Answer:
at d the charge will be 3q and at 3d it will be 9q
Explanation:
for V=Vp-V2d
V=KQ/d=K*6q/2d=3kq/d for potential to 2d at 6q be zero the Vp will equal 3kq/d; hence at d, Q=3q and at 3d, Q=9q
This sequence refer to how neurons send messages electrochemically. When the neurons are at rest, they are at its resting potential at -70 millivolts. If the neurons in the brain send messages, a spike would occur until it reaches the threshold potential at -55 mV. If it reaches the threshold potential, then an action potential occurs.Hence, the sequence would be letter C.
Displacement from the center line for minimum intensity is 1.35 mm , width of the slit is 0.75 so Wavelength of the light is 506.25.
<h3>How to find Wavelength of the light?</h3>
When a wave is bent by an obstruction whose dimensions are similar to the wavelength, diffraction is observed. We can disregard the effects of extremes because the Fraunhofer diffraction is the most straightforward scenario and the obstacle is a long, narrow slit.
This is a straightforward situation in which we can apply the
Fraunhofer single slit diffraction equation:
y = mλD/a
Where:
y = Displacement from the center line for minimum intensity = 1.35 mm
λ = wavelength of the light.
D = distance
a = width of the slit = 0.75
m = order number = 1
Solving for λ
λ = y + a/ mD
Changing the information that the issue has provided:
λ = 1.35 * 10^-3 + 0.75 * 10^-3 / 1*2
=5.0625 *10^-7 = 506.25
so
Wavelength of the light 506.25.
To learn more about Wavelength of the light refer to:
brainly.com/question/15413360
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C6H14
Gaseous state
it's unsaturated hence gaseous
Imagine you were able to throw a ball in a frictionless environment
such as outer space. Once you let go of the ball, it will travel forever
in a straight line, and at a constant speed. (At least until it bumps into
something.)
A car accelerates down the road. The reaction to the tires pushing
on the road is the road pushing on the tires.
