Answer:
32582657
Explanation:
If any number is in the form of 2^{n}-1 is known as Mersenne prime. Here, n is a prime number.
For example:
If n is 3 then the corresponding binary number is as follows:
P=2^{3}-1
P=7
Here, the binary representation of P is (111)₂.
The number of binary digits is 3 which are equal to n.
Consider the given expression
P=2^32582657-1
This is also in the form 2^{n}-1
Here the value of n is 32582657.
Hence, the number of binary digits for the given prime number is 32582657
Answer:
Considering the guidelines of this exercise.
The pieces produced per month are 504 000
The productivity ratio is 75%
Explanation:
To understand this answer we need to analyze the problem. First of all, we can only produce 2 batches of production by the press because we require 3 hours to set it up. So if we rest those 6 hours from the 8 of the shift we get 6, leaving 2 for an incomplete bath. So multiplying 2 batches per day of production by press we obtain 40 batches per day. So, considering we work in this factory for 21 days per month well that makes 40 x 21 making 840 then we multiply the batches for the pieces 840 x 600 obtaining 504000 pieces produced per month. To obtain the productivity ratio we need to divide the standard labor hours meaning 6 by the amount of time worked meaning 8. Obtaining 75% efficiency.
Answer:
DIAMETER = 9.797 m
POWER = 
Explanation:
Given data:
circular windmill diamter D1 = 8m
v1 = 12 m/s
wind speed = 8 m/s
we know that specific volume is given as
where v is specific volume of air
considering air pressure is 100 kPa and temperature 20 degree celcius
v = 0.8409 m^3/ kg
from continuity equation
mass flow rate is given as
the power produced ![\dot W = \dot m \frac{ V_1^2 - V_2^2}{2} = 717.3009 [\frac{12^2 - 8^2}{2} \times \frac{1 kJ/kg}{1000 m^2/s^2}]](https://tex.z-dn.net/?f=%5Cdot%20W%20%3D%20%5Cdot%20m%20%5Cfrac%7B%20V_1%5E2%20-%20V_2%5E2%7D%7B2%7D%20%3D%20717.3009%20%5B%5Cfrac%7B12%5E2%20-%208%5E2%7D%7B2%7D%20%5Ctimes%20%5Cfrac%7B1%20kJ%2Fkg%7D%7B1000%20m%5E2%2Fs%5E2%7D%5D)
Answer:
Explanation:
= Area of section 1 = 
= Velocity of water at section 1 = 100 ft/min
= Specific volume at section 1 = 
= Density of fluid = 
= Area of section 2 = 
Mass flow rate is given by
The mass flow rate through the pipe is
As the mass flowing through the pipe is conserved we know that the mass flow rate at section 2 will be the same as section 1
The speed at section 2 is .
