Answer:
F = 3.86 x 10⁻⁶ N
Explanation:
First, we will find the distance between the two particles:
![r = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2+(z_{2}-z_{1})^2}\\](https://tex.z-dn.net/?f=r%20%3D%20%5Csqrt%7B%28x_%7B2%7D-x_%7B1%7D%29%5E2%2B%28y_%7B2%7D-y_%7B1%7D%29%5E2%2B%28z_%7B2%7D-z_%7B1%7D%29%5E2%7D%5C%5C)
where,
r = distance between the particles = ?
(x₁, y₁, z₁) = (2, 5, 1)
(x₂, y₂, z₂) = (3, 2, 3)
Therefore,
![r = \sqrt{(3-2)^2+(2-5)^2+(3-1)^2}\\r = 3.741\ m\\](https://tex.z-dn.net/?f=r%20%3D%20%5Csqrt%7B%283-2%29%5E2%2B%282-5%29%5E2%2B%283-1%29%5E2%7D%5C%5Cr%20%3D%203.741%5C%20m%5C%5C)
Now, we will calculate the magnitude of the force between the charges by using Coulomb's Law:
![F = \frac{kq_{1}q_{2}}{r^2}\\](https://tex.z-dn.net/?f=F%20%3D%20%5Cfrac%7Bkq_%7B1%7Dq_%7B2%7D%7D%7Br%5E2%7D%5C%5C)
where,
F = magnitude of force = ?
k = Coulomb's Constant = 9 x 10⁹ Nm²/C²
q₁ = magnitude of first charge = 2 x 10⁻⁸ C
q₂ = magnitude of second charge = 3 x 10⁻⁷ C
r = distance between the charges = 3.741 m
Therefore,
![F = \frac{(9\ x\ 10^9\ Nm^2/C^2)(2\ x\ 10^{-8}\ C)(3\ x\ 10^{-7}\ C)}{(3.741\ m)^2}\\](https://tex.z-dn.net/?f=F%20%3D%20%5Cfrac%7B%289%5C%20x%5C%2010%5E9%5C%20Nm%5E2%2FC%5E2%29%282%5C%20x%5C%2010%5E%7B-8%7D%5C%20C%29%283%5C%20x%5C%2010%5E%7B-7%7D%5C%20C%29%7D%7B%283.741%5C%20m%29%5E2%7D%5C%5C)
<u>F = 3.86 x 10⁻⁶ N</u>
Answer:
fo = 378.52Hz
Explanation:
Using Doppler effect formula:
![f'=\frac{C-Vb}{C-Va}*fo](https://tex.z-dn.net/?f=f%27%3D%5Cfrac%7BC-Vb%7D%7BC-Va%7D%2Afo)
where
f' = 392 Hz
C = 340m/s
Vb = 20m/s
Va = 31m/s
Replacing these values and solving for fo:
fo = 378.52Hz
Answer:
![x=65.55m](https://tex.z-dn.net/?f=x%3D65.55m)
Explanation:
Let x be the distance to the shore
From trigonometry properties:
![tan(40^{o} )=\frac{55m}{x} \\x=\frac{55m}{tan(40^{o} )} \\x=65.55m](https://tex.z-dn.net/?f=tan%2840%5E%7Bo%7D%20%29%3D%5Cfrac%7B55m%7D%7Bx%7D%20%5C%5Cx%3D%5Cfrac%7B55m%7D%7Btan%2840%5E%7Bo%7D%20%29%7D%20%5C%5Cx%3D65.55m)
A) 1.55
The speed of light in a medium is given by:
![v=\frac{c}{n}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bc%7D%7Bn%7D)
where
is the speed of light in a vacuum
n is the refractive index of the material
In this problem, the speed of light in quartz is
![v=1.94\cdot 10^8 m/s](https://tex.z-dn.net/?f=v%3D1.94%5Ccdot%2010%5E8%20m%2Fs)
So we can re-arrange the previous formula to find n, the index of refraction of quartz:
![n=\frac{c}{v}=\frac{3\cdot 10^8 m/s}{1.94\cdot 10^8 m/s}=1.55](https://tex.z-dn.net/?f=n%3D%5Cfrac%7Bc%7D%7Bv%7D%3D%5Cfrac%7B3%5Ccdot%2010%5E8%20m%2Fs%7D%7B1.94%5Ccdot%2010%5E8%20m%2Fs%7D%3D1.55)
B) 550.3 nm
The relationship between the wavelength of the light in air and in quartz is
![\lambda=\frac{\lambda_0}{n}](https://tex.z-dn.net/?f=%5Clambda%3D%5Cfrac%7B%5Clambda_0%7D%7Bn%7D)
where
is the wavelenght in quartz
is the wavelength in air
n is the refractive index
For the light in this problem, we have
![\lambda=355 nm\\n=1.55](https://tex.z-dn.net/?f=%5Clambda%3D355%20nm%5C%5Cn%3D1.55)
Therefore, we can re-arrange the equation to find
, the wavelength in air:
![\lambda_0 = n\lambda=(1.55)(355 nm)=550.3 nm](https://tex.z-dn.net/?f=%5Clambda_0%20%3D%20n%5Clambda%3D%281.55%29%28355%20nm%29%3D550.3%20nm)
Answer:
kg
Explanation:
easy question to be honest