No one expected violet & ultraviolet spectral lines to be shifted towards the red.
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.
Answer: displacement of airplane is 172 km in direction 34.2 degrees East of North
Explanation:
In constructing the two displacements it is noticed that the angle between the 75 km vector and the 155 km vector is a right angle (90 degrees).
Hence if the plane starts out at A, it travels to B, 75 km away, then turns 90 degrees to the right (clockwise) and travels to C, 155 km away from B. Angle ABC is 90 degrees, hence we can use Pythagoras theorem to solve for AC
AC2 = AB2 + BC2 ; AC^2 = 752 + 1552 ; from this we get AC = 172 km (3 significant figures)
Angle BAC = Tan-1(155/75) ; giving angle BAC = 64.2 degrees
Hence AC is in a direction (64.2 - 30) = 34.2 degrees East of North
Therefore the displacement of the airplane is 172 km in a direction 34.2 degrees East of North
Answer:
Change in potential energy of the block-spring-Earth
system between Figure 1 and Figure 2 = 1 Nm.
Explanation:
Here, spring constant, k = 50 N/m.
given block comes down eventually 0.2 m below.
here, g = 10 m/s.
let block be at a height h above the ground in figure 1.
⇒In figure 2,
potential energy of the block-spring-Earth
system = m×g×(h - 0.2) + 1/2× k × x². where, x = change in spring length.
⇒ Change in potential energy of the block-spring-Earth
system between Figure 1 and Figure 2 = (m×g×(h - 0.2)) - (1/2× k × x²)
= (1×10×0.2) - (1/2×50×0.2×0.2) = 1 Nm.
