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2 years ago

How do we know the sun rotate

Physics

2 answers:

Dima020 [189]2 years ago

8 0

Answer: One side is blue-shifted and the other side is red-shifted.

Explanation: APEX 12/31/19

scoundrel [369]2 years ago

6 0

Answer:

From the movement of sunspots, Galileo discovered that sun rotate s on its own axis.

Explanation:

All the sunspots are traveling across the Sun's head. This movement is part of the Sun's general rotation of its axis. Observations also suggest that the Sun does not rotate like a solid body, but rotates differently because it is a gas. Actually the Sun is spinning faster at its equator than at at its poles. The Sun rotates once every 24 days at its equator, but only once every 35 days at its poles. We learn this by observing the movement of sunspots and other solar features pass through the Sun.

You might be interested in

What is the difference in KE between a 52.5 kg person running 3.50 m/s and a 0.0200 kg bullet flying 450 m/s?

rusak2 [61]

Answer:

Ek = 1705.28 [J]

Explanation:

In order to solve this problem, we must remember that kinetic energy can be calculated by means of the following equation.

$E_{k}=\frac{1}{2} *m*v^{2}$

where:

m = mass [kg]

v = velocity [m/s]

Ek = kinetic energy [J] (Units of Joules)

For the person running

$E_{k} =\frac{1}{2}*52.2*(3.5)^{2} \\ E_{k} =319.72[J]$

For the bullet

$E_{k} =\frac{1}{2} *m*v^{2}$

$E_{k} =\frac{1}{2} *0.02*(450)^{2} \\E_{k}=2025 [J]$

The difference in Kinetic energy is equal to:

Ek = 2025 - 319.72

Ek = 1705.28 [J]

8 0

2 years ago

Describe how an oscilloscope should be used to measure the frequency of the sound wave from the sonometer

ratelena [41]

Answer:

T = reading (cm) time base (s / cm)

f = 1 / T

Explanation:

An oscilloscope is a piece of equipment that allows you to visualize and measure a wave that reaches you, in the case of having a sonometer this transforms the sound wave into an electrical signal to be introduced through one of the voltage channels of the equipment, on the screen we will see the oscillating alternating signal, if it is fixed we can make the reading, if it is moving the time base and the trigger must be adjusted to stop it.

In the oscilloscope we can read the period of the signal, this is the time it takes for the signal to repeat itself with this value, we can calculate the frequency with the formula, for the reading of the period the distance is measured on the labeled screen and multiplied by the time base

T = reading (cm) time base (s / cm)

f = 1 / T

6 0

2 years ago

An electron accelerates through a 12.5 V potential difference, starting from rest, and then collides with a hydrogen atom, excit

Yanka [14]

Answer:

Initial state Final state

3 ⇒ 2

3 ⇒ 1

2 ⇒ 1

Explanation:

For this exercise we must use Bohr's atomic model

E = - 13.606 / n²

where is the value of 13.606 eV is the energy of the ground state and n is the integer.

The energy acquired by the electron in units of electron volt (eV)

E = e V

E = 12.5 eV

all this energy is used to transfer an electron from the ground state to an excited state

ΔE = 13.6060 (1 / n₀² - 1 / n²)

the ground state has n₀ = 1

ΔE = 13.606 (1 - 1/n²)

1 /n² = 1 - ΔE/13,606

1 / n² = 1 - 12.5 / 13.606

1 / n² = 0.08129

n = √(1 / 0.08129)

n = 3.5

since n is an integer, maximun is

n = 3

because it cannot give more energy than the electron has

From this level there can be transition to reach the base state.

Initial state Final state

3 ⇒ 2

3 ⇒ 1

2 ⇒ 1

8 0

2 years ago

Find the fundamental frequency and the next three frequencies that could cause standing-wave patterns on a string that is 30.0 m

maksim [4K]

Answer:

0.786 Hz, 1.572 Hz, 2.358 Hz, 3.144 Hz

Explanation:

The fundamental frequency of a standing wave on a string is given by

$f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}$

where

L is the length of the string

T is the tension in the string

$\mu$ is the mass per unit length

For the string in the problem,

L = 30.0 m

$\mu=9.00\cdot 10^{-3} kg/m$

T = 20.0 N

Substituting into the equation, we find the fundamental frequency:

$f=\frac{1}{2(30.0)}\sqrt{\frac{20.0}{(9.00\cdot 10^{-3}}}=0.786 Hz$

The next frequencies (harmonics) are given by

$f_n = nf$

with n being an integer number and f being the fundamental frequency.

So we get:

$f_2 = 2 (0.786 Hz)=1.572 Hz$

$f_3 = 3 (0.786 Hz)=2.358 Hz$

$f_4 = 4 (0.786 Hz)=3.144 Hz$

6 0

3 years ago

How should the distance, D, between two point charges, q1 and q2, be changed to double their electric potential energy

Licemer1 [7]

The electrostatic potential energy, U, of one point charge q at position d in the presence of an electric field E is defined as the negative of the work W done by the electrostatic force to bring it from the reference position d to that position

$u \: = \: \frac{kq1q2}{d}$

Thus, to double the electric potential energy U we need to reduce the distance of separation by half (1/2) because they are inversely proportion

6 0

3 years ago