Answer:
according to this question best answer is C
Resistors Working Together.
Resistors are shown coupled in parallel to a voltage source in Figure 10.3.4. When all of the resistors' ends are connected to one another by a continuous wire of minimal resistance and their other ends are also connected to one another by a continuous wire of minimal resistance, the resistors are said to be in parallel. There is a constant potential drop across all resistors. Ohm's law, I=V/R, can be used to determine the current flowing through each resistor while the voltage is constant across each resistor. For instance, the headlights, radio, and other components of an automobile are linked in parallel so that each subsystem can use the entire voltage of the source and function independently. The wiring in your home or any other structure shares the same
The original circuit is shown in part a with two parallel resistors linked to a voltage source, and the equivalent circuit is shown in part b with one equivalent resistor connected to the voltage source.
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#4159
The representation of this problem is shown in Figure 1. So our goal is to find the vector

. From the figure we know that:

From geometry, we know that:

Then using
vector decomposition into components:

Therefore:

So if you want to find out <span>
how far are you from your starting point you need to know the magnitude of the vector

, that is:
</span>

Finally, let's find the <span>
compass direction of a line connecting your starting point to your final position. What we are looking for here is an angle that is shown in Figure 2 which is an angle defined with respect to the positive x-axis. Therefore:
</span>
A force of 43.8 N is required to stretch the spring a distance of 15.5 cm = 0.155 m, so the spring constant <em>k</em> is
43.8 N = <em>k</em> (0.155 m) ==> <em>k</em> = (43.8 N) / (0.155 m) ≈ 283 N/m
The total work done on the spring to stretch it to 15.5 cm from equilibrium is
1/2 (283 N/m) (0.155 m)² ≈ 3.39 J
The total work needed to stretch the spring to 15.5 cm + 10.4 cm = 25.9 cm = 0.259 m from equilibrium would be
1/2 (283 N/m) (0.259 m)² ≈ 9.48 J
Then the additional work needed to stretch the spring 10.4 cm further is the difference, about 6.08 J.