If the frequency of the voltage signal impressed on the parallel plate capacitor is doubled from f to 2f, the amplitude of the m

This is somewhat misleading, and I encountered the same question in my homework. An electric field strength of 1*10^5 N/C is provided, along with a drag force of 7.25*10^-11 N, and the critical detail is that it maintains a constant velocity, indicating that the particle is in equilibrium and not accelerating.
<span>To solve, utilize F=(K*Q1*Q2)/r^2 </span>
<span>You'll want to equate F with the drag force, where the electric field strength translates to (K*Q2)/r^2; substituting the values results in </span>
<span>(7.25*10^-11 N) = (1*10^5 N/C)*Q1 ---> Q1 = 7.25*10^-16 C </span>

Response:

1 slit width = 0.158 mm, slit separation = 0.633 mm, distance between diffraction maxima = 12.7 mm

Explanation:

This involves defining several terms:

λ, the wavelength of the beam

D, the distance from the plane to the slit

x, the distance between minima in the diffraction pattern (in single slit setups)

w, the fringe width for double slit setups

1. In a double slit experiment, the fringe width is also recognized as the distance between Maxima.

Thus, w=λD/d, leading to d=λD/w when rearranged.

So, d= (632.8 x 10^-9 x 1 x 10^3)/1

which equals d= 0.633mm.

For single slit diffraction, minima are defined by a*sinΘ= m*x

where a is the slit width and m is an integer.

For small angles, it simplifies to Θ= (x/D) = (λ/a), allowing us to solve for a:

a = λD/a

a = (632.8x10^-9 x 1)/4

yielding a= 0.158mm.

2. A grating with 20 slits/mm gives d= 1/20mm= 0.05x10^-3.

To find y (the distance between maxima), apply y= λL/d:

Finally, y= (632.8 x 10^-9 x 1)/0.05x10^-3, which results in y= 12.7mm.

Vf=Vi+at=15m/s+(3.5m/s^2)(11s)=53.5m/s

Answer:

             U = 1 / r²

Explanation:

In this problem, the task does not require calculating potential energy via the force equation since these two variables are interconnected.

             

         F = - dU / dr

This derivative represents a gradient, meaning it indicates direction, leading us to write

          dU = - F. dr

The formula for force becomes

         F = B / r³

Now, let’s apply this in the integral:

          ∫ dU = - ∫ B / r³  dr

Here, the force aligns with the displacement, simplifying the scalar product to the product of magnitudes.

Now, we can solve the integrals:

            U - Uo = -B (- / 2r² + 1 / 2r₀²)

To finalize the calculations, a reference point for energy must be designated; commonly, potential energy is set to zero (Uo = 0) at infinity (r = ∞).

             U = B / 2r²

Substituting B = 2, we arrive at:

             U = 1 / r²

According to the second law, heat, often called thermal energy, cannot be entirely turned into work.
The second statement is closely tied to this law.
We can conclude that some energy dissipates while some is used for work.