The closest refractive index to the one observed is that of rayon, about 1.51. Therefore, the fiber is likely from the suspect's shirt.
Answer:
Explanation:
In an electric field E force on charge q
F = Eq , acceleration a = Eq / m
a = 664 x 1.6 x 10⁻¹⁹ / 1.67 x 10⁻²⁷
= 636.16 x 10⁸ m /s²
b )
initial velocity u = 0
final velocity v = 1.46 x 10⁶ m/s
v = u + at
1.46 x 10⁶ = 0 + 636.16 x 10⁸ x t
t = 2.29 x 10⁻⁵ s
c )
s = ut + 1/2 a t²
= 0 + .5 x 636.16 x 10⁸ x ( 2.29 x 10⁻⁵ )²
= 1668 x 10⁻²
= 16.68 m
d )
Kinetic energy = 1/2 m v²
= .5 x 1.67 x 10⁻²⁷ x ( 1.46 x 10⁶ )²
= 1.78 x 10⁻¹⁵ J .
Answer:
Explanation:
Given
Required
Determine the impulse
The impulse is calculated as follows:
Substitute values for Force and Time
<em>Hence, the impulse experienced is 8.0Ns</em>
Answer:
a) 0 < r < R: E = 0, R < r < 2R: E = KQ/r^2, r > 2R: E = 2KQ/r^2
b) See the picture
Explanation:
We can use Gauss's law to find the electric field in all the regions:
EA = qen/e0 where qen is the enclosed charge
Remember that the electric field everywhere outside a sphere is:
E(r) = q/(4*pi*eo*r^2) = Kq/r^2
a)
- For 0 < r < R: There is not enclosed charge because all of it remains on the outer layer of the conducting sphere, therefore E = 0 EA = 0/e0 = 0 E = 0
- For R < r < 2R: Here the enclosed charge is equal Q E = Q/(4*pi*eo*r^2) = KQ/r^2
- For r > 2R: Here the enclosed charge is equal 2Q E = Q/(4*pi*eo*r^2) + Q/(4*pi*eo*r^2) = 2Q/(4*pi*eo*r^2) = 2KQ/r^2
b) At the beginning there is no electric field this is why you see a line in zero, In R the electric field is maximum and then it starts to decrease exponentially with the distance and finally in 2R the field increase a little due to the second sphere to then continue decreasing exponentially with the distance
If the bulbs are in series, then there are no 'branches' ... just
a single current path all the way around.
If the two bulbs have equal resistance, then no matter what their
actual resistance is, the supply voltage will divide equally between
them ... 6 volts across each bulb.