In which direction does the electric field point at a position directly north of a positive charge?
2 answers:
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<h2>Answer: the falling time</h2>
Explanation:
When a body or object falls, basically two forces act on it:
1. The force of air friction, also called<em> </em><u><em>"drag force"</em></u> :
(1)
Where:
is the drag coefficient
is the density of the fluid (air for example)
is the velocity
is the transversal area of the object
So, this force is proportional to the transversal area of the falling element and to the square of the velocity.
2. Its <u>weight </u>due to the gravity force :
(2)
Where:
is the mass of the object
is the acceleration due gravity
So, at the moment <u>when the drag force equals the gravity force, the object will have its terminal velocity:</u>
(3)
(4)
(5) This is the terminal velocity
As we can see, there is no "falling time" in this equation.
Therefore, the terminal velocity is not dependent on the falling time.
The elastic potential energy of a spring is given by
where k is the spring's constant and x is the displacement with respect to the relaxed position of the spring.
The work done by the spring is the negative of the potential energy difference between the final and initial condition of the spring:
In our problem, initially the spring is uncompressed, so
. Therefore, the work done by the spring when it is compressed until 
is
And this value is actually negative, because the box is responsible for the spring's compression, so the work is done by the box.
where k is the spring's constant and x is the displacement with respect to the relaxed position of the spring.
The work done by the spring is the negative of the potential energy difference between the final and initial condition of the spring:
In our problem, initially the spring is uncompressed, so
And this value is actually negative, because the box is responsible for the spring's compression, so the work is done by the box.
Answer:
The ball would have landed 3.31m farther if the downward angle were 6.0° instead.
Explanation:
In order to solve this problem we must first start by doing a drawing that will represent the situation. (See picture attached).
We can see in the picture that the least the angle the farther the ball will go. So we need to find the A and B position to determine how farther the second shot would go. Let's start with point A.
So, first we need to determine the components of the velocity of the ball, like this:
we pick the positive one, so it takes 0.317s for the ball to hit on point A.
so now we can find the distance from the net to point A with this time. We can find it like this:
Once we found the distance between the net and point A, we can similarly find the distance between the net and point B:
t= -0.9159s or t=0.468s
we pick the positive one, so it takes 0.468s for the ball to hit on point B.
so now we can find the distance from the net to point B with this time. We can find it like this:
So once we got the two distances we can now find the difference between them:
so the ball would have landed 3.31m farther if the downward angle were 6.0° instead.
