5. The price of a bulb is 260. With 12% GST, it will cost
291.20. b. 301.20.
c. *270.
d. 286.
Share
5. The price of a bulb is 260. With 12% GST, it will cost
291.20. b. 301.20.
c. *270.
d. 286.
Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
Step-by-step explanation:
Q.2 Fill in the blanks. (04)
1) Slow ____________,the doctor called it!
2) “ What a lot of leaves to ________this fall!”
3) “ How you do go on,Maya,”__________sighed.
4) His little__________motor.
Answer:
Kepler's Third Law
Johannes Kepler was a German astronomer as well as a mathematician. He gave the Three Laws of Planetary Motion.
To solve the question, we will be using the Kepler's Third Law, which states that the square of the orbital time period is proportional to the cube of the semi-major axis of the orbit.
For a circular orbit, the semi-major axis is equal to the radius.
In Mathematical Form,
\Large\boxed{\sf T^2 \propto R^3}T2∝R3
Here, we have the following data:-
• Earth
Orbital Radius = \sf R_1R1 = 149.6 million km
Orbital Period = \sf T_1T1 = 1.00 years
• Mars
Orbital Radius = \sf R_2R2
Orbital Period = \sf T_2T2 = 1.88 years
Using Kepler's Third Law, we can calculate the Orbital Radius of Mars.
\begin{gathered}\sf\displaystyle T^2\propto R^3 \\\\\\ \implies \sf \frac{T_1^2}{T_2^2} = \frac{R_1^3}{R_2^3} \\\\\\ \sf \implies \left(\frac{1.00}{1.88}\right)^2 = \left(\frac{149.6}{R_2}\right)^3 \\\\\\\sf \implies R_2^3 = 149.6^3 \times 1.88^2 \\\\\\ \sf\implies R_2 = \sqrt[3]{149.6^3 \times 3.5344} \\\\\\\sf \implies R_2 = 149.6 \times \sqrt[3]{3.5344} \\\\\\ \sf \implies R_2 \approx 227.878\ \textsf{million km} \\\\\\\implies \boxed{\sf R_2 \approx 227.9 \textsf{ million km}}\end{gathered}T2∝R3⟹T22T12=R23R13⟹(1.881.00)2=(R2149.6)3⟹R23=149.63×1.882⟹R2=3149.63×3.5344⟹R2=149.6×33.5344⟹R2≈227.878 million km⟹R2≈227.9 million km
Thus, The Orbital Radius of Mars is approximately 227.9 million kilometres.
Now, we need the smallest possible distance between Mars and Earth. This can happen when the centre of Sun, Earth and Mars are all in the same line.
This is when Mars will be closest to Earth.
Consider the image attached. The smallest distance between the Earth and Mars will be \sf R_2 - R_1R2−R1
If we call this smallest distance as d, then:
\begin{gathered}\sf d = R_2 - R_1 \\\\\\ \implies \sf d = \textsf{(227.9 - 149.6) million km} \\\\\\ \implies \Large\boxed{\sf d = \textsf{78.3 million km}}\end{gathered}d=R2−R1⟹d=(227.9 - 149.6) million km⟹d=78.3 million km
• Thus, The Smallest Distance between Earth and Mars is about 78.3 million km.