<p>hey guys can somebody help me with these problems please...with the explanations? </p>
<ol>
<li><p>According to the right-hand rule, if a current-carrying wire is grasped in the right hand with the thumb in the direction of the current, the four fingers will curl in the diretion of
a. the curernt's velocity, v
b. the magnetic field, B
c. the magnetic force, F (magnetic)</p></li>
<li><p>A 2.00ohm resistor and a 12.0 ohms resistor are connected in parallel across a 20.0 V battery. What is the current flowing through the 2.00 ohms resistor?</p></li>
<li><p>Two charges of 1.0 microcoulombs and -2.5microcoulombs experience a force of attraction of 1.0 N. What is the separation between charges?</p></li>
<li><p>Three resistors connected in parallel have individual values of 4.0 ohms, 6.0 ohms, and 10.0 ohms. If this combination is connected in series with a 12.0V battery and a 2.0 ohms resistor, what is the current in the 10.0 ohms resistor?</p></li>
</ol>
<p>Thanks so much!</p>
<ol>
<li>field</li>
<li>V = IR, so I = V/R. The circuit is in parallel, so the voltage is the same between the two resistors. I = 20/2, and I = 10 amps.</li>
<li>F = kq1q2/r^2, so r = sqrt(kq1q2/F). r = sqrt(8.99E9 * 1E-6 * 2.5E-6). r = 0.15 meters.</li>
<li>1/4 + 1/6 + 1/10 = 31/60. 60/31 + 2 = 3.94 ohms of resistance. I = V/R, so I = 12/3.94. I = 3.046 amps. V = IR, so V = 3.046 * 2 = 6.092 volts, so the voltage drop across the 2-ohm resistor is 6.092 volts. Therefore, the voltage going across the 10-ohm resistor is 5.908 volts. Since I = V/R, the voltage across the 10-ohm resistor is 5.908 / 10 = .5908 amps.</li>
</ol>
<p>heyy thanks for the reply…
did you get my private message by any chance b/c i think i pressed a wrong button or something lol</p>
<p>can anybody help me with a couple more pleasee with explanations if possible? I really really appreciate your help. </p>
<p><strong><em>| |</em></strong>__ the | | is 60 volts
| |
|<strong>^^</strong><em>^^</em><em>| the first and seonc ^^ are 10 ohms
| |<br>
| </em><em>^^</em> | the ^^ on this line is 8 ohms
|<strong>^^</strong>| |<strong>| the ^^ on this line is 16 ohms
|</strong>^^___| the ^^ on this line is 8 ohms </p>
<ol>
<li><p>What is the equivalent resistance of the resistors in the figure shown above? </p></li>
<li><p>Two charges of 1.0 microcoulombs each are 1.2 cm apart. The magnitude of the electric field midway between these charges is: </p></li>
<li><p>Determine the equivalent capacitance in pF for the network shown when C= 12 pF.
<strong><em>||</em></strong>_______ the || on this line is C
| |
= = both = are C
_____<strong><em>||</em></strong><strong><em>|</em></strong>____| the || on this line is 2C </p></li>
<li><p>Three resistors connected in parallel have individual values of 4 ohms, 6 ohms, and 10 ohms. as shown below. If this combination is connected in series with a 12 V battery and a 2 ohm resistor, what is the current in the 10 ohms resistor?</p></li>
</ol>
<p><strong><em>^^</em></strong>__________ ^^ is 2 ohms
| | | |<br>
= ^^ ^^ ^^ = is 12 V, first ^^ is 4ohms, the 2nd is 6, 3rd is 10ohms
|_______<strong><em>|</em></strong><strong>|</strong>__| </p>
<p>Thanks so much :)</p>
<p>oh no. the lines got messed up. sigh.</p>