Answer:
real, inverted, and smaller than the object
Explanation:
When the object is placed beyond the center of curvature, the image will formed between the focus and the center of curvature. The size of the image is diminished and its nature is real and inverted.
The whole description is shown in the attached figure. It is clear that the size of the image is smaller than the object.
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:
a. P.E = 3430Joules.
b. Workdone = 3430Nm
Explanation:
<u>Given the following data;</u>
Mass = 70kg
Distance = 5m
We know that acceleration due to gravity is equal to 9.8m/s²
To find the potential energy;
Potential energy = mgh
P.E = 70*9.8*5
<em>P.E = 3430J</em>
b. To find the workdone;
Workdone = force * distance
But force = mass * acceleration
Force = 70*9.8
Force = 686 Newton.
Workdone = 686 * 5
<em>Workdone = 3430Nm</em>
Work = force x distance
You can see time doesn’t matter (if we were talking about power, which is the RATE at which work is performed, that would be a different story).
W = 2 x 5 = 10 foot-pounds of work
Foot-pounds are gross units. Better to work in SI units when you can!
Answer: Accoding to research "Triton is unique among all the large moons in the solar system because it orbits Neptune in a direction opposite to the planet's rotation (a "retrograde" orbit). It is unlikely to have formed in this configuration and was probably captured from elsewhere."
Explanation:
