Kepler's third law is used to determine the relationship between the orbital period of a planet and the radius of the planet.
The distance of the earth from the sun is 
.
<h3>
What is Kepler's third law?</h3>
Kepler's Third Law states that the square of the orbital period of a planet is directly proportional to the cube of the radius of their orbits. It means that the period for a planet to orbit the Sun increases rapidly with the radius of its orbit.
Given that Mars’s orbital period T is 687 days, and Mars’s distance from the Sun R is 2.279 × 10^11 m.
By using Kepler's third law, this can be written as,
Substituting the values, we get the value of constant k for mars.
The value of constant k is the same for Earth as well, also we know that the orbital period for Earth is 365 days. So the R is calculated as given below.
Hence we can conclude that the distance of the earth from the sun is .
To know more about Kepler's third law, follow the link given below.
brainly.com/question/7783290.
The time when the particle is at rest is at 1.63 s or 3.36 s.
The velocity is positive at when the time of motion is at 
.
The total distance traveled in the first 10 seconds is 847 m.
<h3>When is a particle at rest?</h3>
- A particle is at rest when the initial velocity of the particle is zero.
The time when the particle is at rest is calculated as follows;
s(t) = 2t³ - 15t² + 33t + 17
The velocity is positive at when the time of motion is as follows;
.
The total distance traveled in the first 10 seconds is calculated as follows;
Learn more about motion of particles here: brainly.com/question/11066673
Answer:
the tension is 18513N
Explanation:
Given that
mass = 1683kg
acceleration = 1.2m/s^2
acceleration due to gravity = 9.8m/s^2
T-mg = ma
T = ma + mg
T = m(a +g)
T = 1683 kg(1.20 m/s2 + 9.8)
T = 1683 (11)
T = 18513N
therefore, the tension is 18513N
Answer: Magnesium Alloy
Explanation:
Cross sectional area A = (pi/4)d sqr
A = pi / 4 x 0.015 sqr
A = 1.76 x 10^-4 m2
Stressed induced in rod S = P/A = 35000/1.76x 10^-4
S= 198.06 M Pa
Reduction in diameter of the titanium alloy = -v (alpha/E) x d
= - 0.33 (( 198.06 X 10^6) / 70 x 10^9 )x 0.015 = - 1.41 x 10^-2 mm
For steel alloy
= - 0.36 (( 198.06 X 10^6) / 150 x 10^9 )x 0.015 = - 1.02 x 10^-2 mm
For magnesium
= 0.27 (( 198.06 X 10^6) / 205 x 10^9 )x 0.015 = 0.391 x 10^-2 mm
The titanium alloy and the steel alloy does not satisfy the second criteria since the reduction in diameter exceeds the allowable limit of 1.2 x 10^2mm
Considering the yield strength of the material, we find the aluminium alloy is not suitable and hence no need to check for the second criteria. By considering the both the given conditions, we find the magnesium alloy is the suitable material.
B. the plane because it is flying and passing through
