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Andru [333]
1 year ago
12

Calculate the average charge on arginine when ph=9.20. (hint : find the average charge for each ionizable group and sum these to

gether.)
Physics
1 answer:
weqwewe [10]1 year ago
4 0
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.
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Sloan [31]

Answer:

6.67ft/s^2

Explanation:

We are given that

Initial velocity=u=18ft/s

Final velocity,v=38ft/s

Time=t=3 s

We have to find the average acceleration over that 3 s period.

We know that

Average acceleration,a=\frac{v-u}{t}{t}

Using the formula

Average acceleration,a=\frac{38-18}{3}ft/s^2

Average acceleration,a=\frac{20}{3}ft/s^2

Average acceleration,a=6.67ft/s^2

Hence, the average acceleration=6.67ft/s^2

5 0
1 year ago
A 4.00-kg mass is attached to a very light ideal spring hanging vertically and hangs at rest in the equilibrium position. The sp
Ahat [919]

Answer:

|v| = 8.7 cm/s

Explanation:

given:

mass m = 4 kg

spring constant k = 1 N/cm = 100 N/m

at time t = 0:

amplitude A = 0.02m

unknown: velocity v at position y = 0.01 m

y = A cos(\omega t + \phi)\\v = -\omega A sin(\omega t + \phi)\\ \omega = \sqrt{\frac{k}{m}}

1. Finding Ф from the initial conditions:

-0.02 = 0.02cos(0 + \phi) => \phi = \pi

2. Finding time t at position y = 1 cm:

0.01 =0.02cos(\omega t + \pi)\\ \frac{1}{2}=cos(\omega t + \pi)\\t=(acos(\frac{1}{2})-\pi)\frac{1}{\omega}

3. Find velocity v at time t from equation 2:

v =-0.02\sqrt{\frac{k}{m}}sin(acos(\frac{1}{2}))

5 0
1 year ago
Read 2 more answers
A semi is traveling down the highway at a velocity of v = 26 m/s. The driver observes a wreck ahead, locks his brakes, and begin
Dovator [93]

Answer:

fcosθ + Fbcosθ  =Wtanθ

Explanation:

Consider the diagram shown in attachment

fx= fcosθ (fx: component of friction force in x-direction ; f: frictional force)

Fbx= Fbcosθ ( Fbx: component of braking force in x-direction ; Fb: braking force)

Wx= Wtanθ (Wx: component of weight in x-direction ; W: Weight of semi)

sum of x-direction forces = 0

fx+ Fbx=Wx

fcosθ + Fbcosθ  =Wtanθ

7 0
1 year ago
You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics
valentina_108 [34]

Answer:

v = 69.82 ms^-1

Explanation:

As we know,

R = vi2 sin2Ꝋ / g  

vi2 =R g / sin2 Ꝋ where R is range R = 52m, Ꝋ = 3 Degrees

vi2 = 52 x 9.8 / sin 2(3) = 4875.227

v = 69.82 ms^-1

3 0
2 years ago
When listening to tuning forks of frequency 256 Hz and 260 Hz, one hears the following number of beats per second. (A) 0 (B) 2 (
Degger [83]

Answer:

(C) 4 beats per second.

Explanation:

As we know that the no of beats can be calculated as.

No. of beats is equal to difference in the tuning forks frequencies.

So,

n= \nu _{1}- \nu _{2}.

Substitute the values of frequencies of 2 tuning forks in the above equation.

n=(260 Hz-256 Hz)\\n=4

Therefore the number of beats per second will be hear by the observer is 4 beats per second.

3 0
1 year ago
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