For this case, let's
assume that the pot spends exactly half of its time going up, and half going
down, i.e. it is visible upward for 0.245 s and downward for 0.245 s. Let us take
the bottom of the window to be zero on a vertical axis pointing upward. All calculations
will be made in reference to this coordinate system. <span>
An initial condition has been supplied by the problem:
s=1.80m when t=0.245s
<span>This means that it takes the pot 0.245 seconds to travel
upward 1.8m. Knowing that the gravitational acceleration acts downward
constantly at 9.81m/s^2, and based on this information we can use the formula:
s=(v)(t)+(1/2)(a)(t^2)
to solve for v, the initial velocity of the pot as it enters
the cat's view through the window. Substituting and solving (note that
gravitational acceleration is negative since this is opposite our coordinate
orientation):
(1.8m)=(v)(0.245s)+(1/2)(-9.81m/s^2)(0.245s)^2
v=8.549m/s
<span>Now we know the initial velocity of the pot right when it
enters the view of the window. We know that at the apex of its flight, the
pot's velocity will be v=0, and using this piece of information we can use the
kinematic equation:
(v final)=(v initial)+(a)(t)
to solve for the time it will take for the pot to reach the
apex of its flight. Because (v final)=0, this equation will look like
0=(v)+(a)(t)
Substituting and solving for t:
0=(8.549m/s)+(-9.81m/s^2)(t)
t=0.8714s
<span>Using this information and the kinematic equation we can find
the total height of the pot’s flight:
s=(v)(t)+(1/2)(a)(t^2) </span></span></span></span>
s=8.549m/s (0.8714s)-0.5(9.81m/s^2)(0.8714s)^2
s=3.725m<span>
This distance is measured from the bottom of the window, and
so we will need to subtract 1.80m from it to find the distance from the top of
the window:
3.725m – 1.8m=1.925m</span>
Answer:
<span>1.925m</span>
Answer:
The answer is A.
Explanation:
Hypertrophy is an increase and growth of muscle cells. Hypertrophy refers to an increase in muscular size achieved through exercise. When you work out, if you want to tone or improve muscle definition, lifting weights is the most common way to increase hypertrophy.
Hope that helped.
The question doesn't give us enough information to answer.
The answer depends on the mass of the object, how long the force
acts on the object, the OTHER forces on the object, and whether the
object is free to move.
-- If you increase the force with which you push on a brick wall,
the amount of work done remains unchanged, namely Zero.
-- If you push on a pingpong ball with a force of 1 ounce for 1 second,
the ball accelerates substantially, it moves a substantial distance, and
so the work done is substantial.
-- But if you push on a battleship, even with a much bigger force ...
let's say 1 pound ... and keep pushing for a month ... the ship accelerates
microscopically, moves a microscopic distance, and the work done by
your force is microscopic.
