The wavelength of the light decreases as it enters into the medium with the greater index of refraction. The wavelength of the light remains constant as it transitions between materials.
Answer:
![\frac{1}{2}\frac{(M_dV_0)^2}{(M_d+M_a)^2} = h](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7B%28M_dV_0%29%5E2%7D%7B%28M_d%2BM_a%29%5E2%7D%20%3D%20h)
Explanation:
First, we will use conservation of the linear momentum:
so:
![M_dv_0 = (M_d+M_a)V_s](https://tex.z-dn.net/?f=M_dv_0%20%3D%20%28M_d%2BM_a%29V_s)
where
is the mass of the dart,
is the speed of the dart just before it strickes the apple,
the mass of the apple and
the velocity of the apple and the dart after the collition.
Then, solving for V_s:
![V_s = \frac{M_dV_0}{M_d+M_a}](https://tex.z-dn.net/?f=V_s%20%3D%20%5Cfrac%7BM_dV_0%7D%7BM_d%2BM_a%7D)
now, using the conservation of energy:
![E_i = E_f](https://tex.z-dn.net/?f=E_i%20%3D%20E_f)
so:
![\frac{1}{2}(M_d+M_a)V_s^2 = (M_a+M_d)gh](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%28M_d%2BM_a%29V_s%5E2%20%3D%20%28M_a%2BM_d%29gh)
where g is the gravity and h how high does the apple move upward.
Now, replacing
and solving for h, we get:
![\frac{1}{2}(M_d+M_a)(\frac{M_dV_0}{M_d+M_a})^2 = (M_a+M_d)h](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%28M_d%2BM_a%29%28%5Cfrac%7BM_dV_0%7D%7BM_d%2BM_a%7D%29%5E2%20%3D%20%28M_a%2BM_d%29h)
![\frac{1}{2}\frac{(M_dV_0)^2}{(M_d+M_a)^2} = h](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7B%28M_dV_0%29%5E2%7D%7B%28M_d%2BM_a%29%5E2%7D%20%3D%20h)
So, the force of gravity that the asteroid and the planet have on each other approximately
![\boxed{\sf{5.43 \times 10^{10} \: N}}](https://tex.z-dn.net/?f=%20%5Cboxed%7B%5Csf%7B5.43%20%5Ctimes%2010%5E%7B10%7D%20%5C%3A%20N%7D%7D%20)
<h3>Introduction</h3>
Hi ! Now, I will help to discuss about the gravitational force between two objects. We already know that <u>gravitational force occurs when two or more objects interact with each other at a certain distance and generally orbit each other to their center of mass</u>. For the gravitational force between two objects, it can be calculated using the following formula :
![\boxed{\sf{\bold{F = G \times \frac{m_1 \times m_2}{r^2}}}}](https://tex.z-dn.net/?f=%20%5Cboxed%7B%5Csf%7B%5Cbold%7BF%20%3D%20G%20%5Ctimes%20%5Cfrac%7Bm_1%20%5Ctimes%20m_2%7D%7Br%5E2%7D%7D%7D%7D%20)
With the following condition :
- F = gravitational force (N)
- G = gravity constant ≈
N.m²/kg²
= mass of the first object (kg)
= mass of the second object (kg)- r = distance between two objects (m)
<h3>Problem Solving</h3>
We know that :
- G = gravity constant ≈
N.m²/kg²
= mass of the first object =
kg.
= mass of the second object =
kg.- r = distance between two objects =
![\sf{8 \times 10^5}](https://tex.z-dn.net/?f=%20%5Csf%7B8%20%5Ctimes%2010%5E5%7D%20)
What was asked :
- F = gravitational force = ... N
Step by step :
![\sf{F = G \times \frac{m_1 \times m_2}{r^2}}](https://tex.z-dn.net/?f=%20%5Csf%7BF%20%3D%20G%20%5Ctimes%20%5Cfrac%7Bm_1%20%5Ctimes%20m_2%7D%7Br%5E2%7D%7D%20)
![\sf{F = 6.67 \times 10^{-11} \times \frac{8.4 \cdot 10^8 \times 6.2 \cdot 10^{23}}{(8 \times 10^5)^2}}](https://tex.z-dn.net/?f=%20%5Csf%7BF%20%3D%206.67%20%5Ctimes%2010%5E%7B-11%7D%20%5Ctimes%20%5Cfrac%7B8.4%20%5Ccdot%2010%5E8%20%5Ctimes%206.2%20%5Ccdot%2010%5E%7B23%7D%7D%7B%288%20%5Ctimes%2010%5E5%29%5E2%7D%7D%20)
![\sf{F = \frac{347.374 \times 10^{-11 + 8 + 23}}{64 \times 10^10}}](https://tex.z-dn.net/?f=%20%5Csf%7BF%20%3D%20%5Cfrac%7B347.374%20%5Ctimes%2010%5E%7B-11%20%2B%208%20%2B%2023%7D%7D%7B64%20%5Ctimes%2010%5E10%7D%7D%20)
![\sf{F \approx 5.43 \times 10^{20 - 10}}](https://tex.z-dn.net/?f=%20%5Csf%7BF%20%5Capprox%20%205.43%20%5Ctimes%2010%5E%7B20%20-%2010%7D%7D%20)
![\boxed{\sf{F \approx 5.43 \times 10^{10} \: N}}](https://tex.z-dn.net/?f=%20%5Cboxed%7B%5Csf%7BF%20%5Capprox%205.43%20%5Ctimes%2010%5E%7B10%7D%20%5C%3A%20N%7D%7D%20)
<h3>Conclusion</h3>
So, the force of gravity that the asteroid and the planet have on each other approximately
![\boxed{\sf{5.43 \times 10^{10} \: N}}](https://tex.z-dn.net/?f=%20%5Cboxed%7B%5Csf%7B5.43%20%5Ctimes%2010%5E%7B10%7D%20%5C%3A%20N%7D%7D%20)
<h3>See More</h3>
Gravity is a thing has depends on ... brainly.com/question/26485200
Answer:
Multiply the pressure in mm Hg by 0.03937 to convert to in Hg. For example, if you have 29 mm Hg, multiply 29 by 0.03937 to get 1.14 in Hg. Divide the pressure of mm Hg by 25.4 to convert to in Hg.
Explanation: