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:
Explanation:
Given that:
The air resistance and friction = 700 N
The gravity caused force = 716 × 9.8 = 7016.8
Total force = (7016.8 + 700) N
Total force = 7716.8 N
∴
Answer:
F - fr = ma
, N - W = 0
Explanation:
In this exercise we are asked to identify the forces that act on the jack, for this we will use Newton's second law
On the x axis
We have two forces: the friction and the force of the girl who pulls the cat
F - fr = ma
On the y axis
There are two forces: normal and weight
N - W = 0
A diagram of these forces can be seen in the attachment
B. trajectory
because it is a set projectile wether or not it was calculated
plz mark brainliest
Answer:
The poles of an electromagnet can even be reversed by reversing the flow of electricity. If a wire carrying an electric current is formed into a series of loops, the magnetic field can be concentrated within the loops. ... All of their little magnetic fields add together, creating a stronger magnetic field.
Explanation:
