Answer:
351 ohm
720 ohm
Explanation:
When c and d are open:
Terminals c and d are open.. If you redraw the circuit as below, you can see that the two resistors in the first column are in parallel as, they are connected together at both pairs of terminals (due to the short).
Hence, we have a pair of parallel resistors:
Req1 = (R1*R2)/ (R1 + R2) = 360*540/(360+540) = 216 ohms
Req2 = (R3*R4)/ (R3 + R4) = 180*540/(180+540) = 135 ohms
Now these two sets are in series with another Hence,
Req = Req1 + Req2 = 216 + 135 = 351 ohms
Answer: 351 ohms
When c and d are shorted:
The current will flow through the least resistant path naturally from resistors R3 and R1 or R4.
Both of these resistor lie in a single path placing the resistors in series to one another, hence
Req = R3 + R1 = 180 + 540 = 720 ohms
Answer:720 ohms
Answer:
Explanation:
according to ohms law we know that
v=IR
given current =2 amps
given resistance =6Ω
so voltage is
v=2*6 =12 V
Answer:
4.37 * 10^-4 J
Explanation:
Energy stored :
mgΔl / 2
m = mass = 10kg ; g = 9.8m/s² ; r = cross sectional Radius = 1cm = 1 * 10-2 m
Δl = mgl / πr²Y
Y = Youngs modulus = Y=3.5 ×10^10 ; l = Length = 1m
Δl = (10 * 9.8 * 1) / π * (1 * 10^-2)²* 3.5 ×10^10
Δl = 98 / 3.5 * π * 10^6
Δl = 0.00000891267
Energy stored :
mgΔl / 2
(10 * 9.8 * 0.00000891267) / 2
= 0.00043672083 J
4.37 * 10^-4 J
Answer: 3 m.
Explanation:
Neglecting the mass of the seesaw, in order the seesaw to be balanced, the sum of the torques created by gravity acting on both children must be 0.
As we are asked to locate Jack at some distance from the fulcrum, we can take torques regarding the fulcrum, which is located at just in the middle of the length of the seesaw.
If we choose the counterclockwise direction as positive, we can write the torque equation as follows (assuming that Jill sits at the left end of the seesaw):
mJill* 5m -mJack* d = 0
60 kg*5 m -100 kg* d =0
Solving for d:
d = 3 m.
