Answer:
At the dimer-dimer interface there might be acting non-covalent forces (van der waals, Hidrogene bridges, hydrophobic forces)
At the monomer-monomer interface there might be covalent forces acting (disulfide bridges).
Explanation:
On the SDS-PAGE application works by disrupting non-covalent bonds in the proteins, and so denaturing them. Therefore, the disulfide bridges won´t be disrupted, so the monomers will remain bounded.
Answer:
a) 
b) 
c)
,
Explanation:
From the question we are told that:
Mass 
Velocity 
Coefficient of Rolling Friction 
a)
Generally the equation for The Propulsion Force is mathematically given by



b)
Therefore Power Required at




c)



Generally the equation for Work-done is mathematically given by



Therefore
Efficiency

Since
Energy in one gallon of gas is

Therefore


Answer:
(a) the particle position = 135 m
(b) the velocity of the particle = 44 m/s
(c) the acceleration of the particle = 50 m/s²
Explanation:
Solution to Question 2.
Given;
velocity of a particle, v = 2 - 4t + 2t³
initial position at t = 0, s₀ = 3 m
(a) the particle position at t = 3 s
s = vt
s = (2 - 4t + 2t³)t
s = 2t - 4t² + 2t⁴
s = s₀ + s₃
s = s₀ + 2(3) - 4(3²) + 2(3⁴)
s = s₀ + 6 - 36 + 162
s = s₀ + 132 m
s = 3m + 132 m
s = 135 m
(b) the velocity of the particle at t = 3 s
v = 2 - 4t + 2t³ = 2 - 4(3) + 2(3)³
v = 44 m/s
(c) the acceleration of the particle at t = 3s
v = 2 - 4t + 2t³

a = -4 + 6(3)²
a = 50 m/s²
Answer:
(a) 
(b) 
Explanation:
<u>Electric Circuits</u>
Suppose we have a resistive-only electric circuit. The relation between the current I and the voltage V in a resistance R is given by the Ohm's law:

(a) The electromagnetic force of the battery is
and its internal resistance is
. Knowing the equivalent resistance of the headlights is
, we can compute the current of the circuit by using the Kirchhoffs Voltage Law or KVL:

Solving for i

i=2.28\ A
The potential difference across the headlight bulbs is


(b) If the starter motor is operated, taking an additional 35 Amp from the battery, then the total load current is 2.28 A + 35 A = 37.28 A. Thus the output voltage of the battery, that is the voltage that the bulbs have is

<u>D. 475 </u>
Explanation:
The numbers do not increase much more then this, so it must be the max