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3 years ago

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An x-ray has a wavelength of 4.18 Å. Calculate the energy (in J) of one photon of this radiation. Enter your answer in scientifi

c notation

1 answer:

Damm [24]3 years ago

4 0

Answer:

E = 4.75 x 10⁻¹⁶ J

Explanation:

given,

wavelength of the x-ray , λ = 4.18 Å

Energy of photon = ?

we know

$E = \dfrac{hc}{\lambda}$

where h is the planks constant

c is the speed of light

h = 6.626 x 10⁻³⁴ m² kg / s

c = 3 x 10⁸ m/s

now,

$E = \dfrac{6.626\times 10^{-34}\times 3 \times 10^8}{4.18\times 10^{-10}}$

E = 4.75 x 10⁻¹⁶ J

hence, the energy of the photon is equal to E = 4.75 x 10⁻¹⁶ J

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To teach you how to find the parameters characterizing an object in a circular orbit around a much heavier body like the earth.

Karolina [17]

Answer:

a) T² = ( $\frac{4\pi ^2}{GM}$ ) r³

b) veloicity the dependency is the inverse of the root of the distance

kinetic energy depends on the inverse of the distance

potential energy dependency is the inverse of distance

angular momentum depends directly on the root of the distance

Explanation:

1) for this exercise we will use Newton's second law

F = ma

in this case the acceleration is centripetal

a = v² / r

the linear and angular variable are related

v = w r

we substitute

a = w² r

force is the universal force of attraction

F = $G \frac{m M}{r^2}$

we substitute

$G \frac{m M}{r^2} = m w^2 r$

w² = $\frac{GM}{r^3}$

angular velocity is related to frequency and period

w = 2π f = 2π / T

we substitute

$( \frac{2\pi }{T} ) = \frac{GM}{r^3}$

the final equation is

T² = () r³

b) the speed of the orbit can be found

v = w r

v = $\sqrt{\frac{GM}{r^3} } \ r$

v = $\sqrt{\frac{GM}{r} }$

in this case the dependency is the inverse of the root of the distance

Kinetic energy

K = ½ M v²

K = ½ M GM / r

K = ½ GM² 1 / r

the kinetic energy depends on the inverse of the distance

Potential energy

U =

U = -G mM / r

dependency is the inverse of distance

Angular momentum

L = r x p

for a circular orbit

L = r p = r Mv

L =

L =

The angular momentum depends directly on the root of the distance

8 0

3 years ago

there is only one basic format for a professional report. a. audience b. All of these should be considered when deciding on a re

Anna007 [38]

Answer:

The answer is letter b. All of these should be considered when deciding on a report format.

Explanation:

A Professional Report is a type of formal document about a topic or information that is intended for a specific audience or purpose. The report's style of writing needs a lot of knowledge from the writer. Oftentimes, it involves the following important elements: <em>Title, Summary, Body, Discussion, Conclusion and Recommendation. </em>

The writer should write according to his target audience and purpose. He also needs to consider the length of his report, as well as the suitable words and sentences that he should use.

Thus, all of the choices are important in writing a professional report. So, the answer is letter b.

5 0

3 years ago

A mass of 1.00×10−6 μkg is the same as

tankabanditka [31]

I'm not sure but I know u is 10^6

6 0

3 years ago

Jupiter is made of gas(like Saturn, Uranus and Neptune). What would happen to the strength of gravity if you

garik1379 [7]

Answer:

a) The strength of gravity decreases if one moved away from Jupiter

b) The strength of gravity increases if one fell into Jupiter

Explanation:

The gravitational attraction is given by Newton law of gravitation as follows;

$Force \ (strength) \ of \ gravity = \dfrac{G \times M \times m}{R^2}$

Where;

G = The universal gravitational constant = 6.67408 × 10⁻¹¹ m³/(kg·s²)

M = The mass of Jupiter

m = The mass of the nearby body

R = The distance between the centers of Jupiter and the body

From the equation, we have that the gravitational strength varies inversely with the square of the separation distance between two bodies

Therefore, as one moves away, R increases, and the strength of gravity reduces

Similarly as the body falls into Jupiter, R, reduces the gravitational strength increases.

7 0

2 years ago

You stand on a straight desert road at night and observe a vehicle approaching. This vehicle is equipped with two small headligh

spayn [35]

To solve this problem we will apply the concepts related to Reyleigh's criteria. Here the resolution of the eye is defined as 1.22 times the wavelength over the diameter of the eye. Mathematically this is,

$\theta = \frac{1.22 \lambda }{D}$

Here,

D is diameter of the eye

$D = \frac{1.22 (539nm)}{5.11 mm}$

$D= 1.287*10^{-4}m$

The angle that relates the distance between the lights and the distance to the lamp is given by,

$Sin\theta = \frac{d}{L}$

For small angle, $sin\theta = \theta$

$sin \theta = \frac{d}{L}$

Here,

d = Distance between lights

L = Distance from eye to lamp

For small angle $sin \theta = \theta$

Therefore,

$L = \frac{d}{sin\theta}$

$L = \frac{0.691m}{1.287*10^{-4}}$

$L = 5367m$

Therefore the distance is 5.367km.

4 0

3 years ago