14 ms is required to reach the potential of 1500 V.
<u>Explanation:</u>
The current is measured as the amount of charge traveling per unit time. So the charge of electrons required for each current is determined as the product of current with time.
As two different current is passing at two different times, the net charge will be the different in current. So,
The electric voltage on the surface of cylinder can be obtained as the ratio of charge to the radius of the cylinder.
Here 
, q is the charge and R is the radius. As 
and R =17 cm = 0.17 m, then the voltage will be
The time is required to find to reach the voltage of 1500 V, so
So, 14 ms is required to reach the potential of 1500 V.
Its very dense. Hey, are you homeschooled?
<u>Yes, work is done when a book falls of the table.</u>
This is because:
When the book falls, it's potential energy is converted into kinetic energy. As it reaches the floor down, this kinetic energy is converted to heat energy and sound energy due to the impact.
When a force is imposed on an object to cause displacement of that object, work is done on that object. For a force to do work on an object, there should be a displacement and this force should cause the displacement. So here, since the book falls from the table and causes the displacement of the book from the table to the floor. It is said that work is done.
Work can be given by the formula:
W = F • d
where F is the force and d is the displacement.
Answer:
θ = 1.591 10⁻² rad
Explanation:
For this exercise we must suppose a criterion when two light sources are considered separated, we use the most common criterion the Rayleigh criterion that establishes that two light sources are separated census the central maximum of one of them coincides with the first minimum of the other source
Let's write the diffraction equation for a slit
a sin θ = m λ
The first minimum occurs for m = 1, also field in these we experience the angles are very small, we can approximate the sin θ = θ
θ = λ / a
In our case, the pupil is circular, so the system must be solved in polar coordinates, so a numerical constant is introduced.
θ = 1.22 λ / D
Where D is the diameter of the pupil
Let's apply this equation to our case
θ = 1.22 600 10⁻⁹ / 0.460 10⁻²
θ = 1.591 10⁻² rad
This is the angle separation to solve the two light sources
Answer:
The magnitude of the electric field at the center of curvature of the arc is 3.87 N/C
Explanation:
Please see the attachments below
