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nata0808 [166]
3 years ago
15

Waldo gets stopped by the police, and the police officer tells Waldo that she caught him on radar doing 50 miles per hour in a 4

0 miles per hour zone.
Waldo responds, " I was traveling 50 miles per hour. I sat at home for 1 hour and then drove for 1 hour before you stopped me. I traveled 50 miles total in that second hour. So, my speed was 25 miles per hour. That's well under the speed limit."
Waldo is wrong, per speeding laws, and does receive a speeding ticket. Explain how Waldo came up with a speed of 25 miles per hour in multiple sentences.
Physics
2 answers:
sergij07 [2.7K]3 years ago
5 0

Answer:

He calculated his average speed.

Explanation:

Waldo calculated an average of his speed in 2 hours rather than his speed whilst driving which means instead of doing 50/1 hour = 50 mph, he did 50/2hours which is how he got 25 mph. Hope this helps!

slava [35]3 years ago
5 0
50/2 hours=25
I think this is correct!
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What is the orbital period of a spacecraft in a low orbit near the surface of mars? The radius of mars is 3.4×106m.
valkas [14]
<h2>Answer: 56.718 min</h2>

Explanation:

According to the Third Kepler’s Law of Planetary motion<em> </em><em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”. </em>

In other words, this law states a relation between the orbital period T of a body (moon, planet, satellite) orbiting a greater body in space with the size a of its orbit.

This Law is originally expressed as follows:

T^{2}=\frac{4\pi^{2}}{GM}a^{3}   (1)

Where;

G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

M=6.39(10)^{23}kg is the mass of Mars

a=3.4(10)^{6}m  is the semimajor axis of the orbit the spacecraft describes around Mars (assuming it is a <u>circular orbit </u>and a <u>low orbit near the surface </u>as well, the semimajor axis is equal to the radius of the orbit)

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}    (2)

T=\sqrt{\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(6.39(10)^{23}kg)}(3.4(10)^{6}m)^{3}}    (3)

T=\sqrt{11581157.44 s^{2}}    (4)

Finally:

T=3403.1099s=56.718min    This is the orbital period of a spacecraft in a low orbit near the surface of mars

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