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Wittaler [7]
4 years ago
10

A tetrameric protein dissociates into dimers when the detergent sodium dodecyl sulfate (SDS) is added to a solution of the prote

in. But the dimers are termed SDS-resistant because they do not further dissociate into monomers in the presence of the detergent. What intermolecular forces might be acting at the dimer-dimer interface? Are the intermolecular forces acting at the monomer-monomer interface different? Explain.
Physics
1 answer:
KatRina [158]4 years ago
5 0

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.

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A car traveling at 28.0 m/s hits a bridge abutment. A passenger in the car, who has a mass of 49.0 kg, moves forward a distance
BaLLatris [955]

The final velocity of the passenger is zero as he is brought to rest by the inflated bag.

V_f = 0

Apply the equation of motion

V_f^2 = V_i^2 +2as

Replacing with our values,

0 = 28^2+2(a)(0.55)

a = \frac{28^2}{2(0.55)}

a = 712.72m/s^2

Calculate the force using the force equation,

F = ma

F = (49kg)(712.72m/s^2)

F = 34.923kN

Therefore the magnitude of force acts on the passenger's upper torso is 34.923kN

3 0
3 years ago
Equation: S=D]
Ahat [919]

The speed of car is 106.67 miles per hour

To solve problems involving speed and distance we are required to first find speed or distance travelled in an hourly timeframe

It is given that car takes 4 hours to cover the distance at a speed of 40 miles per hour

Therefore, total distance travelled by the car= speed x time

                                                                          = 40 miles per hour x 4 hour

                                                                          =160 miles

We are required to find the speed of the car if the distance is to be covered in 1.5 hours

We are required to find the speed by the formula speed= Distance/time

distance= 160 miles     time = 1.5 hour

speed= 160/1.5=106.666=106.67 miles per hour

Hence the speed is 106.67 miles per hour

For further reference:

brainly.com/question/10930186?referrer=searchResults

#SPJ9

7 0
2 years ago
A solid cylinder (1 =1/2mr2 ) with a mass of 4.83 kg and a radius of 0.057 m starts
netineya [11]

Answer:

v = 7.32 m/s

Explanation:

The potential energy will convert to kinetic energy

        ½Iω² + ½mv² = mgh

             Iω² +  mv² = 2mgh

(½mR²)(v/R)² + mv² = 2mgh

           ½mv² + mv² = 2mgh

                 ½v² + v² = 2gh

                      3v²/2 = 2gh

                           v² = 4gh/3

                           v² = 4(9.81)(4.10)/3

                           v² = 53.628

                            v = 7.323114...

                            v = 7.32 m/s

3 0
3 years ago
An elevator motor in a high-rise building can do 3500 kJ of work in 5 min. Find the power developed by the motor. Explain if you
Ilya [14]

Answer:

P = 11666.6 W

Explanation:

Given that,

Work done by the motor, W = 3500 kJ

Time, t = 5 min = 300 s

We need to find the power developed by the motor. Power developed is given by :

P=\dfrac{E}{t}\\\\P=\dfrac{3500\times 10^3}{300}\\\\P=11666.7\ W

So, the required power is 11666.6 W.

4 0
3 years ago
A 700-kg car, driving at 29 m/s, hits a brick wall and rebounds with a speed of 4.5 m/s. what is the car's change in momentum du
Viefleur [7K]

The change in the momentum of the car due to collision between car and wall is \fbox{\begin\ -23450\text{ kg.m}/\text{s}\end{minispace}} or \fbox{\begin\\-2.3450 \times {10^4}\,\text{kg.m}/\text{s}\end{minispace}}.

Further Explanation:

Let us consider the car is moving towards the right direction and it collides with the wall. After the collision, the car will bounce back or car will rebound opposite to the direction of its motion i.e., towards left. Therefore, the final velocity of the car is opposite to the direction of the initial velocity of the car.

Given:

The mass of the car is 700\,{\text{kg}}.

The velocity of the car before collision is 29\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}.

The velocity of the car after collision is 4.5\text{ m}/\text{s}.  

Concept:

The momentum of an object is defined as the product of mass of object and the velocity with which the object is moving.

The initial momentum of the car is:

\fbox{\begin\\{p_i}= m{v_i}\end{minispace}}                                   …… (I)

Here, {p_i} is the initial momentum of the car, m is the mass of the car and {v_i} is the initial velocity of the car.

Substitute 700\,{\text{kg}} for m and 29\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}} for {v_i} in equation (I).  

\begin{gathered}{p_i} = \left( {700\,{\text{kg}}} \right) \cdot \left( {29\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}} \right) \\ = 20300\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/ {\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}} \\ \end{gathered}

The final momentum of the car is defined as the product of mass of car and the velocity of the car after collision or final velocity of the car.

The final momentum of the car is:

\fbox{\begin\\{p_f} = m{v_f}\end{minispace}}                                …… (II)  

Here, {p_f} is the final momentum and {v_f} is the final velocity.

Substitute 700\,{\text{kg}} for m and -4.5\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}} for {v_f} in equation (II).  

\begin{aligned}{p_f}&=\left( {700\,{\text{kg}}} \right)\cdot\left( {- 4.5\,{{\text{m}}\mathord{\left/{\vphantom {{\text{m}}{\text{s}}}} \right.\kern-\nulldelimiterspace}{\text{s}}}}\right)\\&=-3150\,{{{\text{kg}} \cdot {\text{m}}}\mathord{\left/{\vphantom{{{\text{kg}}\cdot {\text{m}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}\\\end{aligned}

The change in the momentum of the car after collision is the difference between the momentum of car before collision and the momentum of car after collision.

The change in momentum of the car is:

\fbox{\begin\Delta p = {p_f} - {p_i}\end{minispace}}                             …… (III)  

Here, \Delta p is the change in the momentum of the car.

Substitute - 3150\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/ {\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}} for {p_f} and 20300\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/ {\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}} for {v_i} in equation (III).

\begin{aligned}\Delta{p}&=-3150\text{ kg}.\text{m}/\text{s}\ -20300\text{ kg}.\text{m}/\text{s}\\&=-23450\text{ kg}.\text{m}/\text{s}\end{aligned}

Thus, the change in the momentum of the car due to collision between car and wall is \fbox{\begin\ -23450\text{ kg.m}/\text{s}\end{minispace}} or\fbox{\begin\\-2.3450 \times {10^4}\,\text{kg.m}/{s}\end{minispace}}.

Learn more:

1. The motion of a body under friction brainly.com/question/4033012/  

2. A ball falling under the acceleration due to gravity brainly.com/question/10934170/

3.Conservation of energy brainly.com/question/3943029/    

Answer Details:

Grade: College

Subject: Physics

Chapter: Kinematics

Keywords:

Change in momentum, collision, initial velocity, final velocity, initial momentum, final momentum,-23450 kgm/s^2, -23450 kgm/s2, -2.3450*10^6 kgm/s^2, -2.3450*10^6 kgm/s2.

7 0
3 years ago
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