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Degger [83]
3 years ago
12

Hemoglobin (Hb) is the O2-carrying protein in our blood. Unlike myoglobin, it has four sites allowing it to bind up to four O2 m

olecules. Additionally, the tendency of O2 to bind to each site increases as the other three sites become occupied (this is an example of a phenomenon known as positive cooperativity). To model this effect, consider a simplified hemoglobin-like protein with two binding sites. When no sites are occupied, the system energy is 0; when only one of the two sites is occupied, the energy is E; and when both sites are occupied, the energy is 2E J. Here J parameterizes the cooperativity, as when both sites are occupied, the energy is not simply the sum of the individual binding energies at each site.
Physics
1 answer:
bogdanovich [222]3 years ago
4 0

Answer:

Yes, the energy is not simply the sum of the individual binding energies at each site, it is the product of energy at each binding site of hemoglobin.

Explanation:

Myoglobin and hemoglobin are two different cells. Myoglobin binds only one oxygen while the hemoglobin has the ability to binds four oxygen atoms at its four sides. Myoglobin present in muscle tissue only while hemoglobin is present in the whole body. Oxyhemoglobin is formed when oxygen binds with hemoglobin cell. This oxygen is take to all cells and energy is released due to the breakdown of glucose molecules with this oxygen.

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A small object with a 5.0-mC charge is accelerating horizontally on a friction-free surface at 0.0050 m/s2 due only to an electr
kolbaska11 [484]

Answer:

0.002 N/C

Explanation:

Parameters given:

Charge of object, q = 5 mC = 5 * 10^{-3} C

Acceleration of object, a = 0.005 m/s^2

Mass of object, m = 2.0 g

The Electric field exerts a particular force on the object, causing it to accelerate (Electrostatic force).

We know that Electrostatic force, F, is given in terms of Electric field, E, as:

F = qE

This means that the object exerts a force of -qE on the Electric force (Action with equal and opposite reaction).

The object also has a force, F, due to its acceleration a. This force is the product of its mass and acceleration. Mathematically:

F = ma

Equating the two forces of the object, we get:

-qE = ma

=> E = \frac{-ma}{q}

Solving for E, we have:

E = \frac{-2 * 10^{-3} * 0.005}{5 * 10^{-3}} \\\\\\E = -0.002 N/C

The magnitude will be:

|E| = |-0.002| N/C = 0.002 N/C

The electric field has a magnitude of 0.002 N/C.

4 0
3 years ago
When a mass of 0.350 kg is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. The mass is now displ
pantera1 [17]

Answer:

The period is T =  0.700 \ s

Explanation:

From the question we are told that  

    The mass is m =  0.350  \ kg

     The extension of the spring is  x =  12.0 \ cm = 0.12 \ m

       

The spring constant for this is mathematically represented as

       k  = \frac{F}{x}

Where F is the force on the spring which is mathematically evaluated as

       F  =  mg  =  0.350 * 9.8

       F  =3.43 \ N

So  

    k  = \frac{3.43 }{ 0.12}

    k  = 28.583 \ N/m

The period of oscillation is mathematically evaluated as

      T =  2 \pi \sqrt{\frac{m}{k} }

substituting values

     T =  2  *  3.142*  \sqrt{\frac{0.35 }{28.583} }

     T =  0.700 \ s

   

7 0
3 years ago
a ballistic pendulum is used to measure the speed of high-speed projectiles. A 6 g bullet A is fired into a 1 kg wood block B su
Galina-37 [17]

Answer:

(a) v-bullet = 399.04 m/s

(b) I = 2.38 kg m/s

(c) T = 2.59 N

Explanation:

(a) To calculate the initial speed of the bullet, you first take into account that the kinetic energy of both wood block and bullet, just after the bullet impacts the block, is equal to the potential gravitational energy of block and bullet when the cord is at 60° respect to the vertical.

The potential energy is given by:

U=(M+m)gh       (1)

U: potential energy

M: mass of the wood block = 1 kg

m: mass of the bullet = 6g = 6.0*10^-3 kg

g: gravitational constant = 9.8m/s^2

h: distance to the ground

The distance to the ground is calculate d by using the information about the length of the cord and the degrees of the cord respect to the vertical:

h=l-lsin\theta\\\\h=2.2m-2,2m\ sin60\°=0.29m

The potential energy is:

U=(1kg+6*10^{-3}kg)(9.8m/s^2)(0.29m)=2.85J

Next, the potential energy is equal to kinetic energy of the block and the bullet at the beginning of its motion:

U=\frac{1}{2}(M+m)v^2\\\\v=\sqrt{2\frac{U}{M+m}}=\sqrt{2\frac{2.85J}{1kg+6*10^{-3}kg}}=2.38\frac{m}{s}

Next, you use the momentum conservation law, in order to calculate the speed of the bullet before the impact:

Mv_1+mv_2=(M+m)v    (2)

v1: initial velocity of the wood block = 0m/s

v2: initial speed of the bullet

v: speed of bullet and block = 2.38m/s

You solve the equation (2) for v2:

M(0)+mv_2=(M+m)v    

v_2=\frac{M+m}{m}v=\frac{1kg+6*10^{-3}kg}{6*10^{-3}kg}(2.38m/s)\\\\v_2=399.04\frac{m}{s}

The speed of the bullet before the impact with the wood block is 399.04 m/s

(b) The impulse is gibe by the change in the velocity of the block, multiplied by the mass of the block:

I=M\Delta v=M(v-v_1)=(1kg)(2.38m/s-0m/s)=2.38kg\frac{m}{s}

The impulse is 2.38 kgm/s

(c) The force on the cord after the impact is equal to the centripetal force over the block and bullet. That is:

T=F_c=(M+m)\frac{v^2}{l}=(1.006kg)\frac{(2.38m/s)^2}{2.2m}=2.59N    

The force on the cord after the impact is 2.59N

4 0
3 years ago
A box has sides of 10 cm, 8.2 cm, and 3.5 cm. What is its volume?
Ksju [112]
The volume would be 287cm³. Multiply all the 3 numbers by each other
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3 years ago
The elements in a column of the periodic table _____.
Shtirlitz [24]
C. have similar properties
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