Answer:
a) 199.04 ohms
b) attached in image
c) -0.696dB
Explanation:
We are given:
Fc = 8Khz = 8000hz
a)Using the formula:
R = 199.04 ohms
b) diagram is attached
c)
At F = 20KHz and Fc= 8KHz we have:
=0.923
|H(F)| in dB = 20log |H(F)|
=20log0.923
= -0.696dB
Answer:
a. 0.4544 N
b.
Explanation:
For computing the normality and molarity of the acid solution first we need to do the following calculations
The balanced reaction
= 0.27264 g
= 0.006816 mol
Now
Moles of needed is
= 0.003408 mol
= 0.333984 g
Now based on the above calculation
a. Normality of acid is
= 0.4544 N
b. And, the acid solution molarity is
= 0.00005112
=
We simply applied the above formulas
Answer & Explanation:
function Temprature
NYC=[33 33 18 29 40 55 19 22 32 37 58 54 51 52 45 41 45 39 36 45 33 18 19 19 28 34 44 21 23 30 39];
DEN=[39 48 61 39 14 37 43 38 46 39 55 46 46 39 54 45 52 52 62 45 62 40 25 57 60 57 20 32 50 48 28];
%AVERAGE CALCULATION AND ROUND TO NEAREST INT
avgNYC=round(mean(NYC));
avgDEN=round(mean(DEN));
fprintf('\nThe average temperature for the month of January in New York city is %g (F)',avgNYC);
fprintf('\nThe average temperature for the month of January in Denvar is %g (F)',avgDEN);
%part B
count=1;
NNYC=0;
NDEN=0;
while count<=length(NYC)
if NYC(count)>avgNYC
NNYC=NNYC+1;
end
if DEN(count)>avgDEN
NDEN=NDEN+1;
end
count=count+1;
end
fprintf('\nDuring %g days, the temprature in New York city was above the average',NNYC);
fprintf('\nDuring %g days, the temprature in Denvar was above the average',NDEN);
%part C
count=1;
highDen=0;
while count<=length(NYC)
if NYC(count)>DEN(count)
highDen=highDen+1;
end
count=count+1;
end
fprintf('\nDuring %g days, the temprature in Denver was higher than the temprature in New York city.\n',highDen);
end
%output
check the attachment for additional Information
Answer:
The question has some details missing : The 35-kg block A is released from rest. Determine the velocity of the 13-kgkg block BB in 4 ss . Express your answer to three significant figures and include the appropriate units. Enter positive value if the velocity is upward and negative value if the velocity is downward.
Explanation:
The detailed steps and appropriate calculation is as shown in the attached file.
