3.1 the only reason i know this is cause i got it wrong
Answer:
The airplane should release the parcel
m before reaching the island
Explanation:
The height of the plane is
, and its speed is v=150 m/s
When an object moves horizontally in free air (no friction), the equation for the y measured with respect to ground is
[1]
And the distance X is
x = V.t [2]
Being t the time elapsed since the release of the parcel
If we isolate t from the equation [1] and replace it in equation [2] we get

Using the given values:

x =
m
<span>These are inert gases, so we can assume they don't react with one another. Because the two gases are also subject to all the same conditions, we can pretend there's only "one" gas, of which we have 0.458+0.713=1.171 moles total. Now we can use PV=nRT to solve for what we want.
The initial temperature and the change in temperature. You can find the initial temperature easily using PV=nRT and the information provided in the question (before Ar is added) and solving for T.
You can use PV=nRT again after Ar is added to solve for T, which will give you the final temperature. The difference between the initial and final temperatures is the change. When you're solving just be careful with the units!
SIDE NOTE: If you want to solve for change in temperature right away, you can do it in one step. Rearrange both PV=nRT equations to solve for T, then subtract the first (initial, i) from the second (final, f):
PiVi=niRTi --> Ti=(PiVi)/(niR)
PfVf=nfRTf --> Tf=(PfVf)/(nfR)
ΔT=Tf-Ti=(PfVf)/(nfR)-(PiVi)/(niR)=(V/R)(Pf/nf-Pi/ni)
In that last step I just made it easier by factoring out the V/R since V and R are the same for the initial and final conditions.</span>
We get the clearest image if there is no magnification. When we have no magnification the image and real object have the same size.
If we look at the diagram that I attached we can see that:

Two triangles that I marked are similar and from this we get:

The image and the object must have the same height so we get:

This tells how far the screen should be from the lens.
The position of the screen on the optical bench is: