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2 years ago

According to the chart which type of electromagnetic wave has a longest wavelength?

Physics

1 answer:

mr Goodwill [35]2 years ago

4 0

Answer:

It's the radio waves having longest wavelength

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Divergent plate boundaries are when plates

umka21 [38]

Explanation:

Move past each other.

Have a good day.

7 0

2 years ago

Read 2 more answers

A disk ring of inner radius a and outer radius b lying on the y − z plane has a uniform surface electric charge density +σ(>

solmaris [256]

Answer:

a) V = k 2π σ (√(b² + x²) - √ (a² + x²)) ,

b) E = - k 2π σ x (1 /√(b² + x²) - 1 /√(a² + x²))

Explanation:

a) The expression for the electric potential is

V = k ∫ dq / r

For this case, consider the disk formed by a series of concentric rings of radius r and width dr, the distance of each ring to point P

R = √(x² + r²)

The charge on a ring is

σ = dq / dA

The area of a ring is

A = π r

dA = 2π r dr

So the charge is

dq = σ 2π r dr

We substitute

V = k σ 2pi ∫ r dr / √(r² + x²)

We integrate

V = k 2π σ √(r² + x²)

We evaluate from the lower limit r = a to the upper limit r = b

V = k 2π σ (√(b² + x²) - √ (a² + x²))

b) the electric field and the potential are related

E = - dV / dx

E = - k 2π σ (1/2 2x /√(b² + x²) - ½ 2x /√(a² + x²))

E = - k 2π σ x (1 /√(b² + x²) - 1 /√(a² + x²))

7 0

3 years ago

What are the chances that radio transmissions from Earth or messages sent on distant space probes will ever be received by livin

spin [16.1K]

Answer:

<u>Explanation:</u>

It is fair to say to a reasonable extent that there are very low chances that radio transmissions from Earth or messages sent on distant space probes will ever be received by living beings .

Bear in mind that for years some scientists have believed without any substantial evidence that there are other living beings in distant space.

4 0

3 years ago

A thin spherical spherical shell of radius R which carried a uniform surface charge density σ. Write an expression for the volum

ozzi

Answer:

Explanation:

From the given information:

We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then

The volume charge distribution relates to the radial direction at r = R

∴

$\rho (r) \ \alpha \ \delta (r -R)$

$\rho (r) = k \ \delta (r -R) \ \ at \ \ (r = R)$

$\rho (r) = 0\ \ since \ r< R \ \ or \ \ r>R---- (1)$

To find the constant k, we examine the total charge Q which is:

$Q = \int \rho (r) \ dV = \int \sigma \times dA$

$Q = \int \rho (r) \ dV = \sigma \times4 \pi R^2$

∴

$\int ^{2 \pi}_{0} \int ^{\pi}_{0} \int ^{R}_{0} \rho (r) r^2sin \theta \ dr \ d\theta \ d\phi = \sigma \times 4 \pi R^2$

$\int^{2 \pi}_{0} d \phi* \int ^{\pi}_{0} \ sin \theta d \theta * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

$(2 \pi)(2) * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

Thus;

$k * 4 \pi \int ^{R}_{0} \delta (r -R) * r^2dr = \sigma \times R^2$

$k * \int ^{R}_{0} \delta (r -R) r^2dr = \sigma \times R^2$

$k * R^2= \sigma \times R^2$

$k = R^2 --- (2)$

Hence, from equation (1), if k = $\sigma$

$\mathbf{\rho (r) = \delta* \delta (r -R) \ \ at \ \ (r=R)}$

$\mathbf{\rho (r) =0 \ \ at \ \ rR}$

To verify the units:

$\mathbf{\rho (r) =\sigma \ * \ \delta (r-R)}$

↓ ↓ ↓

c/m³ c/m³ × 1/m

Thus, the units are verified.

The integrated charge Q

$Q = \int \rho (r) \ dV \\ \\ Q = \int ^{2 \ \pi}_{0} \int ^{\pi}_{0} \int ^R_0 \rho (r) \ \ r^2 \ \ sin \theta \ dr \ d\theta \ d \phi \\ \\ Q = \int ^{2 \pi}_{0} \ d \phi \int ^{\pi}_{0} \ sin \theta \int ^R_{0} \rho (r) r^2 \ dr$