Shear stress created the San Andreas Fault in Southern California. It is an example of a <span>reverse fault.</span>
Answer:
d. We can calculate it by applying Newton's version of Kepler's third law
Explanation:
The measurements of a Star like the Sun have several problems, the first one is distance, but the most important is the temperature since as we get closer all the instruments will melt. This is why all measurements must be indirect because of the effects that these variables create on nearby bodies.
Kepler's laws are deduced from Newton's law of universal gravitation, in these laws the mass of the Sun affects the orbit of the planets since it creates a force of attraction, if measured the orbit and the time it takes to travel it we can know the centripetal acceleration and with it knows the force, from where we clear the mass of the son.
Let's review the statements of the exercise
.a) False. We don't have good enough models for this calculation
.b) False. The size of the sun is very difficult to measure because it is a mass of gas, in addition the density changes strongly with depth
.c) False. The amount of light that comes out of the sun is not all the light produced and is due to quantum effects where the mass of the sun is not taken into account
.d) True. This method has been used to calculate the mass of the sun and the other planets since the variable distance and time are easily measured from Earth
Correct answer is D
The value of normal force as the slider passes point B is
The value of h when the normal force is zero
<h3>How to solve for the normal force</h3>
The normal force is calculated using the work energy principle which is applied as below
K₁ + U₁ = K₂
k represents kinetic energy
U represents potential energy
the subscripts 1,2 , and 3 = a, b, and c
for 1 to 2
K₁ + W₁ = K₂
0 + mg(h + R) = 0.5mv²₂
g(h + R) = 0.5v²₂
v²₂ = 2g(1.5R + R)
v²₂ = 2g(2.5R)
v²₂ = 5gR
Using summation of forces at B
Normal force, N = ma + mg
N = m(a + g)
N = m(v²₂/R + g)
N = m(5gR/R + g)
N = 6mg
for 1 to 3
K₁ + W₁ = K₃ + W₃
0 + mgh = 0.5mv²₃ + mgR
gh = 0.5v²₃ + gR
0.5v²₃ = gh - gR
v²₃ = 2g(h - R)
at C
for normal force to be zero
ma = mg
v²₃/R = g
v²₃ = gR
and v²₃ = 2g(h - R)
gR = 2gh - 2gR
gR + 2gR = 2gh
3gR = 2gh
3R/2 = h
Learn more about normal force at:
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The statement about pointwise convergence follows because C is a complete metric space. If fn → f uniformly on S, then |fn(z) − fm(z)| ≤ |fn(z) − f(z)| + |f(z) − fm(z)|, hence {fn} is uniformly Cauchy. Conversely, if {fn} is uniformly Cauchy, it is pointwise Cauchy and therefore converges pointwise to a limit function f. If |fn(z)−fm(z)| ≤ ε for all n,m ≥ N and all z ∈ S, let m → ∞ to show that |fn(z)−f(z)|≤εforn≥N andallz∈S. Thusfn →f uniformlyonS.
2. This is immediate from (2.2.7).
3. We have f′(x) = (2/x3)e−1/x2 for x ̸= 0, and f′(0) = limh→0(1/h)e−1/h2 = 0. Since f(n)(x) is of the form pn(1/x)e−1/x2 for x ̸= 0, where pn is a polynomial, an induction argument shows that f(n)(0) = 0 for all n. If g is analytic on D(0,r) and g = f on (−r,r), then by (2.2.16), g(z) =
