Answer:
While the water falls v increases and h decreases, so the kinetic energy increases and the gravitational potential energy decreases, and this happens in a way that the total energy is always the same. (If there is no friction)
Explanation:
Answer:
Se the explanation below
Explanation:
We do not feel these forces of these bodies, because they are very small compared to the force of Earth's attraction. Although its mass is greater than that of a human being, its mass is not compared to the Earth's mass. In order to understand this problem we will use numerical data and the universal gravitation formula, to give validity to the explanation.
<u>Force exerted by the Earth on a human being</u>
<u />

Where:
G = universal gravitation constant = 6.673*10^-11 [N*m^2/kg^2]
m1 = mass of the person = 80 [kg]
m2 = mass of the earth 5.97*10^24[kg]
r = distance from the center of the earth to the surface or earth radius = 6371 *10^3 [m]
<u />
Now replacing we have
![F = 6.673*10^{-11} *\frac{80*5.97*10^{24}}{(6371*10^{3})^{2} } \\F = 785[N]](https://tex.z-dn.net/?f=F%20%3D%206.673%2A10%5E%7B-11%7D%20%2A%5Cfrac%7B80%2A5.97%2A10%5E%7B24%7D%7D%7B%286371%2A10%5E%7B3%7D%29%5E%7B2%7D%20%20%7D%20%5C%5CF%20%3D%20785%5BN%5D)
<u>Force exerted by a building on a human being</u>
<u />
Where:
G = universal gravitation constant = 6.673*10^-11 [N*m^2/kg^2]
m1 = mass of the person = 80 [kg]
m2 = mass of the earth 300000 [ton] = 300 *10^6[kg]
r = distance from the building to the person = 2[m]
![F = 6.673*10^{-11}*\frac{80*300*10^6}{2^{2} } \\F= 0.4 [N]](https://tex.z-dn.net/?f=F%20%3D%206.673%2A10%5E%7B-11%7D%2A%5Cfrac%7B80%2A300%2A10%5E6%7D%7B2%5E%7B2%7D%20%7D%20%20%5C%5CF%3D%200.4%20%5BN%5D)
As we can see the force exerted by the Earth is 2000 times greater than that exerted by a building with the proposed data.
Answer:
θ = 29.38°
Explanation:
The centripetal force is given by the formula;
F_c = F_n(sin θ) = mv²/r
Now, the vertical component of the normal force is; F_n(cos θ)
Now, this vertical component is also expressed as; F_n(cos θ) = mg
Thus, the slope is;
F_n(sin θ)/F_n(cos θ) = (mv²/r)/mg
tan θ = v²/rg
v² = rg(tan θ)
The initial speed will be gotten from the relation;
(v_o)² = μ_s(gr)
Plugging rg(tan θ) for (v_o)², we have;
μ_s(gr) = rg(tan θ)
rg will cancel out to give;
μ_s = (tan θ)
Thus, θ = tan^(-1) μ_s
μ_s is coefficient of static friction given as 0.563
θ = tan^(-1) 0.563
θ = 29.38°