Point charges q1=+2.00μC and q2=−2.00μC are placed at adjacent corners of a square for which the length of each side is 5.00 cm.?
Point a is at the center of the square, and point b is at the empty corner closest to q2. Take the electric potential to be zero at a distance far from both charges.
(a) What is the electric potential at point a due to q1 and q2?
(b) What is the electric potential at point b?
(c) A point charge q3 = -6.00 μC moves from point a to point b. How much work is done on q3 by the electric forces exerted by q1 and q2?
Answer:
a) the potential is zero at the center .
Explanation:
a) since the two equal-magnitude and oppositely charged particles are equidistant
b)(b) Electric potential at point b, v = Σ kQ/r
r = 5cm = 0.05m
k = 8.99*10^9 N·m²/C²
Q = -2 microcoulomb
v= (8.99*10^9) * (2*10^-6) * (1/√2m - 1) / 0.0500m
v = -105 324 V
c)workdone = charge * potential
work = -6.00µC * -105324V
work = 0.632 J
<span>Atmospheric refraction is the deviation of light or other electromagnetic wave from a straight line as it passes through the atmosphere due to the variation in air density as a function of height. ... The term also applies to the refraction of sound.</span>
Answer:
T=+1.133N
Explanation:
Tension and weight are forces that have opposite directions
Weight is negative (downward)
W=m*g= 0.11kg*(-9.8m/s^2)
W= -1.078N
Tension is possitive (upward)
The total force will be the sum of both (the difference taking in consideration the direction)
Ft= T+W
Also the total force is the product of the mass due to acceleration:
Ft=m*a
Ft= +0.11kg*0.5m/s^2
Ft=+0.055N (upward)
Tension will be the difference between Ft and W:
T= Ft-W
T=+0.055N-(-1.078N)
T=+1.133N
fluid friction<span> occurs when an object moves through a liquid or gas. the force needed to overcome </span>fluid friction <span>is usually less then that needed to overcome </span>sliding friction<span>. the </span>fluid<span> keeps the surface from making direct contact and thus </span>reducing friction<span>.</span>
I believe the term Frequency is what you are looking for.
