Anything that's not supported and doesn't hit anything, and
doesn't have any air resistance, gains 9.8 m/s of downward
speed every second, on account of gravity. If it happens to
be moving up, then it loses 9.8 m/s of its upward speed every
second, on account of gravity.
(64.2 m/s) - [ (9.8 m/s² ) x (1.5 sec) ]
= (64.2 m/s) - [ 14.7 m/s ]
= 49.5 m/s . (upward)
<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>
Answer:
2.7x10⁻⁸ N/m²
Explanation:
Since the piece of cardboard absorbs totally the light, the radiation pressure can be found using the following equation:

<u>Where:</u>
: is the radiation pressure
I: is the intensity of the light = 8.1 W/m²
c: is the speed of light = 3.00x10⁸ m/s
Hence, the radiation pressure is:

Therefore, the radiation pressure that is produced on the cardboard by the light is 2.7x10⁻⁸ N/m².
I hope it helps you!
Answer:
a) 36 m
b) 64 m
Explanation:
Given:
v₀ = 0 m/2
v = 12 m/s
t = 6 s
Find: Δx
Δx = ½ (v + v₀) t
Δx = ½ (12 m/s + 0 m/s) (6 s)
Δx = 36 m
The track is 100 m, so the sprinter still has to run another 64 m.