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givi [52]
3 years ago
7

Name five characteristics of a star that can be determined by measuring its spectrum. Explain how you would use a spectrum to de

termine these characteristics.
Physics
1 answer:
Ne4ueva [31]3 years ago
6 0

Answer:

Chemical composition, Temperature, Radial velocity, Size or diameter of the star, Rotation.

Explanation:  

Elemental abundances are determined by analyzing the relative strengths of the absorption lines in the spectrum of a star.

The Spectral class to which the star belongs gives the information related to the temperature of the star. It is the spectral lines that determine the spectral class O B A F G K M are the spectral classes.

By measuring the wavelengths of the lines in the star's spectrum gives the radial velocity. Doppler shift is the method used to find the radial velocity.

A star can be classified as a giant or a dwarf . A giant star will have narrow width spectral lines whereas a dwarf star has wider spectral lines.

Broadening of the spectral lines will determine the star's rotation.

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The law of reflection states that if the angle of incidence is 32 degrees, the angle of reflection is ___ degrees.
shusha [124]

The angle of reflection is  equal to angle of incidence so the angle of reflection is also 32°.

6 0
3 years ago
Which term below describes a measurement of how hard an object pushes against a surface?
Scilla [17]

I hope the answer is D. Pressure

8 0
2 years ago
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How much work is done if a 15 newton suitcase is lifted 2 m ?​
nirvana33 [79]

Answer:

196 J

Explanation:

3 0
2 years ago
A cylindrical container closed of both end has a radius of 7cm and height of 6cm A.)find the total surface area of the container
Natalka [10]

Given:

A cylindrical container closed of both end has a radius of 7cm and height of 6cm.

Explanation:

A.) Find the total surface area of the container.

  • A = 2πrh + 2πr²
  • A = 2(3.14)(7)(6) + 2(3.14)(7 × 7)
  • A = 263.76 + 307.72
  • A = 571.48

B.) Find the volume of the container.

  • V = πr²h
  • V = (3.14)(7×7)(6)
  • V = 923.16

Not sure huhuness.

#CarryOnLearning

8 0
3 years ago
Read 2 more answers
Physics B 2020 Unit 3 Test
weqwewe [10]

Answer:

1)

When a charge is in motion in a magnetic field, the charge experiences a force of magnitude

F=qvB sin \theta

where here:

For the proton in this problem:

q=1.602\cdot 10^{-19}C is the charge of the proton

v = 300 m/s is the speed of the proton

B = 19 T is the magnetic field

\theta=65^{\circ} is the angle between the directions of v and B

So the force is

F=(1.602\cdot 10^{-19})(300)(19)(sin 65^{\circ})=8.28\cdot 10^{-16} N

2)

The magnetic field produced by a bar magnet has field lines going from the North pole towards the South Pole.

The density of the field lines at any point tells how strong is the magnetic field at that point.

If we observe the field lines around a magnet, we observe that:

- The density of field lines is higher near the Poles

- The density of field lines is lower far from the Poles

Therefore, this means that the magnetic field of a magnet is stronger near the North and South Pole.

3)

The right hand rule gives the direction of the  force experienced by a charged particle moving in a magnetic field.

It can be applied as follows:

- Direction of index finger = direction of motion of the charge

- Direction of middle finger = direction of magnetic field

- Direction of thumb = direction of the force (for a negative charge, the direction must be reversed)

In this problem:

- Direction of motion = to the right (index finger)

- Direction of field = downward (middle finger)

- Direction of force = into the screen (thumb)

4)

The radius of a particle moving in a magnetic field is given by:

r=\frac{mv}{qB}

where here we have:

m=6.64\cdot 10^{-22} kg is the mass of the alpha particle

v=2155 m/s is the speed of the alpha particle

q=2\cdot 1.602\cdot 10^{-19}=3.204\cdot 10^{-19}C is the charge of the alpha particle

B = 12.2 T is the strength of the magnetic field

Substituting, we find:

r=\frac{(6.64\cdot 10^{-22})(2155)}{(3.204\cdot 10^{-19})(12.2)}=0.366 m

5)

The cyclotron frequency of a charged particle in circular motion in a magnetic field is:

f=\frac{qB}{2\pi m}

where here:

q=1.602\cdot 10^{-19}C is the charge of the electron

B = 0.0045 T is the strength of the magnetic field

m=9.31\cdot 10^{-31} kg is the mass of the electron

Substituting, we find:

f=\frac{(1.602\cdot 10^{-19})(0.0045)}{2\pi (9.31\cdot 10^{-31})}=1.23\cdot 10^8 Hz

6)

When a charged particle moves in a magnetic field, its path has a helical shape, because it is the composition of two motions:

1- A uniform motion in a certain direction

2- A circular motion in the direction perpendicular to the magnetic field

The second motion is due to the presence of the magnetic force. However, we know that the direction of the magnetic force depends on the sign of the charge: when the sign of the charge is changed, the direction of the force is reversed.

Therefore in this case, when the particle gains the opposite charge, the circular motion 2) changes sign, so the path will remains helical, but it reverses direction.

7)

The electromotive force induced in a conducting loop due to electromagnetic induction is given by Faraday-Newmann-Lenz:

\epsilon=-\frac{N\Delta \Phi}{\Delta t}

where

N is the number of turns in the loop

\Delta \Phi is the change in magnetic flux through the loop

\Delta t is the time elapsed

From the formula, we see that the emf is induced in the loop (and so, a current is also induced) only if \Delta \Phi \neq 0, which means only if there is a change in magnetic flux through the loop: this occurs if the magnetic field is changing, or if the area of the loop is changing, or if the angle between the loop and the field is changing.

8)

The flux is calculated as

\Phi = BA sin \theta

where

B = 5.5 T is the strength of the magnetic field

A is the area of the coil

\theta=18^{\circ} is the angle between the  direction of the field and the plane of the loop

Here the loop is rectangular with lenght 15 cm and width 8 cm, so the area is

A=(0.15 m)(0.08 m)=0.012 m^2

So the flux is

\Phi = (5.5)(0.012)(sin 18^{\circ})=0.021 Wb

See the last 7 answers in the attached document.

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5 0
3 years ago
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