Answer: To detect and correct errors, additional bits are added to the data bits at the time of transmission. The additional bits are called parity bits. They allow detection or correction of errors. The data bits along with the parity bits form a code word.
Explanation:
The pressure drop of air in the bed is 14.5 kPa.
<u>Explanation:</u>
To calculate Re:
From the tables air property
Ideal gas law is used to calculate the density:
ρ = 
ρ = 1.97 Kg / 
ρ = 
R = 
= 8.2 × 
/ 28.97×
R = 2.83 × atm / K Kg
q is expressed in the unit m/s
q = 1.24 m/s
Re = 
Re = 2278
The Ergun equation is used when Re > 10,
= 4089.748 Pa/m
ΔP = 4089.748 × 3.66
ΔP = 14.5 kPa
Answer and Explanation:
clear all; close all;
N=512;
t=(1:N)/N;
fs=1000;
f=(1:N)*fs/N;
x= sin(2*pi*200*t) + sin(2*pi*400*t);
y= sin(2*pi*200*t) + sin(2*pi*900*t);
for n = 1:20
a(n) = (2/N)*sum(x.*(cos(2*pi*n*t)))
b(n) = (2/N)*sum(x.*(sin(2*pi*n*t)))
c(n) = sqrt(a(n).^2+b(n).^2)
theta(n) =-(360/(2*pi))*atan(b(n)./a(n));
end
plot(f(1:20),c(1:20),'rd');
disp([a(1:4),b(1:4),c(1:4),theta(1:4)])
Answer: Option D) 298 g/mol is the correct answer
Explanation:
Given that;
Mass of sample m = 13.7 g
pressure P = 2.01 atm
Volume V = 0.750 L
Temperature T = 399 K
Now taking a look at the ideal gas equation
PV = nRT
we solve for n
n = PV/RT
now we substitute
n = (2.01 atm x 0.750 L) / (0.0821 L-atm/mol-K x 399 K
)
= 1.5075 / 32.7579
= 0.04601 mol
we know that
molar mass of the compound = mass / moles
so
Molar Mass = 13.7 g / 0.04601 mol
= 297.7 g/mol ≈ 298 g/mol
Therefore Option D) 298 g/mol is the correct answer
Answer:
K
Explanation:
For installations where the nonlinear load is huge, most consulting engineers will specify K-rated transformers.
