Answer:
The team's model is incorrect because they provide the same materials to build the same product.
Explanation:
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.
![T^2 \propto R^3](https://tex.z-dn.net/?f=T%5E2%20%5Cpropto%20R%5E3)
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,
![T^2 \propto R^3](https://tex.z-dn.net/?f=T%5E2%20%5Cpropto%20R%5E3)
![T^2 = kR^3](https://tex.z-dn.net/?f=T%5E2%20%3D%20kR%5E3)
Substituting the values, we get the value of constant k for mars.
![687^2 = k\times (2.279 \times 10^{11})^3](https://tex.z-dn.net/?f=687%5E2%20%3D%20k%5Ctimes%20%282.279%20%5Ctimes%2010%5E%7B11%7D%29%5E3)
![k = 3.92 \times 10^{-29}](https://tex.z-dn.net/?f=k%20%3D%203.92%20%5Ctimes%2010%5E%7B-29%7D)
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.
![365^3 = 3.92\times 10^{-29} R^3](https://tex.z-dn.net/?f=365%5E3%20%3D%203.92%5Ctimes%2010%5E%7B-29%7D%20R%5E3)
![R^3 = 3.39 \times 10^{33}](https://tex.z-dn.net/?f=R%5E3%20%3D%203.39%20%5Ctimes%2010%5E%7B33%7D)
![R= 1.50 \times 10^{11}\;\rm m](https://tex.z-dn.net/?f=R%3D%201.50%20%5Ctimes%2010%5E%7B11%7D%5C%3B%5Crm%20m)
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.
Answer:43.34 m
Explanation:
Given
acceleration(a)![=2 m/s^2](https://tex.z-dn.net/?f=%3D2%20m%2Fs%5E2)
Initial Velocity(u)=0 m/s
After 6 s fuel runs out
Velocity after 6 s
v=u+at
![v=0+2\times 6=12 m/s](https://tex.z-dn.net/?f=v%3D0%2B2%5Ctimes%206%3D12%20m%2Fs)
After this object will start moving under gravity
height reached in first 6 s
![s=ut+\frac{at^2}{2}](https://tex.z-dn.net/?f=s%3Dut%2B%5Cfrac%7Bat%5E2%7D%7B2%7D)
![s=0+\frac{2\times 6^2}{2}](https://tex.z-dn.net/?f=s%3D0%2B%5Cfrac%7B2%5Ctimes%206%5E2%7D%7B2%7D)
s=36 m
After fuel run out distance traveled in upward direction is
![v^2-u^2=2as_0](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2as_0)
here v=0
u=12 m/s
![a=9.8 m/s^2](https://tex.z-dn.net/?f=a%3D9.8%20m%2Fs%5E2)
![0-12^2=2(-9.8)(s)](https://tex.z-dn.net/?f=0-12%5E2%3D2%28-9.8%29%28s%29)
![s_0=\frac{144}{2\times 9.8}=7.34 m](https://tex.z-dn.net/?f=s_0%3D%5Cfrac%7B144%7D%7B2%5Ctimes%209.8%7D%3D7.34%20m)
![s+s_0=36+7.34=43.34 m](https://tex.z-dn.net/?f=s%2Bs_0%3D36%2B7.34%3D43.34%20m)
These tools enable the specific execution of a task or a group of tasks allowing the fulfillment of specific objectives within different stages of product development