The intensity of the electric field is 30,000 N/C
Explanation:
The strength of the electric field produced by a single-point charge is given by the equation
where:
is the Coulomb's constant
q is the magnitude of the charge
r is the distance from the charge
In this problem, we have:
is the magnitude of the charge
r = 3 cm = 0.03 m is the distance at which we are calculating the field intensity
Substituting, we find:
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The magnitude of the E-field decreases as the square of the distance from the charge, just like gravity.
Location ' x ' is √(2² + 3²) = √13 m from the charge.
Location ' y ' is √ [ (-3)² + (-2)² ] = √13 m from the charge.
The magnitude of the E-field is the same at both locations.
The direction is also the same at both locations ... it points toward the origin.
When you shine a lite through a prism is reflects out light through all of the edges and causes light separation. Or just simply shine a laser through the edge of a sideways piece of glass.
I hope that this was helpful for you.
Answer:
Explanation:
initial velocity, u = 0
final velocity, v = 60 mph = 26.8 m/s
time t = 10 s
Let a be the acceleration and s be he distance traveled.
Use first equation of motion
v = u + a t
26.8 = 0 + a x 10
a = 2.68 m/s
Use second equation of motion
s = ut + 1/2 at²
s = 0 + 0.5 x 2.68 x 10 x 10
s = 134 m
As, 1 m = 3.28 ft
So, s = 134 x 3.28 ft
s = 439.6 ft
