What is the frequency if a wave that pases a given pount 22 times in 2 seconds

In a chemical reaction blank are the substances left over

Zepler [3.9K]

Answer: In a chemical reaction blank are the substances left over

Explanation: The substance left over after a reaction takes place

3 0

3 years ago

Two charges are located in the x–y plane. If q1 = -2.90 nC and is located at x = 0.00 m, y = 0.840 m and the second charge has m

Lunna [17]

Answer:

Epx= - 21.4N/C

Epy= 19.84N/C

Explanation:

Electric field theory

The electric field at a point P due to a point charge is calculated as follows:

E= k*q/r²

E= Electric field in N/C

q = charge in Newtons (N)

k= electric constant in N*m²/C²

r= distance from load q to point P in meters (m)

Equivalences

1nC= 10⁻⁹C

known data

q₁=-2.9nC=-2.9 *10⁻⁹C

q₂=5nC=5 *10⁻⁹C

r₁=0.840m

$r_{2} =\sqrt{1^{2} +0.8^{2} } =\sqrt{1.64}$

$sin\beta =\frac{0.8}{\sqrt{1.64} } =0.6246$

$cos\beta =\frac{1}{\sqrt{1.64} } =0.7808$

Calculation of the electric field at point P due to q1

Ep₁x=0

$Ep_{1y} =\frac{k*q_{1} }{r_{1}^{2} } =\frac{8.99*10^{9}*2.9*10^{-9} }{0.84^{2} } =36.95\frac{N}{C}$

Calculation of the electric field at point P due to q2

$Ep_{2x} =-\frac{k*q_{2} *cos\beta }{r_{2}^{2} } =-\frac{8.99*10^{9}*5*10^{-9} *0.7808 }{(\sqrt{1.64})^{2} } =-21.4\frac{N}{C}$

$Ep_{2y} =-\frac{k*q_{2} *sin\beta }{r_{2}^{2} } =-\frac{8.99*10^{9}*5*10^{-9} *0.6242 }{(\sqrt{1.64})^{2} } =-17.11\frac{N}{C}$

Calculation of the electric field at point P(0,0) due to q1 and q2

Epx= Ep₁x+ Ep₂x==0 - 21.4N/C =- 21.4N/C

Epy= Ep₁y+ Ep₂y=36.95 N/C-17.11N =19.84N/C

7 0

3 years ago

The energy of the electron in Hydrogen atom can be shown to be given by En = -13.6/n^2 eV, where n represents the principal quan

Tpy6a [65]

Answer:

This represents radiation in ultra-violet region .

Explanation:

Energy of the orbit where n = 3 is given as follows

$E_3 =- \frac{13.6 }{3^2}$

$E_3 =- \frac{13.6 }{9}$ = -1.511 eV

Energy of the orbit where n = 1 is given as follows

$E_1 =- \frac{13.6 }{1^2}$

$E_1 =- \frac{13.6 }{1}$ = 13.6 eV

Difference of [tex]E_3 and [tex]E_1 = - 1.511+ 13.6

= 12.089 eV.

The wavelength of light having this energy in nm is given by the expression as follows

Wavelength in nm = 1244 / energy in eV

= 1244 / 12.089

= 102.90 nm

This represents radiation in ultra-violet region .

8 0

3 years ago

An inventor develops a stationary cycling device by which an individual, while pedaling, can convert all of the energy expended

mr Goodwill [35]

Answer:

$Q=94185\ J$

Explanation:

Given:

mass of water, $m=0.3\ kg$

initial temperature of water, $T_i=20^{\circ}C$

final temperature of water, $T_f=95^{\circ}C$

specific heat of water, $c=4186\ J.kg^{-1}.K^{-1}$

<u>Now the amount of heat energy required:</u>

$Q=m.c.\Delta T$

$Q=0.3\times 4186\times (95-20)$

$Q=94185\ J$

Since all of the mechanical energy is being converted into heat, therefore the same amount of mechanical energy is required.

4 0

3 years ago

Consider two diffraction gratings. One grating has 3000 lines per cm, and the other one has 6000 lines per cm. Both gratings are

Usimov [2.4K]

Answer:

<em>The 6000 lines per cm grating, will produces the greater dispersion .</em>

Explanation:

A diffraction grating is an optical component with a periodic (usually one that has ridges or rulings on their surface rather than dark lines) structure that splits and diffracts light into several beams travelling in different directions.

The directions of the light beam produced from a diffraction grating depend on the spacing of the grating, and also on the wavelength of the light.

For a plane diffraction grating, the angular positions of principle maxima is given by

(a + b) sin ∅n = nλ

where

a+b is the distance between two consecutive slits

n is the order of principal maxima

λ is the wavelength of the light

From the equation, we can see that without sin ∅ exceeding 1, increasing the number of lines per cm will lead to a decrease between the spacing between consecutive slits.

In this case, light of the same wavelength is used. If λ and n is held constant, then we'll see that reducing the distance between two consecutive slits (a + b) will lead to an increase in the angle of dispersion sin ∅. So long as the limit of sin ∅ not greater that one is maintained.

7 0

3 years ago