Tectonic plates are large segments of the earth's crust that move slowly. suppose one such plate has an average speed of 4.8 cm
1 answer:
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To solve for force, you need to get the product of mass and acceleration.
F = ma
Your given is:
m = 150g
a = ?
v = 50 m/s
t = 20ms
As you can see, you do not have acceleration yet. But if you read the problem you can come up with the formula of acceleration.
Acceleration is the change in velocity over a period of time.
a = change in velocity/time
To get the change in velocity, you get the difference between the initial velocity and final velocity:
The ball was moving initially at a velocity of 50 m/s and it came to a stop. This is your clue. If a ball comes to a stop then that means that the final velocity of the ball is 0 m/s.
So we can put it into our formula now:
WAIT! As you can see, the units do not match. We have ms and s into our equation and that means you cannot proceed till they are the same. First we need to convert ms to s.
x 
= 
So your new time is 0.02s. Now we put this time into the formula:
As you can see our acceleration is a negative value, this indicates that it decelerated or slowed down which makes sense because it was brought to a stop.
So now we have our acceleration. Now using this, we can get our force.
Before we start doing this, you need to take note that the unit of force is N, but when you expand it, it is
but as you can see our mass given is in grams. So again, before you put them into the equation we need to change it into kg first.
Our new mass is 0.150kg.
To make things clearer, let us write down all our new values:
m = 0.150kg
a = -2,500
Now that all our units match, we can put that into our formula:
The value again is negative because it is going against the initial direction of the ball. But if your instructor just wants to get the value of force or the magnitude of the force, just disregard the sign.
F = ma
Your given is:
m = 150g
a = ?
v = 50 m/s
t = 20ms
As you can see, you do not have acceleration yet. But if you read the problem you can come up with the formula of acceleration.
Acceleration is the change in velocity over a period of time.
a = change in velocity/time
To get the change in velocity, you get the difference between the initial velocity and final velocity:
The ball was moving initially at a velocity of 50 m/s and it came to a stop. This is your clue. If a ball comes to a stop then that means that the final velocity of the ball is 0 m/s.
So we can put it into our formula now:
WAIT! As you can see, the units do not match. We have ms and s into our equation and that means you cannot proceed till they are the same. First we need to convert ms to s.
So your new time is 0.02s. Now we put this time into the formula:
As you can see our acceleration is a negative value, this indicates that it decelerated or slowed down which makes sense because it was brought to a stop.
So now we have our acceleration. Now using this, we can get our force.
Before we start doing this, you need to take note that the unit of force is N, but when you expand it, it is
Our new mass is 0.150kg.
To make things clearer, let us write down all our new values:
m = 0.150kg
a = -2,500
Now that all our units match, we can put that into our formula:
The value again is negative because it is going against the initial direction of the ball. But if your instructor just wants to get the value of force or the magnitude of the force, just disregard the sign.
Answer: 17.83 AU
Explanation:
According to Kepler’s Third Law of Planetary motion <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>
(1)
Talking in general, this law states a relation between the <u>orbital period</u> of a body (moon, planet, satellite, comet) orbiting a greater body in space with the <u>size</u>
of its orbit.
However, if is measured in <u>years</u>, and
is measured in <u>astronomical units</u> (equivalent to the distance between the Sun and the Earth:
), equation (1) becomes:
(2)
This means that now both sides of the equation are equal.
Knowing and isolating
from (2):
(3)
(4)
Finally:
(5)