<span>Normal
fault. Normal (extensional ) fault is a
displacement of a rock as a result of rock-mass movement and occurs when the
crust is stretching. Because of the stretching the thickness of the crust is
reduced and the crust or horizontally extended. </span>
The formula for this problem that we will be using is:
F * cos α = m * g * μs where:F = 800m = 87g = 9.8
cos α = m*g*μs/F= 87*9.8*0.55/800= 0.59 So solving the alpha, find the arccos above.
α = arccos 0.59 = 54 ° is the largest value of alpha
Answer:
3) Ep = 13243.5[J]
4) v = 17.15 [m/s]
Explanation:
3) In order to solve this problem, we must use the principle of energy conservation. That is, the energy will be transformed from potential energy to kinetic energy. We can calculate the potential energy with the mass and height data, as shown below.
m = mass = 90 [kg]
h = elevation = 15 [m]
Potential energy is defined as the product of mass by gravity by height.
![E_{p}=m*g*h\\E_{p}=90*9.81*15\\E_{p}=13243.5[J]](https://tex.z-dn.net/?f=E_%7Bp%7D%3Dm%2Ag%2Ah%5C%5CE_%7Bp%7D%3D90%2A9.81%2A15%5C%5CE_%7Bp%7D%3D13243.5%5BJ%5D)
This energy will be transformed into kinetic energy.
Ek = 13243.5 [J]
4) The velocity can be determined by defining the kinetic energy, as shown below.
![E_{k}=\frac{1}{2} *m*v^{2} \\v = \sqrt{\frac{2*E_{k} }{m} }\\ v= \sqrt{\frac{2*13243.5 }{90} }\\v=17.15[m/s]](https://tex.z-dn.net/?f=E_%7Bk%7D%3D%5Cfrac%7B1%7D%7B2%7D%20%2Am%2Av%5E%7B2%7D%20%20%5C%5Cv%20%3D%20%5Csqrt%7B%5Cfrac%7B2%2AE_%7Bk%7D%20%7D%7Bm%7D%20%7D%5C%5C%20v%3D%20%5Csqrt%7B%5Cfrac%7B2%2A13243.5%20%7D%7B90%7D%20%7D%5C%5Cv%3D17.15%5Bm%2Fs%5D)
Answer:

Explanation:
GIVEN
diameter = 15 fm =
m
we use here energy conservation

there will be some initial kinetic energy but after collision kinetic energy will zero

on solving these equations we get kinetic energy initial
J ..............(i)
That is, the alpha particle must be fired with 35.33 MeV of kinetic energy. An alpha particle with charge q = 2 e
and gains kinetic energy K =e∆V ..........(ii)
by accelerating through a potential difference ∆V
Thus the alpha particle will
just reach the
nucleus after being accelerated through a potential difference ∆V
equating (i) and second equation we get
e∆V = 35.33 Me V

There was no net force on the stuffed toy, because the kids might have the same strength, The same force is on both sides of it. T<span>hey cancel each other out. They exert a force on the stuffed toy equal in strength but opposite in direction. The forces are balanced and the stuffed toy does not move. </span>Its like a game of tug-o-war, but you and I have the same strength. the rope would be still and not moving.