The answer is wind forces and Earth’s rotation
Answer:
option (B)
Explanation:
Intensity of unpolarised light, I = 25 W/m^2
When it passes from first polarisr, the intensity of light becomes
![I'=\frac{I_{0}}{2}=\frac{25}{2}=12.5 W/m^{2}](https://tex.z-dn.net/?f=I%27%3D%5Cfrac%7BI_%7B0%7D%7D%7B2%7D%3D%5Cfrac%7B25%7D%7B2%7D%3D12.5%20W%2Fm%5E%7B2%7D)
Let the intensity of light as it passes from second polariser is I''.
According to the law of Malus
![I'' = I' Cos^{2}\theta](https://tex.z-dn.net/?f=I%27%27%20%3D%20I%27%20Cos%5E%7B2%7D%5Ctheta)
Where, θ be the angle between the axis first polariser and the second polariser.
![I'' = 12.5\times Cos^{2}15](https://tex.z-dn.net/?f=I%27%27%20%3D%2012.5%5Ctimes%20Cos%5E%7B2%7D15)
I'' = 11.66 W/m^2
I'' = 11.7 W/m^2
B: Energy lose
i say this because in order to change they lose energy.
Answer:
Opposition of passing a electric circuit
Answer:
B. 0.16 m
Explanation:
The vertical distance by which the player will miss the target is equal to the vertical distance covered by the dart during its motion.
Since the dart is thrown horizontally, the initial vertical velocity is zero:
![v_y = 0](https://tex.z-dn.net/?f=v_y%20%3D%200)
While the horizontal velocity is
![v_x = 15 m/s](https://tex.z-dn.net/?f=v_x%20%3D%2015%20m%2Fs)
The horizontal distance covered is
![d_x = 2.7 m](https://tex.z-dn.net/?f=d_x%20%3D%202.7%20m)
Since the dart moves by uniform motion along the horizontal direction, the time it takes for covering this distance is
![t=\frac{d_x}{v_x}=\frac{2.7 m}{15 m/s}=0.18 s](https://tex.z-dn.net/?f=t%3D%5Cfrac%7Bd_x%7D%7Bv_x%7D%3D%5Cfrac%7B2.7%20m%7D%7B15%20m%2Fs%7D%3D0.18%20s)
along the vertical direction, the motion is a uniformly accelerated motion with constant downward acceleration g=9.8 m/s^2, so the vertical distance covered is given by
![d_y = \frac{1}{2}gt^2=\frac{1}{2}(9.8 m/s^)(0.18 s)^2=0.16 m](https://tex.z-dn.net/?f=d_y%20%3D%20%5Cfrac%7B1%7D%7B2%7Dgt%5E2%3D%5Cfrac%7B1%7D%7B2%7D%289.8%20m%2Fs%5E%29%280.18%20s%29%5E2%3D0.16%20m)