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MrRa [10]
3 years ago
12

A graduate student has done a careful analysis of the spectrum of a star. While she has found lines from many elements, there wa

s not a trace of the element helium in the spectra she has been analyzing. From this she can now conclude:_______
Physics
1 answer:
disa [49]3 years ago
7 0

Explanation:

The lines in the spectra of star is due to absorption of light emitted by star by different elements  present in atmosphere of the star .

If some element is present in the outer atmosphere of he star , it will absorb a particular portion of light , for which there will be black line in the spectrum .

These are called Fraunhoffer's lines .

Now there is no such line at the place of Helium in the light spectrum . This means that Helium element is not present in the outer atmosphere of the star.

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A plane traveling 1001 meters over the ocean at a speed of 135 meters/sec will drop a box of supplies for shipwreck victims belo
Lubov Fominskaja [6]

Answer:

600 metera

Explanation:

sana makatulong

8 0
3 years ago
In an experiment, a rectangular block with height h is allowed to float in two separate liquids. In the first liquid, which is w
Amiraneli [1.4K]

Answer:

The relative density of the second liquid is 7.

Explanation:

From archimede's principle we know that the force that a liquid exerts on a object equals to the weight of the liquid that the object displaces.

Let us assume that the volume of the object is 'V'

Thus for the liquid in which the block is completely submerged

The buoyant force should be equal to weight of liquid

Mathematically

F_{buoyant}=Weight\\\\\rho _{1}\times V\times g=m\times g\\\\\therefore \rho _{1}=\frac{m}{V}...............(i)

Thus for the liquid in which the block is 1/7 submerged

The buoyant force should be equal to weight of liquid

Mathematically

F'_{buoyant}=Weight\\\\\rho _{2}\times \frac{V}{7}\times g=m\times g\\\\\therefore \rho _{2}=\frac{7m}{V}.................(ii)

Comparing equation 'i' and 'ii' we see that

\rho_{2}=7\times \rho _{1}

Since the first liquid is water thus \rho _{1}=1gm/cm^3

Thus the relative density of the second liquid is 7.

6 0
3 years ago
Read 2 more answers
What is the frequency and wavelength, in nanometers, of photons capable of just ionizing nitrogen atoms?
nika2105 [10]

Answer:

The frecuency and wavelength of a photon capable to ionize the nitrogen atom are ν = 3.394×10¹⁵ s⁻¹  and λ = 88.31 nm.

Explanation:It is possible to know what are the frequency and wavelength of a photon capable to ionize the nitrogen atom using the equation of the energy of a photon described below.

E = hc/λ  (1)

Where h is the Planck constant, c is the speed of light and λ is the wavelength of the photon.

But first, it is neccesary to know the ionization energy of the nitrogen atom. The ionization energy is the energy needed to remove an electron from an atom, for the Nitrogen atom it will lose an electron of its outer orbit from the nucleus, farther snuff, so the electric force is weaker. Experimentally, it is known that it has a value of 14.04 eV. This value is easy to found in a periodic table.

So the nitrogen atom will need a photon with the energy of 14.04 eV to remove the electron from its outer orbit.

Replacing the Planck constant, the speed of light and the energy of the photon in the equation 1, the wavelength can be calculated:

λ = hc/E  (2)

Where h = 6.626×10⁻³⁴ J.s and c = 3.00×10⁸ m/s

But the Planck constant can be expressed in electron volts:

1 eV = 1.602 x 10⁻¹⁹ J

h = 6.626x10⁻³⁴ J/1.602x10⁻¹⁹ J . eV .s

h= 4.136x10⁻¹⁵ eV.s

Now, it is convenient to express the speed of light in nanometers:

1nm = 1x10⁻⁹ m

c = 3.00x10⁸ m/ 1x10⁻⁹ m

c = 3x10¹⁷ nm/s

Substituting in equation 2:

λ =  (4.136x10⁻¹⁵ eV.s)(3x10¹⁷ nm/s)/14.04 eV

λ = 1240 eV. nm/ 14.04 eV

λ = 88.31 nm

The frenquency is calculated using the equation 2 in the following way:

E = hν  (3)

Where ν is the frecuency

ν = E/h

ν = 14.04 eV/4.136×10⁻¹⁵ eV.s

ν = 3.394×10¹⁵ s-1

So the frecuency of a photon, capable to ionize the nitrogen atom, will be 3.394×10¹⁵ s⁻¹ and its wavelength 88.31 nm.

4 0
4 years ago
A student uses a spring loaded launcher to launch a marble vertically in the air. The mass of the marble is
GarryVolchara [31]

Answer:

Part a)

When spring compressed by 2 cm

H = 1.47 m

Part b)

When spring is compressed by 4 cm

H = 5.94 m

Explanation:

Part a)

As we know that the spring is compressed and released

so here spring potential energy is converted into gravitational potential energy at its maximum height

So we will have

\frac{1}{2}kx^2 = mg(H + x)

0.5(220)(0.02)^2 = 0.003(9.81)(H + 0.02)

so we have

H = 1.47 m

Part b)

Similarly when spring is compressed by 4 cm

then we have

\frac{1}{2}kx^2 = mg(H + x)

0.5(220)(0.04)^2 = 0.003(9.81)(H + 0.04)

so we have

H = 5.94 m

8 0
4 years ago
What is the wavelength of a bullet that is 0.450 g traveling at 2000. m/s?
ziro4ka [17]
Λ= 7.36E - 34 m

Use the Broglie wavelength equation. 

λ= { h/ mv} * SQRT {1- (v^2/ c^2)}
7 0
3 years ago
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