The short answer is that the displacement is equal tothe area under the curve in the velocity-time graph. The region under the curve in the first 4.0 s is a triangle with height 10.0 m/s and length 4.0 s, so its area - and hence the displacement - is
1/2 • (10.0 m/s) • (4.0 s) = 20.00 m
Another way to derive this: since velocity is linear over the first 4.0 s, that means acceleration is constant. Recall that average velocity is defined as
<em>v</em> (ave) = ∆<em>x</em> / ∆<em>t</em>
and under constant acceleration,
<em>v</em> (ave) = (<em>v</em> (final) + <em>v</em> (initial)) / 2
According to the plot, with ∆<em>t</em> = 4.0 s, we have <em>v</em> (initial) = 0 and <em>v</em> (final) = 10.0 m/s, so
∆<em>x</em> / (4.0 s) = (10.0 m/s) / 2
∆<em>x</em> = ((4.0 s) • (10.0 m/s)) / 2
∆<em>x</em> = 20.00 m
Answer:
A
Explanation:
Iron and gadlinium are both very easily made into magnetic substances. Cobalt is also capable of being magnetized. Aluminum, put in an alloy, can make a magnetic substance, but
Aluminum by itself is not able to be magnetized.
<span>Final Velocity = Vf = 0 m/s --------------> (Vf = 0 because ball's speed at its max height is 0)
Initial Velocity = Vi = ?
Total time (upward & downward) = 8.0 seconds
* Time upward = 4 seconds & ................( As time for ball upward & downward is equal )
* Time downward = 4 seconds..
Gravitational Acceleration = g = -9.8 m/s²
Use Equation;
Vf = Vi - gt
0 = Vi - 9.8 * 4
0 = Vi - 39.2
39.2 = Vi
=> Vi = Initial Velocity = 39.2 m/s</span>
There's so much going on here, in a short period of time.
<u>Before the kick</u>, as the foot swings toward the ball . . .
-- The net force on the ball is zero. That's why it just lays there and
does not accelerate in any direction.
-- The net force on the foot is 500N, originating in the leg, causing it to
accelerate toward the ball.
<u>During the kick</u> ... the 0.1 second or so that the foot is in contact with the ball ...
-- The net force on the ball is 500N. That's what makes it accelerate from
just laying there to taking off on a high arc.
-- The net force on the foot is zero ... 500N from the leg, pointing forward,
and 500N as the reaction force from the ball, pointing backward.
That's how the leg's speed remains constant ... creating a dent in the ball
until the ball accelerates to match the speed of the foot, and then drawing
out of the dent, as the ball accelerates to exceed the speed of the foot and
draw away from it.
That is the answer to the question
I hope this helps you.
Thank you for your question
