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ira [324]
3 years ago
13

A parallel beam of light of wavelength 4.5 x 10^-7 m is incident on a pair of slits that are 5.0 x 10^-4 m apart. The interferen

ce pattern is focused on a screen 2.0 m from the slits. The separation between two adjacent bright fringes is most nearly:_________
Physics
1 answer:
RSB [31]3 years ago
3 0

Answer:

1.8x10⁻³m

Explanation:

From the question above, the following information was used to solve the problem.

wavelength λ = 4.5x10⁻⁷m

Length L = 2.0 meters

distance d = 5 x 10₋⁴m

ΔY = λL/d

= 4.5x10⁻⁷m (2) / 5 x 10₋⁴m

= 0.00000045 / 0.0005

= 0.0000009/0.0005

= 0.0018

= 1.8x10⁻³m

from the solution above The separation between two adjacent bright fringes is most nearly 1.8x10⁻³m

thank you!

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in a certain experiment, a radio transmitter emits sinusoidal electromagnetic waves of frequency 105.0 mhz in opposite direction
Romashka [77]

As a result, the hollow is 10.90 meters long and the distance between the nodal planes is 1.36 meters.

<h3>Explain electromagnetic waves.</h3>

The oscillations between an electric field and a magnetic field produce waves known as electromagnetic waves, or EM waves.

By definition, we understand that the frequency equals,

f = c/λ

where,

λ = wavelength

c= Speed of light

λ = 2L / n

While the wavelength is equal to,

Where,

L = Length

n = Number of antinodes/nodes

PART A) We know that the first component's wavelength is 110 MHz, so

λ = c/ f

λ = 3*10^8 / 11*10^6

λ = 1.36m

Therefore the distance between the nodal planes is 1.36m

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λ` = 2l /n

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7 0
1 year ago
MATHPHYSSSSSSSS PLEASEEEEEE IM SORRY YOU PROBABLY HATE ME
inysia [295]

Answer:

3.1 m/s

Explanation:

First, find the time it takes for the cat to land.  Take down to be positive.

Given:

Δy = 0.61 m

v₀ = 0 m/s

a = 9.81 m/s²

Find: t

Δy = v₀ t + ½ at²

(0.61 m) = (0 m/s) t + ½ (9.81 m/s²) t²

t = 0.353 s

Now find the horizontal velocity needed to travel 1.1 m in that time.

Given:

Δx = 1.1 m

a = 0 m/s²

t = 0.353 s

Find: v₀

Δx = v₀ t + ½ at²

(1.1 m) = v₀ (0.353 s) + ½ (0 m/s²) (0.353 s)²

v₀ = 3.1 m/s

3 0
3 years ago
A uniformly charged, one-dimensional rod of length L has total positive charge Q. Itsleft end is located at x = ????L and its ri
GREYUIT [131]

Answer:

|\vec{F}| = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))

Explanation:

The force on the point charge q exerted by the rod can be found by Coulomb's Law.

\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\^r

Unfortunately, Coulomb's Law is valid for points charges only, and the rod is not a point charge.

In this case, we have to choose an infinitesimal portion on the rod, which is basically a point, and calculate the force exerted by this point, then integrate this small force (dF) over the entire rod.

We will choose an infinitesimal portion from a distance 'x' from the origin, and the length of this portion will be denoted as 'dx'. The charge of this small portion will be 'dq'.

Applying Coulomb's Law:

d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qdq}{x + x_0}(\^x)

The direction of the force on 'q' is to the right, since both charges are positive, and they repel each other.

Now, we have to write 'dq' in term of the known quantities.

\frac{Q}{L} = \frac{dq}{dx}\\dq = \frac{Qdx}{L}

Now, substitute this into 'dF':

d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qQdx}{L(x+x_0)}(\^x)

Now we can integrate dF over the rod.

\vec{F} = \int{d\vec{F}} = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}\int\limits^{L}_0 {\frac{1}{x+x_0}} \, dx = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))(\^x)

4 0
3 years ago
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