To measure the strength of an earthquake, you can use either a Richter scale or Mercalli scale. Richter scale uses the amplitued of the wave and the distance from the source. Mercalli scale uses observations of people and is not considered to be scientific as Richter scale.
Answer:
22 N applied force
Explanation: Since they are both pushing the wagon in the same direction the force adds up.

Explanation:
The acceleration due to gravity g is defined as

and solving for R, we find that

We need the mass M of the planet first and we can do that by noting that the centripetal acceleration
experienced by the satellite is equal to the gravitational force
or

The orbital velocity <em>v</em> is the velocity of the satellite around the planet defined as

where <em>r</em><em> </em>is the radius of the satellite's orbit in meters and <em>T</em> is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

Solving for <em>M</em>, we get

Putting this expression back into Eqn(1), we get




Answer:
The balloon would still move like a rocket
Explanation:
The principle of work of this system is the Newton's third law of motion, which states that:
"When an object A exerts a force on an object B (action), object B exerts an equal and opposite force (reaction) on object A"
In this problem, we can identify the balloon as object A and the air inside the balloon as object B. As the air goes out from the balloon, the balloon exerts a force (backward) on the air, and as a result of Newton's 3rd law, the air exerts an equal and opposite force (forward) on the balloon, making it moving forward.
This mechanism is not affected by the presence or absence of surrounding air: in fact, this mechanism also works in free space, where there is no air (and in fact, rockets also moves in space using this system, despite the absence of air).
A. Bones provide protection for organs in the body