Ok, I think this is right but I am not sure:
Q = ϵ
0AE
A= π π
r^2
=(8.85x10^-12 C^2/Nm^2)
( π π (0.02m)^2)
(3x10^6 N/C) =3.3x10^-8 C = 33nC N = Q/e = (3.3x10^-8 C)/(1.60x10^-19 C/electron) = 2.1x10^11 electrons
Answer:
a. N = 2.49W b. 0.40
Explanation:
a. What is the magnitude of the normal force FNFN between a rider and the wall, expressed in terms of the rider's weight W?
Since the normal force equals the centripetal force on the rider, N = mrω² where r = radius of cylinder = 3.05 m and ω = angular speed of cylinder = 0.450 rotations/s = 0.450 × 2π rad/s = 2.83 rad/s
Now N = mrω² = m(3.05 m) × (2.83 rad/s)² = 24.43m
The rider's weight W = mg = 9.8m
The ratio of the normal force to the rider's weight is
N/W = 24.43m/9.8m = 2.49
So the normal force expressed in term's of the rider's weight is
N = 2.49W
b. What is the minimum coefficient of static friction µsμs required between the rider and the wall in order for the rider to be held in place without sliding down?
The frictional force, F on the rider by the wall of the cylinder equals the weight, W of the rider. F = W.
Since the frictional force F = μN, where μ = coefficient of static friction between rider and wall of cylinder and N = normal force between rider and wall of cylinder.
So, the normal force equals
N = F/μ = W/μ = mg/μ = mrω²
μ = mg/mrω²
= W/N
= 9.8m/24.43m
= 0.40
<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:
the efficiency of hydralic is 79.88%
Explanation:
convert mm to m
1mm = (1/1000)m
diameter of pipe upsteam
d₁= 90mm= 0.09m
diameter of pipe downsteam
d₂= 30mm = 0.03m
finding velocity of upsteam
recall Q=A₁V₁
V₁=Q/A₁
V₁=3.14m/s
velocity of downsteam
V₂= Q/A₂
V₂= 28.29m/s
mass flow rate
m= ρQ
ρ is the density of water
m = 1000× 0.02
m= 20kg/s
the efficiency of hydralic is 79.88%
Arginine is a basic aminoacid, because it has two amino groups and one acid
group.
At a low pH, every ionizable group is protoned. At a little higher pH, the
acid group looses its proton. A little higher pH, one amino group looses its
proton. At a very high pH, all ionizable groups are not protoned.
Pkas
<span>
<span><span>
<span>
pka1 = 1.82
</span>
<span>
pka2 = 8.99
</span>
<span>
pka3 = 12.48
</span>
</span>
</span></span>
So 9.20 is higher tan the second pKa and lower than the third pka. This
means the acid has already lost its proton, and one of the aminos too, but the
second amino hasn’t. When an acid is not protoned, it has a negative charge.
When an amino is not protoned, it’s neutral. When an amino is protoned, it has
a positive charge. So this amnino acid has one positive charge (one of the aminos) and one negative
charge (the acid), what makes it neutral.
