
chip load P = 0.25 mm/tooth 
depth = 5 mm 
solution
we know velocity that is express as
velocity V =  D N    ........1
 D N    ........1
N =  
   
N = 278.5 rpm 
and 
now we get feed rate in milling operation is 
F  n P N 
F = 5 × 0.25 × 278.5 
F = 348.15 mm/min 
and
now we get actual machine time to make 1 pass across surface of worl is 
machine time =  
   
machine time =  
  
machine time = 0.1723 min = 10.34 seconds
and 
max material removal rate during these cut is 
material removal rate = A × d × N 
material removal rate = 400 × 60 × 5  × 278.5 
material removal rate = 33420000 mm³/min