Answer:
The Richter scale measures the largest wiggle (amplitude) on the recording, but other magnitude scales measure different parts of the earthquake. The USGS currently reports earthquake magnitudes using the Moment Magnitude scale, though many other magnitudes are calculated for research and comparison purposes.
It would have to be 36,719 Km high in order to be to be in geosynchronous orbit.
To find the answer, we need to know about the third law of Kepler.
<h3>What's the Kepler's third law?</h3>
- It states that the square of the time period of orbiting planet or satellite is directly proportional to the cube of the radius of the orbit.
- Mathematically, T²∝a³
<h3>What's the radius of geosynchronous orbit, if the time period and altitude of ISS are 90 minutes and 409 km respectively?</h3>
- The time period of geosynchronous orbit is 24 hours or 1440 minutes.
- As the Earth's radius is 6371 Km, so radius of the ISS orbit= 6371km + 409 km = 6780km.
- If T1 and T2 are time period of geosynchronous orbit and ISS orbit respectively, a1 and a2 are radius of geosynchronous orbit and ISS orbit, as per third law of Kepler, (T1/T2)² = (a1/a2)³
- a1= (T1/T2)⅔×a2
= (1440/90)⅔×6780
= 43,090 km
- Altitude of geosynchronous orbit = 43,090 - 6371= 36,719 km
Thus, we can conclude that the altitude of geosynchronous orbit is 36,719km.
Learn more about the Kepler's third law here:
brainly.com/question/16705471
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The seventh planet from the sun is Uranus.
Answer:
Explanation:
Hello! To solve this problem we must be clear about the concept of energy conservation, and kinetic energy with the following sentence
The kinetic energy of the two cars (v = 1.2m / S) plus the kinetic energy of the third car (v = 3.5m / S) must be equal to the kinetic energy of the three cars together.
The kinetic energy is calculated by the following equation.
m= mass of the cars=26500kg
V=speed
E=kinetic energy
taking into account the above, the following equation is inferred
1= the cars are separated
2=
the cars are togheter
E1=E2
where
m= mass of each car
V1= 1.2m/s
Va=3.5,m/S
m= mass of each car
V=speed (in m/s) of the three coupled cars after the first couples with the other two
Solving
the speed of the three coupled cars after the first couples with the other two is 2.245m/s
Yes D is definitely the answer
<em>V</em><em>=</em><em>I×</em><em>R</em>
<em>V</em><em>=</em><em>0</em><em>.</em><em>0</em><em>2</em><em>5</em><em>×</em><em>3</em><em>6</em>
<em>V</em><em>=</em><em>0</em><em>.</em><em>9</em><em> </em><em>sa</em><em>me</em><em> </em><em>as</em><em> </em><em>0</em><em>.</em><em>9</em><em>0</em>
