The potential difference between the two points if the work done on the charge is +3.4×10^7 joules would be A. 10volts.
Notice that
<em>B</em> = 4<em>i</em> + 6<em>j</em> - 2<em>k</em> = 2 (2<em>i</em> + 3<em>j</em> - <em>k</em>) = 2<em>A</em>
so both vectors point in the same direction and the angle between them is (A) 0°.
Work done = Potential Energy = (mg)h
mg = weight of the ball
= 1000 N * 2.5 m = 2500 Joules.
The work done during the 5 second is 2500 Joules.
Answer:
+1.46×10¯⁶ C
Explanation:
From the question given above, the following data were obtained:
Charge 1 (q₁) = +26.3 μC = +26.3×10¯⁶ C
Force (F) = 0.615 N
Distance apart (r) = 0.750 m
Electrical constant (K) = 9×10⁹ Nm²/C²
Charge 2 (q₂) =?
The value of the second charge can be obtained as follow:
F = Kq₁q₂ / r²
0.615 = 9×10⁹ × 26.3×10¯⁶ × q₂ / 0.750²
0.615 = 236700 × q₂ / 0.5625
Cross multiply
236700 × q₂ = 0.615 × 0.5625
Divide both side by 236700
q₂ = (0.615 × 0.5625) / 236700
q₂ = +1.46×10¯⁶ C
NOTE: The force between them is repulsive as stated from the question. This means that both charge has the same sign. Since the first charge has a positive sign, the second charge also has a positive sign. Thus, the value of the second charge is +1.46×10¯⁶ C
B. all of nature follows uniform, unchaining laws.
