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NeTakaya
2 years ago
13

How do you know that waves sent from the sun to earth are not mechanical waves? Explain

Physics
1 answer:
Anna71 [15]2 years ago
3 0
Mechanical waves require matter to travel medium. If there is no substance or matter to travel through, mechanical waves cannot propagate.

Since space is mostly vacuum, there is no medium for waves to travel through: waves sent by the sun (electromagnetic waves) are not mechanical waves
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Cells a and f shows an early and late stage of the same phase of mitosis . what phase is it?
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Calculate the minimum amount of work needed to move a 500-kg rocket payload from Earth's surface to the ISS in orbit 400,000m ab
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Answer:

2.000.000.000

Explanation:

Apply the formula:

Work = Force . Displacement

W = 500.10 . 400.000            (the 10 come from gravity)

W = 5000 . 400.000

W = 2.000.000.000 Joules

I think it is that, can be wrong.

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3 years ago
When electromagnetic fields interact with charged particles,
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a

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they became stronger

5 0
3 years ago
The intensity of light from a star (its brightness) is the power it outputs divided by the surface area over which it’s spread:
kow [346]

Answer:

\frac{d_{1}}{d_{2}}=0.36

Explanation:

1. We can find the temperature of each star using the Wien's Law. This law is given by:

\lambda_{max}=\frac{b}{T}=\frac{2.9x10^{-3}[mK]}{T[K]} (1)

So, the temperature of the first and the second star will be:

T_{1}=3866.7 K

T_{2}=6444.4 K

Now the relation between the absolute luminosity and apparent brightness  is given:

L=l\cdot 4\pi r^{2} (2)

Where:

  • L is the absolute luminosity
  • l is the apparent brightness
  • r is the distance from us in light years

Now, we know that two stars have the same apparent brightness, in other words l₁ = l₂

If we use the equation (2) we have:

\frac{L_{1}}{4\pi r_{1}^2}=\frac{L_{2}}{4\pi r_{2}^2}

So the relative distance between both stars will be:

\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{L_{1}}{L_{2}} (3)

The Boltzmann Law says, L=A\sigma T^{4} (4)

  • σ is the Boltzmann constant
  • A is the area
  • T is the temperature
  • L is the absolute luminosity

Let's put (4) in (3) for each star.

\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{A_{1}\sigma T_{1}^{4}}{A_{2}\sigma T_{2}^{4}}

As we know both stars have the same size we can canceled out the areas.

\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{T_{1}^{4}}{T_{2}^{4}}

\frac{d_{1}}{d_{2}}=\sqrt{\frac{T_{1}^{4}}{T_{2}^{4}}}

\frac{d_{1}}{d_{2}}=\sqrt{\frac{T_{1}^{4}}{T_{2}^{4}}}

\frac{d_{1}}{d_{2}}=0.36

I hope it helps!

5 0
3 years ago
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