First is nebula, second is low mass red star, third is turn into white dwarfs
Answer:
Constant-velocity reference frames
Explanation:
i think. sorry if its wrong!
To solve this problem we will apply the concepts related to the conservation of momentum. Recall that the momentum is the product between the mass and the speed of the bodies. The mathematical expression for this problem could be given as
![m_1v_1 = m_2v_2](https://tex.z-dn.net/?f=m_1v_1%20%3D%20m_2v_2)
Here,
= Mass of canoe 1
= Mass of canoe 2
= Velocity of canoe 1
= Velocity of canoe 2
Replacing with our values we have that,
![(320kg)(0.56m/s) = m_2(0.45m/s)](https://tex.z-dn.net/?f=%28320kg%29%280.56m%2Fs%29%20%3D%20m_2%280.45m%2Fs%29)
![m_2 = \frac{(320kg)(0.56m/s)}{ (0.45m/s)}](https://tex.z-dn.net/?f=m_2%20%3D%20%5Cfrac%7B%28320kg%29%280.56m%2Fs%29%7D%7B%20%280.45m%2Fs%29%7D)
![m_2 = 3.9*10^2 kg](https://tex.z-dn.net/?f=m_2%20%3D%203.9%2A10%5E2%20kg)
Therefore the mass of the canoe 2 is 390kg.
<em>NOTE: The omission of the negative speed symbol is given by the reference system and to facilitate the reading of the calculations.</em>
Answer:
B. The highest temperature in Seattle last summer
Explanation:
D. The average amount of precipitation in Seattle during,March
Answer:
The value is ![\alpha = 0.002 Np/m](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%200.002%20Np%2Fm)
Explanation:
From the question we are told that
The first amplitude of the wave is ![E_{max}1 = 98.02 \ V/m](https://tex.z-dn.net/?f=E_%7Bmax%7D1%20%3D%20%2098.02%20%5C%20%20V%2Fm)
The first depth is ![D_1 = 10 \ m](https://tex.z-dn.net/?f=D_1%20%3D%20%2010%20%5C%20%20m)
The second amplitude is
The second depth is ![D_2 = 100 \ m](https://tex.z-dn.net/?f=D_2%20%3D%20100%20%5C%20m)
Generally from the spatial wave equation we have
![v(x) = Ae^{-\alpha d}cos(\beta x + \phi_o)](https://tex.z-dn.net/?f=v%28x%29%20%3D%20%20Ae%5E%7B-%5Calpha%20d%7Dcos%28%5Cbeta%20x%20%20%2B%20%5Cphi_o%29)
=> ![\frac{v(x)}{v(x)} =\frac{ Ae^{-\alpha d}cos(\beta x + \phi_o)}{ Ae^{-\alpha d}cos(\beta x + \phi_o)}](https://tex.z-dn.net/?f=%5Cfrac%7Bv%28x%29%7D%7Bv%28x%29%7D%20%3D%5Cfrac%7B%20%20Ae%5E%7B-%5Calpha%20d%7Dcos%28%5Cbeta%20x%20%20%2B%20%5Cphi_o%29%7D%7B%20Ae%5E%7B-%5Calpha%20d%7Dcos%28%5Cbeta%20x%20%20%2B%20%5Cphi_o%29%7D)
So considering the ratio of the equation for the two depth
![\frac{A}{A_S} = \frac{e^{-D_1 \alpha }}{e^{-D_2 \alpha }}](https://tex.z-dn.net/?f=%5Cfrac%7BA%7D%7BA_S%7D%20%20%3D%20%20%5Cfrac%7Be%5E%7B-D_1%20%5Calpha%20%7D%7D%7Be%5E%7B-D_2%20%5Calpha%20%7D%7D)
=> ![\frac{98.02}{81.87} = \frac{e^{-10 \alpha }}{e^{-100 \alpha }}](https://tex.z-dn.net/?f=%5Cfrac%7B98.02%7D%7B81.87%7D%20%20%3D%20%20%5Cfrac%7Be%5E%7B-10%20%5Calpha%20%7D%7D%7Be%5E%7B-100%20%5Calpha%20%7D%7D)
=> ![\alpha = \frac{0.18}{90}](https://tex.z-dn.net/?f=%5Calpha%20%20%3D%20%20%5Cfrac%7B0.18%7D%7B90%7D)
=> ![\alpha = 0.002 Np/m](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%200.002%20Np%2Fm)