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i did it and got it right guys i hope it helps a lot
decomposing water does not require High activation energy
Answer:
The gravitational potential energy of the nickel at the top of the monument is 8.29 J.
Explanation:
We can find the gravitational potential energy using the following formula.
Identifying given information.
The nickel has a mass 
, and it is a the top of Washington Monument.
The Washington Monument has a height of 
, thus we need to find the equivalence in meters using unit conversion in order to find the gravitational potential energy.
Converting from feet to meters.
Using the conversion factor 1 m = 3.28 ft, we have
That give u s
Finding Gravitational Potential Energy.
We can replace the height and mass on the formula
And we get
The gravitational potential energy of the nickel at the top of the monument is 8.29 J.
Answer:
First, the different indices of refraction must be taken into account (in different media): for example, the refractive index of light in a vacuum is 1 (since vacuum = c). The value of the refractive index of the medium is a measure of its "optical density": Light spreads at maximum speed in a vacuum but slower in others transparent media; therefore in all of them n> 1. Examples of typical values of are those of air (1,0003), water (1.33), glass (1.46 - 1.66) or diamond (2.42).
The refractive index has a maximum value and a minimum value, which we can calculate the minimum value by means of the following explanation:
The limit or minimum angle, α lim, is defined as the angle of refraction from which the refracted ray disappears and all the light is reflected. As in the maximum value of angle of refraction, from which everything is reflected, is βmax = 90º, we can know the limit angle (the minimum angle that we would have to have to know the minimum index of refraction) by Snell's law:
βmax = 90º ⇒ n 1x sin α (lim) = n 2 ⇒ sin α lim = n 2 / n 1
Explanation:
When a light ray strikes the separation surface between two media different, the incident beam is divided into three: the most intense penetrates the second half forming the refracted ray, another is reflected on the surface and the third is breaks down into numerous weak beams emerging from the point of incidence in all directions, forming a set of stray light beams.
