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 frequency of the oscillation in hertz is calculated to be 0.00031 Hz.
The frequency of a wave is defined as the number of cycles completed per second while the period refers to the time taken to complete a cycle. The frequency is the inverse of period.
So;
Period(T) = 54 minutes or 3240 seconds
Frequency (f) = T-1 = 1/T = 1/3240 seconds = 0.00031 Hz
Learn more: brainly.com/question/14588679
Stress required to cause slip on in the direction [ 1 1 0 ] is 7.154 MPa
<u>Explanation:</u>
Given -
Stress Direction, A = [1 0 0 ]
Slip plane = [ 1 1 1]
Normal to slip plane, B = [ 1 1 1 ]
Critical stress, Sc = 2.92 MPa
Let the direction of slip on = [ 1 1 0 ]
Let Ф be the angle between A and B
cos Ф = A.B/ |A| |B| = [ 1 0 0 ] [1 1 1] / √1 √3
cos Ф = 1/√3
σ = Sc / cosФ cosλ
For slip along [ 1 1 0 ]
cos λ = [ 1 1 0 ] [ 1 0 0 ] / √2 √1
cos λ = 1/√2
Therefore,
σ = 2.92 / 1/√3 1/√2
σ = √6 X 2.92 MPa = 2.45 X 2.92 = 7.154MPa
Therefore, stress required to cause slip on in the direction [ 1 1 0 ] is 7.154MPa
Answer:
The answer is A, B, C and D
Explanation:
(is that how it works?)
Energy of a random atomic and molecular is called molecules energy
