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2 years ago

Select the star life cycle that is accurate.

Physics

1 answer:

Veseljchak [2.6K]2 years ago

8 0

Answer:

B according to the picture

Explanation:

Star of 3 stellar masses => red supergiant=>supernova=>neutron star

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To model this process, assume two charged spherical conductors are connected by a long conducting wire and a 1.20-mC charge is p

il63 [147K]

Answer:

Part a: The electric potential of each sphere is 1.35x10⁸V

Part b: The electric field at the surface of sphere 1 and 2 is 2.25x10⁹ N/C and 6.75x10⁹ N/C respectively

Explanation:

As the complete question is not given, the similar question is attached herewith. The values are used as indicated in the given question

Let r_1 = 6 cm=0.06 m

r2 = 2 cm = 0.02 m

Q = 1.2 mC

Let q1 and q2 are the charges on each sphere.

q1 + q2 = 1.2 mC -------(1)

In the equilibrium, V1 = V2

k*q1/r1 = k*q2/r2

q1/0.06 = q2/0.02

q1/q2 = 0.06/0.02

q1/q2 = 3 ---------(2)

On solving equation 1 and 2

we get

q1 = 0.9 mC

q2 = 0.3 mC

V1 = k*q1/r1 = (9*10^9*0.9*10^-3)/0.06 = 1.35*10^8 Volts

V2 = k*q2/r2 = 9*10^9*0.3*10^-3/0.02 = 1.35*10^8 Volts

So the electric potential of each sphere is 1.35x10⁸V

Part b

Now the electric potential is given as

E1 = k*q1/r1^2 = 9*10^9*0.9*10^-3/0.06^2 = 2.25*10^9 N/C

E2 = k*q2/r2 = 9*10^9*0.3*10^-3/0.02^2 = 6.75*10^9 N/C

So the electric field at the surface of sphere 1 and 2 is 2.25x10⁹ N/C and 6.75x10⁹ N/C respectively

7 0

3 years ago

HELP MEEEEE HURRRYYYY EASYY

alisha [4.7K]

Answer:

Explanation:

thats what i think it depends on the amount of force someone dropped the ball with.

7 0

3 years ago

Using what you know about the periodic table, choose all of the correct answers,

Klio2033 [76]

Answer:

What are the choices?

7 0

3 years ago

Ejection of Electrons from Hydrogen by Incident Photons Light of wavelength 80 nm is incident on a sample of hydrogen gas, resul

timofeeve [1]

Answer:

a) $K_{max}$ = 1.9 eV = 3.04 10⁻¹⁹ J,b ) This means that some electrons are at the first excited level of the hydrogen atom, which is highly likely as the temperature rises.

Explanation:

a) To calculate the maximum kinetic energy of the expelled electrons let's use the relationships of the photoelectric effect

$K_{max}$ = h f - Φ

Where K is the kinetic energy, h the Planck constant that is worth 6.63 10⁻³⁴ Js, f the frequency and Φ the work function

The speed of light is related to wavelength and frequency

c = λ f

Let's analyze the work function, it is the energy needed to start an electron from a metal, in this case to start an electron from a hydrogen atom its fundamental energy is needed, so

Φ= E₀ = 13.6 eV

let's replace and calculate the energy of the incident photon

E = h c / λ

E = 6.63 10⁻³⁴ 3 10⁸/80 10⁻⁹

E = 2,486 10⁻¹⁸ J

Let's reduce to eV

E = 2,486 10⁻¹⁸ (1 eV / 1.6 10⁻¹⁹)

E = 15.5 eV

Now we can calculate the kinetic energy

$K_{max}$ = h c / f - fi

$K_{max}$ = 15.5 -13.6

$K_{max}$ = 1.9 eV

b) Extra energy = 10.2 eV

The total kinetic energy of electrons is

Total kinetic energy = 1.9 +10.2 = 12.1 eV

For the calculation we are assuming that all the electors are in the hydrogen base state, but for temperatures greater than 0K some electors may be in some excited state, so less energy is needed to tear them out of hydrogen atom.

Let's analyze this possibility

ΔE = E photon - Total kinetic energy electron

ΔE = 15.5 - 12.1

ΔE = 3.4 eV

If we use the Bohr ratio for the hydrogen atom

$E_{n}$ = 13.606 / n2

n = √ 13.606 / En

n = √ (13606 / 3.4)

n = 2

This means that some electrons are at the first excited level of the hydrogen atom, which is highly likely as the temperature rises.

8 0

3 years ago

Knowing the constant g what will the gravitational force between two masses be if the gravitational force between them is 36n an

charle [14.2K]

The gravitational force between two masses is given by:
$F=G \frac{m_1 m_2}{r^2}$
where
G is the gravitational constant
m1 and m2 are the two masses
r is the separation between the two masses

We see that the force is proportional to the inverse of the square of the distance: $F \sim \frac{1}{r^2}$
therefore, if the distance is tripled:
r'=3r
The force decreases by a factor 1/9:
$F \sim \frac{1}{(3r)^2}= \frac{1}{9} \frac{1}{r^2}$

Since the original force was 36 N, the new force will be
$F' = \frac{1}{9} (36 N)= 4 N$

6 0

3 years ago