Answer:
a.) -147V
b.) -120V
c.) 51V
Explanation:
a.) Equation for potential difference is the integral of the electrical field from a to b for the voltage V_ba = V(b)-V(a).
b.) The problem becomes easier to solve if you draw out the circuit. Since potential at Q is 0, then Q is at ground. So voltage across V_MQ is the same as potential at V_M.
c.) Same process as part b. Draw out the circuit and you'll see that the potential a point V_N is the same as the voltage across V_NP added with the 2V from the other box.
Honestly, these things take practice to get used to. It's really hard to explain this.
Answer:
32000 bits/seconds
Explanation:
Given that :
there are 16 signal combinations (states) = 2⁴
bits n = 4
and a baud rate (number of signals/second) = 8000/second
Therefore; the number of bits per seconds can be calculated as follows:
Number of bits per seconds = bits n × number of signal per seconds
Number of bits per seconds = 4 × 8000/second
Number of bits per seconds = 32000 bits/seconds
Answer:
Eutectic product in Fe-C system is called Ledeburite-C.
Answer:
The tension in the rope at the lowest point is 270 N
Explanation:
Given;
weight of the ball, W = 150 N
length of the rope, r = 4 m
velocity of the ball, v = 5.6 m/s
When the ball passes through the lowest point, the tension on the rope is the sum of weight of the ball and centripetal force.
T = W + F
Centripetal force, F = mv²/r
where;
m is the mass of the ball
m = W/g
m = 150 / 9.8 = 15.306 kg
Centripetal force, F = mv²/r
F = (15.306 x 5.6²)/4
F = 120 N
T = W + F
T = 150 + 120
T = 270 N
Therefore, the tension in the rope at the lowest point is 270 N
Answer:
X_cp = c/2
Explanation:
We are given;
Chord = c
Angle of attack = α
p u (s) = c 1
p1(s)=c2,
and c2 > c1
First of all, we need to find the resultant normal force on the plate and the total moment about leading edge.
I've attached the solution
