What would happen if mass is continually added to a 1.4 solar mass neutron star?

What type of stress is shown in the following diagram?

leva [86]

Here stress is parallel to the surface of the body. So it's a Shear stress.

7 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.

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

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

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

I hope it helps!

5 0

3 years ago

A rigid tank contains 7 kg of an ideal gas at 5 atm and 30c. a valve is opened, and half of mass of the gas can escape. the fin

Readme [11.4K]

The equation of state for an ideal gas is
$pV=nRT$
where p is the gas pressure, V the volume, n the number of moles, R the gas constant and T the temperature.

The equation of state for the initial condition of the gas is
$p_1 V_1 = n_1 R T_1$ (1)
While the same equation for the final condition is
$p_2 V_2 = n_2 R T_2$ (2)

We know that in the final condition, half of the mass of the gas is escaped. This means that the final volume of the gas is half of the initial volume, and also that the final number of moles is half the initial number of moles, so we can write:
$V_1 = 2 V_1$
$n_1 = 2 n_2$
If we substitute these relationship inside (1), and we divide (1) by (2), we get
$\frac{p_1}{p_2} = \frac{T_1}{T_2}$

And since the initial temperature of the gas is $T_1 = 30 C=303 K$ , we can find the final temperature of the gas:
$T_2 = T_1 \frac{p_2}{p_1}=(303 K) \frac{1.5 atm}{5.0 atm}=90.9 K$

6 0

3 years ago

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A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 4.0 m/s. The car is a distanc

Brums [2.3K]

For this specific problem, the maximum value for d is 52m. I am hoping that this answer has satisfied your query about and it will be able to help you, and if you’d like, feel free to ask another question.

4 0

2 years ago

What is matter and explain

Mice21 [21]

Matter is any substance that has mass and takes up space by having volume.

3 0

3 years ago

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