Answer:
Explanation:
1. We can find the temperature of each star using the Wien's Law. This law is given by:
(1)
So, the temperature of the first and the second star will be:
![T_{1}=3866.7 K](https://tex.z-dn.net/?f=T_%7B1%7D%3D3866.7%20K)
![T_{2}=6444.4 K](https://tex.z-dn.net/?f=T_%7B2%7D%3D6444.4%20K)
Now the relation between the absolute luminosity and apparent brightness is given:
(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}](https://tex.z-dn.net/?f=%5Cfrac%7BL_%7B1%7D%7D%7B4%5Cpi%20r_%7B1%7D%5E2%7D%3D%5Cfrac%7BL_%7B2%7D%7D%7B4%5Cpi%20r_%7B2%7D%5E2%7D)
So the relative distance between both stars will be:
(3)
The Boltzmann Law says,
(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}}](https://tex.z-dn.net/?f=%5Cleft%28%5Cfrac%7Bd_%7B1%7D%7D%7Bd_%7B2%7D%7D%5Cright%29%5E%7B2%7D%3D%5Cfrac%7BA_%7B1%7D%5Csigma%20T_%7B1%7D%5E%7B4%7D%7D%7BA_%7B2%7D%5Csigma%20T_%7B2%7D%5E%7B4%7D%7D)
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}}](https://tex.z-dn.net/?f=%5Cleft%28%5Cfrac%7Bd_%7B1%7D%7D%7Bd_%7B2%7D%7D%5Cright%29%5E%7B2%7D%3D%5Cfrac%7BT_%7B1%7D%5E%7B4%7D%7D%7BT_%7B2%7D%5E%7B4%7D%7D)
![\frac{d_{1}}{d_{2}}=\sqrt{\frac{T_{1}^{4}}{T_{2}^{4}}}](https://tex.z-dn.net/?f=%5Cfrac%7Bd_%7B1%7D%7D%7Bd_%7B2%7D%7D%3D%5Csqrt%7B%5Cfrac%7BT_%7B1%7D%5E%7B4%7D%7D%7BT_%7B2%7D%5E%7B4%7D%7D%7D)
I hope it helps!
what?theres no picture. what does this mean?
Answer:
W = 0.562 J
Explanation:
Initial potential energy of two point charges is given as
![U_i = \frac{kq_1q_2}{r_1}](https://tex.z-dn.net/?f=U_i%20%3D%20%5Cfrac%7Bkq_1q_2%7D%7Br_1%7D)
here we have
![q_1 = 3.10 \mu C](https://tex.z-dn.net/?f=q_1%20%3D%203.10%20%5Cmu%20C)
![q_2 = -4.50 \mu C](https://tex.z-dn.net/?f=q_2%20%3D%20-4.50%20%5Cmu%20C)
![r_1 = 0.140 m](https://tex.z-dn.net/?f=r_1%20%3D%200.140%20m)
now we have
![U_i = \frac{(9\times 10^9)(3.10 \mu C)(-4.50 \mu C)}{0.140}](https://tex.z-dn.net/?f=U_i%20%3D%20%5Cfrac%7B%289%5Ctimes%2010%5E9%29%283.10%20%5Cmu%20C%29%28-4.50%20%5Cmu%20C%29%7D%7B0.140%7D)
![U_i = -0.897 J](https://tex.z-dn.net/?f=U_i%20%3D%20-0.897%20J)
now at final position the distance between two charges is given as
![r_2 = \sqrt{0.265^2 + 0.265^2} = 0.375 m](https://tex.z-dn.net/?f=r_2%20%3D%20%5Csqrt%7B0.265%5E2%20%2B%200.265%5E2%7D%20%3D%200.375%20m)
Now final energy is given as
![U_f =\frac{kq_1q_2}{r_2}](https://tex.z-dn.net/?f=U_f%20%3D%5Cfrac%7Bkq_1q_2%7D%7Br_2%7D)
![U_f = \frac{(9\times 10^9)(3.10 \mu C)(-4.50 \mu C)}{0.375}](https://tex.z-dn.net/?f=U_f%20%3D%20%5Cfrac%7B%289%5Ctimes%2010%5E9%29%283.10%20%5Cmu%20C%29%28-4.50%20%5Cmu%20C%29%7D%7B0.375%7D)
![U_f = -0.335 J](https://tex.z-dn.net/?f=U_f%20%3D%20-0.335%20J)
Now the work done to change the position of charge is given as
![W = U_f - U_i](https://tex.z-dn.net/?f=W%20%3D%20U_f%20-%20U_i)
![W = (-0.335) - (-0.897) = 0.562 J](https://tex.z-dn.net/?f=W%20%3D%20%28-0.335%29%20-%20%28-0.897%29%20%3D%200.562%20J)
Answer:
Not sure which statements were give to you for this question, but the vehicle's acceleration from noon to 2:00 PM was <em>zero</em>, the vehicle had a <em>positive</em> acceleration between 2:00 pm and 3:00 pm, and the vehicle had a <em>negative</em> acceleration at 4:30 PM.
Explanation:
From 12:00 PM to 2:00 PM, the vehicle traveled at a constant velocity of 70 MPH, meaning there wasn't any change to the speed. The velocity <em>remained the same</em>. An hour later, the velocity of the vehicle <em>increased</em> to 80 MPH, and finally at 4:30 PM, the velocity of the car <em>decreased </em>and was at 40 MPH.
Answer:
A solid Hope this helped, Have a Great Day!!
Explanation:
Hope this helped, Have a Great Day!!