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.
1) weight of the box: 980 N
The weight of the box is given by:

where m=100.0 kg is the mass of the box, and
is the acceleration due to gravity. Substituting in the formula, we find

2) Normal force: 630 N
The magnitude of the normal force is equal to the component of the weight which is perpendicular to the ramp, which is given by

where W is the weight of the box, calculated in the previous step, and
is the angle of the ramp. Substituting, we find

3) Acceleration: 
The acceleration of the box along the ramp is equal to the component of the acceleration of gravity parallel to the ramp, which is given by

Substituting, we find

Answer:
a) Fₓ = 23.5 N
b) Net force = Fₓ
Explanation:
An image of the question as described is attached to this solution.
From the image attached, the forces acting on the box include the weight of the box, the normal reaction of the surface on the box, the applied force on the box and the Frictional force opposing the motion of the box (which is negligible and equal to 0)
a) From the diagram, the horizontal component of the force is
Fₓ = 25 cos 20° = 23.49 N = 25 N
b) Again, from the diagram attached, doing a force balance on the box, in the horizontal direction, we obtain
Net force = Fₓ - Frictional force
But frictional force is 0 N
Net force = Fₓ
Hope this Helps!!!
The Young modulus is given by:

where
F is the force applied

is the initial length of the wire

is the cross-sectional area of the wire

is the stretch of the wire
The wire in the problem stretches by

of its length, this means

We can also calculate the area of the wire; its radius is in fact half the diameter:

and so the area is

We know the force applied to the wire, F=20 N, so now we have everything to calculate the Young modulus:
In Millikan oil drop experiment, when the switch is opened and by altering supply the charge of electron is determined.
Explanation:
Millikan's oil drop experiment is held to determine the terminal velocity and charge of the oil drop.
Firstly without any supply of voltage when an oil drop is sprinkled and these droplets gather electrons together and gives negative charge as they pass through air.
By applying and altering voltage applied on the plates, drop can be suspended in air. Millikan observed one drop after another, varying the voltage and noting the effect. After many repetitions he concluded that charge could assume only certain fixed values.
After conducting many times he concluded 1.602176487 ×10−19 C as the charge of an electron.