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3 years ago

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Ax = 4.3 m

1 answer:

Dimas [21]3 years ago

7 0

Answer:

the answer will be option no b plss mark me brainliest

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Problem 1: Two sources emit waves that are coherent, in phase, and have wavelengths of 26.0 m. Do the waves interfere constructi

Anton [14]

1) Destructive interference

The condition for constructive interference to occur is:

$\delta = m\lambda$ (1)

where

$\delta =|d_1 -d_2|$ is the path difference, with

$d_1$ is the distance of the point from the first source

$d_2$ is the distance of the point from the second source

m is an integer number

$\lambda$ is the wavelength

In this problem, we have

$d_1 = 78.0 m\\d_2 = 143 m\\\lambda=26.0 m$

So let's use eq.(1) to see if the resulting m is an integer

$\delta =|78.0 m-143 m|=65 m\\m=\frac{\delta }{\lambda}=\frac{65 m}{26.0 m}=2.5$

It is not an integer so constructive interference does not occur.

Let's now analyze the condition for destructive interference:

$\delta = (m+\frac{1}{2})\lambda$ (2)

If we apply the same procedure to eq.(2), we find

$m=\frac{\delta}{\lambda}-\frac{1}{2}=\frac{65.0 m}{26.0 m}-0.5=2$

which is an integer: so, this point is a point of destructive interference.

2) Constructive interference

In this case we have

$d_1 = 91.0 m\\d_2 =221.0 m$

So the path difference is

$\delta =|91.0 m-221.0 m|=130.0 m$

Using the condition for constructive interference:

$m=\frac{\delta }{\lambda}=\frac{130.0 m}{26.0 m}=5$

Which is an integer, so this is a point of constructive interference.

3) Destructive interference

In this case we have

$d_1 = 44.0 m\\d_2 =135.0 m$

So the path difference is

$\delta =|44.0 m-135.0 m|=91.0 m$

Using the condition for constructive interference:

$m=\frac{\delta }{\lambda}=\frac{91.0 m}{26.0 m}=3.5$

This is not an integer, so this is not a point of constructive interference.

So let's use now the condition for destructive interference:

$m=\frac{\delta}{\lambda}-\frac{1}{2}=\frac{91.0 m}{26.0 m}-0.5=3$

which is an integer: so, this point is a point of destructive interference.

3 0

3 years ago

Radiant energy comes from the _______.

snow_lady [41]

Answer:

C. Sun

Explanation:

7 0

3 years ago

Read 2 more answers

In a semiclassical model of the hydrogen atom, the electron orbits the proton at a distance of 0.053 nm. Part A What is the elec

Bezzdna [24]

Answer with Explanation:

We are given that

$r=0.053 nm=0.053\times 10^{-9} m$

$1 nm=10^{-9} m$

Charge on proton,q= $1.6\times 10^{-19} C$

a.We have to find the electric potential of the proton at the position of the electron.

We know that the electric potential

$V=\frac{kq}{r}$

Where $k=9\times 10^9$

$V=\frac{9\times 10^9\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}$

$V=27.17 V$

B.Potential energy of electron,U= $\frac{kq_e q_p}{r}$

Where

$q_e=-1.6\times 10^{-19} c=$ Charge on electron

$q_p=q=1.6\times 10^{-19} C$ =Charge on proton

Using the formula

$U=\frac{9\times 10^9\times (-1.6\times 10^{-19}\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}$

$U=-4.35\times 10^{-18} J$

8 0

3 years ago

The Moon orbits Earth in an average of p = 27.3 days at an average distance of a =384,000 kilometers. Using Newton’s version of

Korvikt [17]

Answer:

The mass of the earth, $M=6.023\times 10^{24}\ kg$

Explanation:

It is given that,

Time taken by the moon to orbit the earth, $T=27.3\ days=2358720\ m$

Distance between moon and the earth, $r=384000\ km=384\times 10^6\ m$

We need to find the mass of the Earth using Kepler's third law of motion as :

$T^2=\dfrac{4\pi^2}{GM}r^3$

$M=\dfrac{4\pi^2r^3}{T^2G}$

$M=\dfrac{4\pi^2\times (384\times 10^6)^3}{(2358720)^2\times 6.67\times 10^{-11}}$

$M=6.023\times 10^{24}\ kg$

So, the mass of the earth is $6.023\times 10^{24}\ kg$ . Hence, this is the required solution.

7 0

3 years ago

Winner of five swimming gold medals at the 1988 olympics

Fed [463]

Matt Biondi..?
(I don’t know if it’s right, sorry if it is wrong)
:)

6 0

3 years ago