Answer:
The correct option is;
Still constant
Explanation:
The relative refractive index ₁n₂ between the two medium can be as follows;
Therefore, given that the speed of light in medium 1 is constant and the speed of light on medium 2 is also constant, the relative refractive index ₁n₂ = c₁/c₂ is always constant.
Answer:
Mount Everest formed from a tectonic smashup between the Indian and Eurasian tectonic plates tens of millions of years ago.
Explanation:
Answer:
6.67×10¯⁹ A
Explanation:
From the question given above, the following data were obtained:
Quantity of electricity (Q) = 2 μC
Time (t) = 5 mins
Current (I) =?
Next, we shall convert 2 μC to C. This can be obtained as follow:
1 μC = 1×10¯⁶ C
Therefore,
2 μC = 2 μC × 1×10¯⁶ C / 1 μC
2 μC = 2×10¯⁶ C
Next, we shall convert 5 mins to seconds. This can be obtained as follow:
1 min = 60 secs
Therefore,
5 min = 5 min × 60 sec / 1 min
5 mins = 300 s
Finally, we shall determine the current in the circuit. This can be obtained as follow:
Quantity of electricity (Q) = 2×10¯⁶ C
Time (t) = 300 s
Current (I) =?
Q = It
2×10¯⁶ = I × 300
Divide both side by 300
I = 2×10¯⁶ / 300
I = 6.67×10¯⁹ A
Thus, the current in the circuit is 6.67×10¯⁹ A
The molar mass of ammonium sulphate [(NH4)2SO4] is 132.17 g (option E). Details about molar mass can be found below.
<h3>How to calculate molar mass?</h3>
The molar mass of a substance can be calculated by adding the atomic masses of the elements in the compound.
According to this question, the atomic mass of nitrogen is given as 14.01, hydrogen is 1.01, sulfur is 32.07, and oxygen is 16.00.
The molar mass of ammonium sulphate is as follows:
[(NH4)2SO4] = [14.01 + 1(4)]2 + 32.07 + 16.00(4)
= 36.02 + 32.07 + 64
= 132.09
Therefore, the molar mass of ammonium sulphate [(NH4)2SO4] is 132.17 g.
Learn more about molar mass at: brainly.com/question/12127540
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Answer:
The velocity will be v = 22.1[m/s]
Explanation:
We can solve this problem by using the principle of energy conservation, where potential energy is converted to kinetic energy. For this problem we will take the point with maximum potential energy when the body is 25 [m] high. By the time the height is zero, the potential energy will have been transformed into kinetic energy, and we can find the velocity of the body.
![Ep = m*g*h\\where:\\m = mass = 88.2[kg]\\h = elevation = 25[m]\\g = gravity = 9.81 [m/s^2]\\Ep = 88.2*25*9.81 = 21631.05[J]\\](https://tex.z-dn.net/?f=Ep%20%3D%20m%2Ag%2Ah%5C%5Cwhere%3A%5C%5Cm%20%3D%20mass%20%3D%2088.2%5Bkg%5D%5C%5Ch%20%3D%20elevation%20%3D%2025%5Bm%5D%5C%5Cg%20%3D%20gravity%20%3D%209.81%20%5Bm%2Fs%5E2%5D%5C%5CEp%20%3D%2088.2%2A25%2A9.81%20%3D%2021631.05%5BJ%5D%5C%5C)
Now we know that the energy will be transformed.
![Ek=Ep\\Ek=0.5*m*v^{2} \\where:\\v=velocity [m/s]\\v=\sqrt{\frac{Ek}{0.5*m} } \\v=\sqrt{\frac{21631.05}{0.5*88.2} } \\v=22.14[m/s]](https://tex.z-dn.net/?f=Ek%3DEp%5C%5CEk%3D0.5%2Am%2Av%5E%7B2%7D%20%5C%5Cwhere%3A%5C%5Cv%3Dvelocity%20%5Bm%2Fs%5D%5C%5Cv%3D%5Csqrt%7B%5Cfrac%7BEk%7D%7B0.5%2Am%7D%20%7D%20%5C%5Cv%3D%5Csqrt%7B%5Cfrac%7B21631.05%7D%7B0.5%2A88.2%7D%20%7D%20%5C%5Cv%3D22.14%5Bm%2Fs%5D)
