The Apollo 11 mission is famous for putting the first man on the moon. Jenny informed her friends that the Apollo 11 lunar module was on the surface of the moon for $8 \times 10^7$ milliseconds. What would be the most appropriate unit of time for Jenny to use instead of milliseconds?
Added by Dana G.
Close
Step 1
1 second = 1000 milliseconds, so (8 × 10^7) ms ÷ 1000 = 8 × 10^4 s = 80,000 seconds. Show more…
Show all steps
Your feedback will help us improve your experience
Suzanne Harwood and 86 other Algebra educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Apollo 11 On July $19,1969,$ Apollo $11^{\prime}$ s revolution around the Moon was adjusted to an average orbit of $111 \mathrm{km} .$ The radius of the Moon is $1785 \mathrm{km},$ and the mass of the Moon is $7.3 \times 10^{22} \mathrm{kg}.$ a. How many minutes did Apollo 11 take to orbit the Moon once? b. At what velocity did Apollo 11 orbit the Moon?
Gravitation
Using the Law of Universal Gravitation
The Apollo 11 spacecraft that landed on the moon in 1969 traveled there at a speed relative to the earth of $1.08 \times 10^{4} \mathrm{~m} / \mathrm{s}$. To an observer on the earth, how much longer than his own day was a day on the spacecraft?
Pranay S.
Recommended Textbooks
Elementary and Intermediate Algebra
Algebra and Trigonometry
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD