A hot iron is turned off and cools down to room temperature. The iron cools because:

Which element in the 18th column of the periodic table (shown below) has the largest radius?

Alex17521 [72]

Answer: Rn :)))) no explanation needed

3 0

2 years ago

. How does the area of a rectangle relate to the lengths of each side?

iragen [17]

Area is 2-dimensional which means it has a length and a width area is measured in square units such as square inches, square feet or square meters. To find the area of a rectangle multiply the length by the width.

3 0

3 years ago

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a runner increases her speed from 3.1 m/s to 3.5 m/s during the last 15 seconds of her run, what was her acceleration during her

Rama09 [41]

Acceleration = velocity/time
A= 3.5m/s/15s
A= 0.23m/s^2

5 0

3 years ago

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Free point cuz i cewl like dat‍♀️‍♂️

DedPeter [7]

Yes thank u teehee

.................... x

8 0

2 years ago

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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.

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

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

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

I hope it helps!

5 0

3 years ago