The whole question is talking about the amplitude of a wave
that's transverse and wiggling vertically.
Equilibrium to the crest . . . that's the amplitude.
Crest to trough . . . that's double the amplitude.
Trough to trough . . . How did that get in here ? Yes, that's
the wavelength, but it has nothing to do
with vertical displacement.
Frequency . . . that's how many complete waves pass a mark
on the ground every second. Doesn't belong here.
Notice that this has to be a transverse wave. If it's a longitudinal wave,
like sound or a slinky, then it may not have any displacement at all
across the direction it's moving.
It also has to be a vertically 'polarized' wave. If it's wiggling across
the direction it's traveling BUT it's wiggling side-to-side, then it has
no vertical displacement. It still has an amplitude, but the amplitude
is all horizontal.
The sun <u><em>appears</em></u> brighter than any other star.
(It isn't really, but it looks that way because it's much much much much much much closer to us than any other star.)
Concave makes things smaller and convex makes things bigger
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:
Travelled 18 km, they are 6 km from home.
Explanation:
12/2 (halfway) is 6km. So, 6 + 12 would be 18 km, total amount travelled. The total distance of the trip would be 24 km (12 km out, 12km back) if they travelled 12+6 (18km) then they only have 6 km more to go.
