All electric circuits have devices that are run by ____

If two point sources of light are being imaged by this telescope, what is the maximum wavelength λ at which the two can be resol

Mrrafil [7]

Answer:

The maximum wavelength is 492 nm.

Explanation:

Given that,

Angular separation $\theta=3.0\times10^{-5}\ rad$

Suppose a telescope with a small circular aperture of diameter 2.0 cm.

We need to calculate the maximum wavelength

Using formula of angular separation

$\sin\theta=\dfrac{1.22\lambda}{d}$

$\lambda=\dfrac{d\sin\theta}{1.22}$

Put the value into the formula

$\lambda=\dfrac{2.0\times\sin(3\times10^{-5})}{1.22}$

For small angle $\sin\theta\approx\theta$

$\lambda=\dfrac{0.02\times3\times10^{-5}}{1.22}$

$\lambda=4.92\times10^{-7}\ m$

$\lambda=492\ nm$

Hence, The maximum wavelength is 492 nm.

5 0

2 years ago

Please select the word from the list that best fits the definition Visualizes where items are located

Amanda [17]

Answer:I’m pretty sure it’s spatial

Explanation:

3 0

3 years ago

Read 2 more answers

3 small identical balls have charges -3×10^-12 ,8×10^-12 and 4×10^-12 find the charge

Norma-Jean [14]

Answer:

3×10^-12 C

Explanation:

The total of the three charges is ...

(-3 +8 +4)×10^-12 C = 9×10^-12 C

Assuming the charge is equally distributed between the balls when they are brought in contact, the charge on each ball will be ...

(9/3)×10^-12 C = 3×10^-12 C

7 0

2 years ago

What is the potential difference across a parallel-plate capacitor whose plates are separated by a distance of 4.0 mm where each

suter [353]

The potential difference across the parallel plate capacitor is 2.26 millivolts

<h3>Capacitance of a parallel plate capacitor</h3>

The capacitance of the parallel plate capacitor is given by C = ε₀A/d where

ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m,
A = area of plates and
d = distance between plates = 4.0 mm = 4.0 × 10⁻³ m.

<h3>Charge on plates</h3>

Also, the surface charge on the capacitor Q = σA where

σ = charge density = 5.0 pC/m² = 5.0 × 10⁻¹² C/m² and
a = area of plates.

<h3>The potential difference across the parallel plate capacitor</h3>

The potential difference across the parallel plate capacitor is V = Q/C

= σA ÷ ε₀A/d

= σd/ε₀

Substituting the values of the variables into the equation, we have

V = σd/ε₀

V = 5.0 × 10⁻¹² C/m² × 4.0 × 10⁻³ m/8.854 × 10⁻¹² F/m

V = 20.0 C/m × 10⁻³/8.854 F/m

V = 2.26 × 10⁻³ Volts

V = 2.26 millivolts

So, the potential difference across the parallel plate capacitor is 2.26 millivolts

Learn more about potential difference across parallel plate capacitor here:

brainly.com/question/12993474

7 0

2 years ago

Joseph studied whether different materials can block certain electromagnetic waves by testing television reception in different

Sedbober [7]

Joseph's experiment could be improved by using the same antenna at each part of the house during each trial instead of using different antenna. By doing so, he can obtain accurate results how is the signal in different part of the house under the same conditions (despite the location). So, he will see the dependence of the signal on the location. If he uses different antenna, than this antenna can also have influence of the signal.

5 0

3 years ago

All electric circuits have devices that are run by _____ energy.