Answer:
mass*velocity=1.5*10^4 * 15
= 22.5*10^4
By definition, the refractive index is
n = c/v
where c = 3 x 10⁸ m/s, the speed of light in vacuum
v = the speed of light in the medium (the liquid).
The frequency of the light source is
f = (3 x 10⁸ m/s)/(495 x 10⁻⁹ m) = 6.0606 x 10¹⁴ Hz
Because the wavelength in the liquid is 434 nm = 434 x 10⁻⁹ m,
v = (6.0606 x 10¹⁴ 1/s)*(434 x 10⁻⁹ m) = 2.6303 x 10⁸ m/s
The refractive index is (3 x 10⁸)/(2.6303 x 10⁸) = 1.1406
Answer: a. 1.14
Answer:
W = 907963.50 J = 907.96 J
Explanation:
Note: Refer to the figure attached
Now, from the figure we have similar triangles ΔAOB and ΔCOD
we have

or

Now, the work done to empty the tank can be given as:

or

or

or

or
![W = 6773.76\pi[\frac{5}{3}x^3-\frac{1}{4}x^4]^4_0](https://tex.z-dn.net/?f=W%20%3D%206773.76%5Cpi%5B%5Cfrac%7B5%7D%7B3%7Dx%5E3-%5Cfrac%7B1%7D%7B4%7Dx%5E4%5D%5E4_0%20)
or
![W = 6773.76\pi[\frac{128}{3}]](https://tex.z-dn.net/?f=W%20%3D%206773.76%5Cpi%5B%5Cfrac%7B128%7D%7B3%7D%5D%20)
or
W = 907963.50 J = 907.96 J