The answer is
D. in a straightforward, objective manner
charge stored in the capacitor=3.29 x 10⁻⁴ C
Explanation:
we use the formula
Q= C V
Q= charge
C= capacitor=25.3 μF= 25.3 x 10⁻⁶ F
V= voltage= 13 V
Q=(25.3 x 10⁻⁶ ) (13)
Q= 3.29 x 10⁻⁴ C
Answer:
180 miles/hr
Explanation:
15 min=1/4 hr
s= distance/time
s=45miles/1/4 hr
s=45miles x 4/1=180 miles/hr
Let l = Q/L = linear charge density. The semi-circle has a length L which is half the circumference of the circle. So w can relate the radius of the circle to L by
<span>C = 2L = 2*pi*R ---> R = L/pi </span>
<span>Now define the center of the semi-circle as the origin of coordinates and define a as the angle between R and the x-axis. </span>
<span>we can define a small charge dq as </span>
<span>dq = l*ds = l*R*da </span>
<span>So the electric field can be written as: </span>
<span>dE =kdq*(cos(a)/R^2 I_hat + sin(a)/R^2 j_hat) </span>
<span>dE = k*I*R*da*(cos(a)/R^2 I_hat + sin(a)/R^2 j_hat) </span>
<span>E = k*I*(sin(a)/R I_hat - cos(a)/R^2 j_hat) </span>
<span>E = pi*k*Q/L(sin(a)/L I_hat - cos(a)/L j_hat)</span>
-- From January 15 to February 6 is a period of 22 days.
-- The period of the full cycle of moon phases is 29.53 days.
-- So those dates represent (22/29.53) = 74.5% of a full cycle of phases.
-- That's almost exactly 3/4 of a full cycle, so on February 6, the moon would be almost exactly at <em>Third Quarter</em>. That's the <em>left half of a disk </em>(viewed from the northern hemisphere).
