Answer:
The maximum data rate supported by this line is 39900 bps
Explanation:
The maximum data rate supported by this line can be obtained using the formula below
c = W*log2(S/N+1)
where;
c is the maximum data rate supported by the line
W is the bandwidth = 4kHz
S/N+1 is the signal to noise ratio = 1001
c = 4*log2(1001)
c = 39868.9 ≅ 39900 bps
Therefore, the maximum data rate supported by this line is 39900 bps
Answer:
the pressure at B is 527psf
Explanation:
Angular velocity, ω = v / r
ω = 20 /1.5
= 13.333 rad/s
Flow equation from point A to B
![P_A+rz_A-\frac{1}{2} Pr_A^2w^2=P_B+rz_B-\frac{1}{2} pr^2_Bw^2\\\\P_B = P_A + r(z_A-z_B)+\frac{1}{2} pw^2[(r_B^2)-(r_A)^2]\\\\P_B = [25 +(0.8+62.4)(0-1)+\frac{1}{2}(0.8\times1.94)\times(13.333)^2[2.5^2-1.5^2] ]\\\\P_B = 25 - 49.92+551.79\\\\P_B = 526.87psf\\\approx527psf](https://tex.z-dn.net/?f=P_A%2Brz_A-%5Cfrac%7B1%7D%7B2%7D%20Pr_A%5E2w%5E2%3DP_B%2Brz_B-%5Cfrac%7B1%7D%7B2%7D%20pr%5E2_Bw%5E2%5C%5C%5C%5CP_B%20%3D%20P_A%20%2B%20r%28z_A-z_B%29%2B%5Cfrac%7B1%7D%7B2%7D%20pw%5E2%5B%28r_B%5E2%29-%28r_A%29%5E2%5D%5C%5C%5C%5CP_B%20%3D%20%5B25%20%2B%280.8%2B62.4%29%280-1%29%2B%5Cfrac%7B1%7D%7B2%7D%280.8%5Ctimes1.94%29%5Ctimes%2813.333%29%5E2%5B2.5%5E2-1.5%5E2%5D%20%20%5D%5C%5C%5C%5CP_B%20%3D%2025%20-%2049.92%2B551.79%5C%5C%5C%5CP_B%20%3D%20526.87psf%5C%5C%5Capprox527psf)
the pressure at B is 527psf
Newton' 1st Law of Motion
A soccer ball will not move until a player kicks it
A bowling ball hits the pins sending the pins flying for a STRIKE!
Newton's 2nd Law of Motion
it takes less force to move a ping pong ball then a bowling ball.
f - ma
Newton's 3rd Law of Motion
a fireman turns on his house and is knocked backward
If air is let out of a balloon quickly, air pushes down & balloon goes up
