A point charge q1 = 4.50 nC is located on the x-axis at x = 1.95 m , and a second point charge q2 = -6.80 nC is on the y-axis at
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Answer:
Terminal velocity of object = 12.58 m/s
Explanation:
We know that the terminal velocity is attained when drag force and gravitational force are of the same magnitude.
Gravitational force = mg = 80 * 9.8 = 784 N
Drag force =
Equating both, we have
So v = 12.58 m/s or v = -15.58 m/s ( not possible)
So terminal velocity of object = 12.58 m/s
Answer:
<em>a) 318.2 W/m^2</em>
<em>b) 2.5 x 10^-4 J</em>
<em>c) 1.55 x 10^-8 v/m</em>
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Explanation:
Power of laser P = 1 mW = 1 x 10^-3 W
exposure time t = 250 ms = 250 x 10^-3 s
If beam diameter = 2 mm = 2 x 10^-3 m
then
cross-sectional area of beam A = = (3.142 x
)/4
A = 3.142 x 10^-6 m^2
a) Intensity I = P/A
where P is the power of the laser
A is the cros-sectional area of the beam
I = ( 1 x 10^-3)/(3.142 x 10^-6) = <em>318.2 W/m^2</em>
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b) Total energy delivered E = Pt
where P is the power of the beam
t is the exposure time
E = 1 x 10^-3 x 250 x 10^-3 = <em>2.5 x 10^-4 J</em>
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c) The peak electric field is given as
E =
where I is the intensity of the beam
E is the electric field
c is the speed of light = 3 x 10^8 m/s
= 8.85 x 10^9 m kg s^-2 A^-2
E = = <em>1.55 x 10^-8 v/m</em>