Which of the following is an example of a latitude line

The ph balance is maintained by which organ?

Lena [83]

The Kidneys And Lungs.

7 0

3 years ago

Calculate the energy, wavelength, and frequency of the emitted photon when an electron moves from an energy level of -3.40 eV to

jasenka [17]

Answer:

(a) The energy of the photon is 1.632 x $10^{-8}$ J.

(b) The wavelength of the photon is 1.2 x $10^{-17}$ m.

(c) The frequency of the photon is 2.47 x $10^{25}$ Hz.

Explanation:

Let;

$E_{1}$ = -13.60 ev

$E_{2}$ = -3.40 ev

(a) Energy of the emitted photon can be determined as;

$E_{2}$ - $E_{1}$ = -3.40 - (-13.60)

= -3.40 + 13.60

= 10.20 eV

= 10.20(1.6 x $10^{-9}$ )

$E_{2}$ - $E_{1}$ = 1.632 x $10^{-8}$ Joules

The energy of the emitted photon is 10.20 eV (or 1.632 x $10^{-8}$ Joules).

(b) The wavelength, λ, can be determined as;

E = (hc)/ λ

where: E is the energy of the photon, h is the Planck's constant (6.6 x $10^{-34}$ Js), c is the speed of light (3 x $10^{8}$ m/s) and λ is the wavelength.

10.20(1.6 x $10^{-9}$ ) = (6.6 x $10^{-34}$ * 3 x $10^{8}$ )/ λ

λ = $\frac{1.98*10^{-25} }{1.632*10^{-8} }$

= 1.213 x $10^{-17}$

Wavelength of the photon is 1.2 x $10^{-17}$ m.

(c) The frequency can be determined by;

E = hf

where f is the frequency of the photon.

1.632 x $10^{-8}$ = 6.6 x $10^{-34}$ x f

f = $\frac{1.632*10^{-8} }{6.6*10^{-34} }$

= 2.47 x $10^{25}$ Hz

Frequency of the emitted photon is 2.47 x $10^{25}$ Hz.

6 0

3 years ago

The Earth moving round the Sun in a circular orbit is acted upon by a

zalisa [80]

No I do not agree. It is because work is done when force acting o a body displace or covers certain displacement in the direction of force applied .

3 0

3 years ago

A car of mass m=1000kg is traveling at speed v and brakes. The skid marks are 20m long and the coefficient of kinetic friction i

scoray [572]

Answer:

v = 14 m/s

= 31.3 mph

The answer would be the same if the mass of the car were 2000 kg

Explanation:

Let V be the final velocity of the car after skidding, and v be the initial velocity of the car. Let a be the acceleration of the car and Δx be the distance the car travels after applying brakes (length of the skid marks). Let Fk be the force of friction between the tyres and the road. Let N be the normal force exerted on the car and μ be the co efficient of kinetic friction.

V^2 = v^2 + 2×a×Δx

Now V, the final velocity is zero as the car stops

0 = v^2 + 2×a×Δx

v^2 = -2×a×Δx

v =√-2×a×Δx .....*

Now applying Newton's Second Law

Fnet = m×a

-Fk = m×a

-μ×N = m×a

-μ×m×g = m×a (The mass cancels out)

a = -μ×g

Substituting the value of a back to equation *

v = √-2×(-μ×g)×Δx

v = √-2×(-0.5×9.8)×20

v = 14 m/s

Therefore the speed the car was travelling with v = 14 m/s

which is equal to 31.3 mph

Now if you were to change the mass of the car to 2000 kg the value for v would still be the same. As it is seen above mass cancels out so it does not influence or affect the value of the velocity obtained.

8 0

3 years ago

Which of these forms of radiation passes most easily through the disk of the Milky Way?

Rama09 [41]

<h2>Answer: 3 - infrared light</h2>

Explanation:

<u>There are certain areas of the Milky Way that cannot be observed using the visible range of the electromagnetic spectrum</u> (this includes blue light and red light). This is because these areas are covered or hidden behind columns of interstellar dust and dark matter.

However, using infrared light and sometimes radio waves, it is possible to observe the galaxy better, because this light manages to pass through all that interstellar dust.

4 0

3 years ago