Answer: N has to be lesser than or equal to 1666.
Explanation:
Cost of parts N in FPGA = $15N
Cost of parts N in gate array = $3N + $20000
Cost of parts N in standard cell = $1N + $100000
So,
15N < 3N + 20000 lets say this is equation 1
(cost of FPGA lesser than that of gate array)
Also. 15N < 1N + 100000 lets say this is equation 2
(cost of FPGA lesser than that of standardcell)
Now
From equation 1
12N < 20000
N < 1666.67
From equation 2
14N < 100000
N < 7142.85
AT the same time, Both conditions must hold true
So N <= 1666 (Since N has to be an integer)
N has to be lesser than or equal to 1666.
Answer:
Explanation:
var generator = new Random(1);
// Now the nextGaussian() function returns a normal distribution of random numbers with the following parameters: a mean of zero and a standard deviation of one
var draw = function() {
var num = generator.nextGaussian();
var standardDeviation = 60;
var mean = 2003;
// Multiply by the standard deviation and add the mean.
var x = standardDeviation * num + mean;
noStroke();
fill(214, 159, 214, 10);
ellipse(x, 200, 16, 16); };
Hope this will be helpful
Answer:
Class of fit:
Interference (Medium Drive Force Fits constitute a special type of Interference Fits and these are the tightest fits where accuracy is important).
Here minimum shaft diameter will be greater than the maximum hole diameter.
Medium Drive Force Fits are FN 2 Fits.
As per standard ANSI B4.1 :
Desired Tolerance: FN 2
Tolerance TZone: H7S6
Max Shaft Diameter: 3.0029
Min Shaft Diameter: 3.0022
Max Hole Diameter:3.0012
Min Hole Diameter: 3.0000
Max Interference: 0.0029
Min Interference: 0.0010
Stress in the shaft and sleeve can be considered as the compressive stress which can be determined using load/interference area.
Design is acceptable If compressive stress induced due to selected dimensions and load is less than compressive strength of the material.
Explanation:
Answer:
a. V = 109.64 × 10⁵ ft/min
b. Mw = 654519.54 kg/hr
Explanation:
Given Parameters
mass flow rate of water, Mw = 90000g/min = 6607.33 kg/s
inlet temperature of water, T1 = 84 F = 28.89 C
outlet temperature of water, T2 = 68 F = 20 C
specific heat capacity of water, c = 4.18kJ/kgK
rate of heat remover from water, Qw is given by
Qw = 6607.33[28.89 - 20] * 4.18
Qw = 245529.545kw
For air, inlet condition
DBT = 70 F hi = 43.43 kJ/kg
WBT = 60 F wi = 0.00874 kJ/kg
u1 = 0.8445 m/kg
oulet condition,
DBT = 70 F RH = 100.1
h1 = 83.504kJ/kg
Wo = 0.222kJ/kg
check the attached file for complete solution
Answer:
T=151 K, U=-1.848*10^6J
Explanation:
The given process occurs when the pressure is constant. Given gas follows the Ideal Gas Law:
pV=nRT
For the given scenario, we operate with the amount of the gas- n- calculated in moles. To find n, we use molar mass: M=102 g/mol.
Using the given mass m, molar mass M, we can get the following equation:
pV=mRT/M
To calculate change in the internal energy, we need to know initial and final temperatures. We can calculate both temperatures as:
T=pVM/(Rm); so initial T=302.61K and final T=151.289K
Now we can calculate change of U:
U=3/2 mRT/M using T- difference in temperatures
U=-1.848*10^6 J
Note, that the energy was taken away from the system.
