1.47x10^5 Joules
The gravitational potential energy will be the mass of the object, multiplied by the height upon which it can drop, multiplied by the local gravitational acceleration. And since it started at the top of a 60.0 meter hill, halfway will be at 30.0 meters. So
500 kg * 30.0 m * 9.8 m/s^2 = 147000 kg*m^2/s^ = 147000 Joules.
Using scientific notation and 3 significant figures gives 1.47x10^5 Joules.
Answer:
the answer is c
Explanation:
it's c because the moon has to be a full moon to be a solar eclipse when the sun moon and earth line up
Answer: conducir la política, acciones y asuntos de (un estado, organización o personas).
Answer:
The Moon's distance from the Earth varies during its orbit
Explanation:
The correct statement is ,The Moon's distance from the Earth varies during its orbit.
Important point regarding moon:
1 .Moon is a natural satellite of the earth.
2. Moon is the fifth largest satellite in solar system.
3.Second densest satellite in solar system.
4.Moon rotates about earth.
5.Moon is an astronomical body .
Answer:
a

b

c
Explanation:
From the question we are told that
The angle of incidence is 
The refractive index of water is 
Generally Snell's law is mathematically represented as

Here
is the refractive index of air with value 
is the angle of refraction
So
![\theta _2 = sin^{-1}[\frac{n_1 * sin(\theta _1)}{n_2} ]](https://tex.z-dn.net/?f=%5Ctheta%20_2%20%20%3D%20%20sin%5E%7B-1%7D%5B%5Cfrac%7Bn_1%20%2A%20sin%28%5Ctheta%20_1%29%7D%7Bn_2%7D%20%5D)
=> ![\theta _2 = sin^{-1}[\frac{1.3 * sin(10)}{1} ]](https://tex.z-dn.net/?f=%5Ctheta%20_2%20%20%3D%20%20sin%5E%7B-1%7D%5B%5Cfrac%7B1.3%20%2A%20sin%2810%29%7D%7B1%7D%20%5D)
=> 
Given that the angle should not be greater than
then the angle of incidence will be
![\theta _1 = sin^{-1}[\frac{n_2 * sin(\theta _2)}{n_1} ]](https://tex.z-dn.net/?f=%5Ctheta%20_1%20%20%3D%20%20sin%5E%7B-1%7D%5B%5Cfrac%7Bn_2%20%2A%20sin%28%5Ctheta%20_2%29%7D%7Bn_1%7D%20%5D)
=> ![\theta _1 = sin^{-1}[\frac{1 * sin(45)}{1.3} ]](https://tex.z-dn.net/?f=%5Ctheta%20_1%20%20%3D%20%20sin%5E%7B-1%7D%5B%5Cfrac%7B1%20%2A%20sin%2845%29%7D%7B1.3%7D%20%5D)
=> 
Generally for critical angle is mathematically represented as
![\theta_c = sin^{-1}[\frac{n_2}{n_1} ]](https://tex.z-dn.net/?f=%5Ctheta_c%20%20%3D%20%20sin%5E%7B-1%7D%5B%5Cfrac%7Bn_2%7D%7Bn_1%7D%20%5D)
=>
=>