Jeff uses all of his weight to lift the 600 lbs. C is your answer i do believe
Ye sit does. But since people weren’t there to hear it, doesn’t mean that it didn’t make a sound.
Answer:
The rifle barrel must be pointed at a height of 4.45cm above the target so that the bullet hits dead center.
Explanation:
First, we need to sketch the situation so we can have a better idea of what the problem looks like (Refer to uploaded picture).
So as you may see in the drawing, when pointing the rifle to the target, we can see it as a triangle, but in reality, the bullet will have a parabolic trajectory. Both points of view will help us determine what the height must be. In order to find it, we need to first determine at what angle the bullet should be shot. In order to do so we can use the range formula, which looks like this:

Where R is the range of the bullet (this is how far it goes before it has the
same height it was shot from), v is the original speed of the bullet, θ is the angle at which the bullet is shot and g is the acceleration of gravity.
We can solve this equation for theta, so we get:




so now we can substitute the given data:

so we get:
θ=0.05614°
once we get the angle, we can look at the triangle diagram. From the drawing we can see that we can use the tan function to find the height:

so we can solve this for h, so we get:

which yields:

or

Answer:
D. location
Explanation:
The gravitational field strength of Earth is determined by the virtue of the location within the Earth's gravitational field.
That's why all objects regardless of their mass, shape, and size free fall towards the Earth with an acceleration equal to the acceleration at that location in the absence of air resistance.
According to the gravitational force between two bodies, the force experienced by one body due to the other is independent of its own mass.
The gravitational force is given by equation
F = GMm/r²
If F is the force acting on the smaller body of mass 'm', then
F = ma
Therefore, the equation becomes,
ma = GMm/r²
a = GM/r²
The value of 'a' changes with respect to the value of 'r' such that if 'r' is the radius of the Earth, then the acceleration at a height 'h' from Earth surface is given by
a = GM/(r+h)²
Here it is clear that the acceleration at any point is only the inherent property of the Earth itself.
The gravitational field strength of Earth is determined by the virtue of the location within the Earth's gravitational field.