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3 years ago

Explain how astronomers might use spectroscopy to determine the composition and temperature of a star.

1 answer:

4 0

Everything starts from spectroscopy. Astronomers only have concentrated information at wavelengths that are emitted from the stars. What they do with this information is to obtain the frequency range of the stars and through spectroscopes they are responsible for dividing the radiation beams and determining the coincidence with the emission of those same waves, of chemical elements. From these observation techniques it is possible to obtain the composition and according to the color, obtaining characteristics such as temperature. The spectrum of stars consists of dark and bright lines called Fraunhofer lines. This spectrum is compared to the spectrum of different elements to find the composition of the stars. This is possible because the elements emit or absorb only specific wavelengths.

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Calculate the ratio of the resistance of 12.0 m of aluminum wire 2.5 mm in diameter, to 30.0 m of copper wire 1.6 mm in diameter

alukav5142 [94]

Answer: 0.258

Explanation:

The resistance $R$ of a wire is calculated by the following formula:

$R=\rho\frac{l}{s}$ (1)

Where:

$\rho$ is the resistivity of the material the wire is made of. For aluminium is $\rho_{Al}=2.65(10)^{-8}m\Omega$ and for copper is $\rho_{Cu}=1.68(10)^{-8}m\Omega$

$l$ is the length of the wire, which in the case of aluminium is $l_{Al}=12m$ , and in the case of copper is $l_{Cu}=30m$

$s$ is the transversal area of the wire. In this case is a circumference for both wires, so we will use the formula of the area of the circumference:

$s=\pi{(\frac{d}{2})}^{2}$ (2) Where $d$ is the diameter of the circumference.

For aluminium wire the diameter is $d_{Al}=2.5mm=0.0025m$ and for copper is $d_{Cu}=1.6mm=0.0016m$

So, in this problem we have two transversal areas:

For aluminium:

$s_{Al}=\pi{(\frac{d_{AL}}{2})}^{2}=\pi{(\frac{0.0025m}{2})}^{2}$

$s_{Al}=0.000004908m^{2}$ (3)

For copper:

$s_{Cu}=\pi{\frac{(d_{Cu}}{2})}^{2}=\pi{(\frac{0.0016m}{2})}^{2}$

$s_{Cu}=0.00000201m^{2}$ (4)

Now we have to calculate the resistance for each wire:

Aluminium wire:

$R_{Al}=2.65(10)^{-8}m\Omega\frac{12m}{0.000004908m^{2}}$ (5)

$R_{Al}=0.0647\Omega$ (6) Resistance of aluminium wire

Copper wire:

$R_{Cu}=1.68(10)^{-8}m\Omega\frac{30m}{0.00000201m^{2}}$ (6)

$R_{Cu}=0.250\Omega$ (7) Resistance of copper wire

At this point we are able to calculate the ratio of the resistance of both wires:

$Ratio=\frac{R_{Al}}{R_{Cu}}$ (8)

$\frac{R_{Al}}{R_{Cu}}=\frac{0.0647\Omega}{0.250\Omega}$ (9)

Finally:

$\frac{R_{Al}}{R_{Cu}}=0.258$ This is the ratio

3 0

3 years ago

when you see something, your brain processes the visual information so that you are aware of it. Most of this processing occurs

Natasha_Volkova [10]

Front love of the brain
it focused on learning

5 0

3 years ago

Read 2 more answers

Ina shoots a large marble (Marble A, mass: 0.08 kg) at a smaller marble (Marble B, mass: 0.05 kg) that is sitting still. Marble

Neporo4naja [7]

Answer:

2.4 m/s

Explanation:

Momentum is conserved.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(0.08 kg)(0.5 m/s) + (0.05 kg)(0 m/s) = (0.08 kg)(-0.1 m/s) + (0.05 kg) v

0.04 kg m/s = -0.08 kg m/s + (0.05 kg) v

0.12 kg m/s = (0.05 kg) v

v = 2.4 m/s

4 0

3 years ago

If you measure the amount of work accomplished in a particular time interval, u have measured-

serious [3.7K]

Answer:

The power

Explanation:

We know that the work definition is given by the following expression:

W = F * d

where:

F = force [Newtons] [N]

d = distance [meters] [m]

W = work [Joules]

And the expression that defines the work done by unit of time is called - Power, therefore:

P = W/t

where:

P = power [watts] [w]

W = work [Joules] [J]

t = time [seconds] [s]

6 0

3 years ago

if you have to apply 40 n of force on a crowbar to lift a 400 n rock what is the actual machanical andvatage of the crow bar

Eduardwww [97]

M.A. = 400/40

M.A. = 10

Hope this helps!

7 0

3 years ago