Where does the suns rays strike earth most direcly and least directly

Ocean waves pass through two small openings, 20.0 m apart, in a breakwater. You're in a boat 70.0 m from the breakwater and init

Klio2033 [76]

Answer:

λ = 5.65m

Explanation:

The Path Difference Condition is given as:

δ= $(m+\frac{1}{2})\frac{lamda}{n}$ ;

where lamda is represent by the symbol (λ) and is the wavelength we are meant to calculate.

m = no of openings which is 2

∴δ= $\frac{3*lamda}{2}$

n is the index of refraction of the medium in which the wave is traveling

To find δ we have;

δ= $\sqrt{70^2+(33+\frac{20}{2})^2 }-\sqrt{70^2+(33-\frac{20}{2})^2 }$

δ= $\sqrt{4900+(\frac{66+20}{2})^2}-\sqrt{4900+(\frac{66-20}{2})^2}$

δ= $\sqrt{4900+(\frac{86}{2})^2 }-\sqrt{4900+(\frac{46}{2})^2 }$

δ= $\sqrt{4900+43^2}-\sqrt{4900+23^2}$

δ= $\sqrt{4900+1849}-\sqrt{4900+529}$

δ= $\sqrt{6749}-\sqrt{5429}$

δ= 82.15 -73.68

δ= 8.47

Again remember; to calculate the wavelength of the ocean waves; we have:

δ= $\frac{3*lamda}{2}$

δ= 8.47

8.47 = $\frac{3*lamda}{2}$

λ = $\frac{8.47*2}{3}$

λ = 5.65m

3 0

4 years ago

Why do radio waves have the lowest frequency of all the waves on the electromagnetic spectrum?

lara [203]

Explanation:

Electromagnetic waves are the waves which are created as the result of the electrical waves which are perpendicular to each other and also perpendicular to the direction of propagation.

Electromagnetic spectrum is range of the frequencies and their respective wavelengths of the various type of the electromagnetic radiation.

In order of the increasing frequency and the photon energy and the decreasing wavelength the spectrum are:

radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays and gamma rays.

The energy of the radio waves photons is the lowest of all the other waves in the electromagnetic spectrum.

Also, $E={h\times \nu}$

Where,

h is Plank's constant having value $6.626\times 10^{-34}\ Js$

Thus, energy is directly proportional to the frequency. The radio waves have the lowest frequency.

8 0

4 years ago

The question ask:

abruzzese [7]

Answer:

$\theta=28.07^{\circ}$

Explanation:

Speed of van, v = 28 m/s

Radius of unbanked curve, r = 150 m

Let $\theta$ is the angle with the vertical. In case of banking of road,

$T\ cos\theta=mg$ .............(1)

And

$T\ sin\theta=\dfrac{mv^2}{r}$ ..........(2)

From equation (1) and (2) :

$tan\theta=\dfrac{v^2}{rg}$

$tan\theta=\dfrac{(28)^2}{150\times 9.8}$

$\theta=28.07^{\circ}$

So, the string makes an angle of 28.07 degrees with the vertical. Hence, this is the required solution.

4 0

3 years ago

A 1.0-m-tall vertical tube is filled with 20°C water. A tuning fork vibrating at 580 Hz is held just over the top of the tube as

Mademuasel [1]

Answer:

water heights of the tube are 0.851 m , 0.553 m, 0.255 m

Explanation:

given data

frequency = 580 Hz

temperature = 20°C

tube = 1 m

to find out

water heights of the tube

solution

we will apply here formula for length that is

length L = v ( 2n -1 ) / 4f

here v is velocity o sound that is 343.2 m/s

so for n = 1

L = 343.2 ( 2(1) -1 ) / 4(580) = 0.147931 m

for n = 2

L = 343.2 ( 2(2) -1 ) / 4(580) = 0.443793 m

for n = 3

L = 343.2 ( 2(3) -1 ) / 4(580) = 0.739655 m

for n = 4

L = 343.2 ( 2(4) -1 ) / 4(580) = 1.035517 m is greater than 1

and so here height is measured less than 1 m

so water heights of the tube are 1 m - 0.147931 m , 1 m - 0.443793 m, 1 m - 0.739655 m

so water heights of the tube are 0.851 m , 0.553 m, 0.255 m

3 0

3 years ago

The electric flux through a square-shaped area of side 5 cm near a very large, thin, uniformly-charged sheet is found to be 3\ti

deff fn [24]

Answer:

Explanation:

Given

side of square shape $a=5\ cm$

Electric flux $\phi =3\times 10^{-5}\ N.m^2/C$

Permittivity of free space $\epsilon_0=8.85\times 10^{-12} \frac{C^2}{N.m^2}$

Flux is given by

$\phi =EA\cos \theta$

where E=electric field strength

A=area

$\theta$ =Angle between Electric field and area vector

$E=\frac{\phi }{A\cos (0)}$

$E=\frac{3\times 10^{-5}}{25\times 10^{-4}\times \cos(0)}$

$E=0.012\ N/C$

and Electric field by a uniformly charged sheet is given by

$E=\frac{\sigma }{2\epsilon_0}$

where $\sigma$ =charge density

$=\frac{\sigma }{\epsilon_0}$

$\sigma =0.012\times 8.85\times 10^{-12}$

$\sigma =2.12\times 10^{-13}\ C/m^2$

5 0

3 years ago