Each plate of a parallel-plate capacator is a square with side length r, and the plates are separated by a distance d. The capac

Answer:

Lower than

Explanation:

When the cube is placed on the raft, the displaced water equals the combined weight of both the cube and raft. In contrast, when submerged in water, the displaced water corresponds to the raft's weight plus the cube's volume. Since the cube's volume is smaller than what is needed to displace its weight with water, the water level is lower.

Answer:

   C = 4,174 10³ V / m^{3/4},  E = 7.19 10² / ∛x,    E = 1.5  10³ N/C

Explanation:

In this problem, we are tasked with determining the constant value and the generated electric field.

We will begin with computing the constant C:

           V = C

           C = V / x^{4/3}

            C = 220 / (11 10⁻²)^{4/3}

            C = 4,174 10³ V / m^{3/4}

Next, we will find the electric field by utilizing the formula:

            V = E dx

             E = dx / V

             E = ∫ dx / C x^{4/3}

            E = 1 / C  x^{-1/3} / (- 1/3)

            E = 1 / C (-3 / x^{1/3})

We consider the evaluation from the lower limit x = 0 where E = E₀ = 0 to the upper limit x = x, resulting in E = E:

            E = 3 / C     (0- (-1 / x^{1/3}))

            E = 3 / 4,174 10³   (1 / x^{1/3})

           E = 7.19 10² / ∛x

Substituting x = 0.110 cm:

          E = 7.19 10² /∛0.11

          E = 1.5  10³ N/C

The magnetic field is calculated to be -6.137 × T. Explanation: Given the radio wave wavelength of λ = 0.3 m and an intensity of I = 45 W/m² at times t = 0 and t = 1.5 ns, we determine Bz at the origin. We use the intensity formula relating to the electric field, which incorporates the known intensity of 45, the speed of light c = 3 × m/s², and ∈o as 8.85 × C²/N.m², leading us to E = 184.15. Consequently, applying the equations, we find B = -6.137 × T at the z-axis.

Answer:

K = 1.525 10⁻⁹ x⁴ + 4.1 10⁶ x

Explanation:

To calculate the kinetic energy variation, we can utilize the work-energy theorem.

W = ΔK

∫ F .dx = K - K₀

If the object starts from rest, then K₀ = 0.

So, ∫ F dx cos θ = K.

As the force and displacement directions align, the angle is zero, and hence the cosine is 1.

Now we can substitute and perform integration:

α ∫ x³ dx + β ∫ dx = K.

Thus, α x⁴ / 4 + β x = K.

Next, we evaluate from the limits F = 0 to F:

α (x⁴ / 4 - 0) + β (x - 0) = K.

Consequently, K = αX⁴ / 4 + β x.

This results in K = 1.525 10⁻⁹ x⁴ + 4.1 10⁶ x.

To finalize the computation, we need to ascertain the displacement.

ke prop to v^2

ke1/v1^2=ke2/v2^2

400/50x50=joules/100x100

400x2x2

1600j