Answer:
It would be -8.6753 x 10 to the 6th power
Explanation
CR < CY < CB
<h3>Which factors affect the critical angle for a given pair of media?</h3>
The factors which affect the critical angle are
(a) The colour (or wavelength) of light
(b) The temperature
(i) Effect of colour of light: The critical angle for a pair of media is less for the violet light and more for the red light. Thus the critical angle increases with the increase in wavelength of light.
(ii) Effect of temperature: The critical angle increases with increase in temperature because on increasing temperature of medium, its refractive index decreases.
According to the question,
μ 1 sinCR =1
μ 2 sinCY =1
μ 3 sinCB =1
μ 1 > μ 2 and μ 2 > μ 3
⟹μ 1 > μ 2 > μ 3
CR < CY < CB
Thus,
The critical angle increases with the increase in wavelength of light.
Learn more about wavelength of light here:
brainly.com/question/27557868
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Answer:
a) velocity v = 322.5m/s
b) time t = 19.27s
Explanation:
Note that;
ads = vdv
where
a is acceleration
s is distance
v is velocity
Given;
a = 6 + 0.02s
so,

Remember that
![v = \frac{ds}{dt} \\\frac{ds}{v} = dt\\\int\limits^s_0 {\frac{ds}{\sqrt{12s+0.02s^{2} } } } \, ds = \int\limits^t_0 {} \, dt \\t= (5\sqrt{2} ) ln \frac{| [s + 300 + \sqrt{(s^{2} + 600s)} ] |}{300} .......2](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7Bds%7D%7Bdt%7D%20%5C%5C%5Cfrac%7Bds%7D%7Bv%7D%20%3D%20dt%5C%5C%5Cint%5Climits%5Es_0%20%7B%5Cfrac%7Bds%7D%7B%5Csqrt%7B12s%2B0.02s%5E%7B2%7D%20%7D%20%7D%20%7D%20%5C%2C%20ds%20%3D%20%5Cint%5Climits%5Et_0%20%7B%7D%20%5C%2C%20dt%20%5C%5Ct%3D%20%20%285%5Csqrt%7B2%7D%20%29%20ln%20%20%5Cfrac%7B%7C%20%5Bs%20%2B%20300%20%2B%20%5Csqrt%7B%28s%5E%7B2%7D%20%20%2B%20600s%29%7D%20%5D%20%7C%7D%7B300%7D%20.......2)
substituting s = 2km =2000m, into equation 1
v = 322.5m/s
substituting s = 2000m into equation 2
t = 19.27s
Answer:
the energy when it reaches the ground is equal to the energy when the spring is compressed.
Explanation:
For this comparison let's use the conservation of energy theorem.
Starting point. Compressed spring
Em₀ = K_e = ½ k x²
Final point. When the box hits the ground
Em_f = K = ½ m v²
since friction is zero, energy is conserved
Em₀ = Em_f
1 / 2k x² = ½ m v²
v =
x
Therefore, the energy when it reaches the ground is equal to the energy when the spring is compressed.
The answer is Strontium(Sr)