In the equation for the gravitational force between two objects, which quantity must be squared?
1 answer:
Answer:
d
Explanation:
The quantity that must be squared in the equation of gravitational force is distance d.
According to the universal gravitational law, the square of the distance between two objects is inversely proportional to the force of gravity.
- Therefore, the quantity to be squared is d
The formula is given as:
Fg =
So d is the quantity that must be squared
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The strength of the gravitational field is given by:
where
G is the gravitational constant
M is the Earth's mass
r is the distance measured from the centre of the planet.
In our problem, we are located at 300 km above the surface. Since the Earth radius is R=6370 km, the distance from the Earth's center is:
And now we can use the previous equation to calculate the field strength at that altitude:
And we can see this value is a bit less than the gravitational strength at the surface, which is
.
where
G is the gravitational constant
M is the Earth's mass
r is the distance measured from the centre of the planet.
In our problem, we are located at 300 km above the surface. Since the Earth radius is R=6370 km, the distance from the Earth's center is:
And now we can use the previous equation to calculate the field strength at that altitude:
And we can see this value is a bit less than the gravitational strength at the surface, which is
Answer:
A.) The arrow`s range is 624,996 m
B.) The arrow`s range is 846.887 m, when the horse is galloping
Explanation:
We have a case of oblique movement. In these cases the movement in the X axis is a Uniform Rectelinear Movement (URM), and a Uniform Accelerated Movement (UAM) in the Y axis.
By the way, the equations that we use for the X axis will be from URM, and those for the Y axis wiil be from UAM.
<u>Equations</u>
X axis:
Y axis:
A.) First, it is necessary to know t, total time.
To figure out t value, we use UAM, since time is determined by this movement.
Now, at the end of the movement, , then
Caculate the segcond degree equation to obtain the two possible values for t:
But, in physics, time it could not be negative, so we take
Caculate now:
B.) Now, the narrow has an additional speed, that could be sum to the speed due to the bow.
Using the same procedure that item A, caculate X
First, we need to know the new time
And we obtain:
One more time, we take the positive time:
Finally: