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kotykmax [81]
3 years ago
12

A particular molecule has a small molecular mass of approximately 1000 daltons. What should be done to make this molecule more a

ntigenic?
Physics
1 answer:
barxatty [35]3 years ago
7 0

Answer:

Bind it to a large protein

Explanation:

An antigen is a molecule that binds to Ag-specific receptors, but cannot necessarily induce an immune response in the body by itself. Antigens are proteins , peptides (amino acid chains) and

polysaccharides (chains of monosaccharides/simple sugars) but

lipids and nucleic acids become antigens only when combined with proteins and polysaccharides. [4] In general, saccharides and lipids (as opposed to peptides) qualify as antigens but not as immunogens since they cannot elicit an immune response on their own. Furthermore, for a peptide to induce an immune response it must be a large enough size, thus binding to proteins

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A 3.7-kg object is acted on by two forces. One of the forces is 11 N acting toward the
vitfil [10]

Answer:

F₂ = -7.3 N

Explanation:

Given that,

The mass of an object, m₁ = 3.7 kg

First force, F₁ = 11 N

The net acceleration of the object is 1 m/s².

We know that,

F₁+F₂ = ma

11+F₂ = (3.7)(1)

F₂ = 3.7-11

F₂ = -7.3 N

so, the other force is 7.3 N and it is acting in west direction.

7 0
2 years ago
We can model the motion of a bumblebee's wing as simple harmonic motion. A bee beats its wings 250 times per second, and the win
zlopas [31]

Answer:

The amplitude of the wing tip's motion is 1.6 mm.

Explanation:

Given that,

Beat = 250 /s

Speed = 2.5 m/s

We need to calculate the amplitude of the wing tip's motion

Using the equation for the maximum velocity

v_{max}=2\pi f A

A=\dfrac{v}{2\pi f}

Where,

v = speed

f = frequency

A = amplitude

Put the value into the formula

A=\dfrac{2.5}{2\pi\times250}

A=0.00159 = 1.59\times10^{-3}\ m

A=1.6\ mm

Hence, The amplitude of the wing tip's motion is 1.6 mm.

3 0
3 years ago
Read 2 more answers
a nonrelativistic proton is confined to a length of 2.0 pm on the x-axis. what is the kinetic energy of the proton if its speed
Elina [12.6K]

Answer:

K = 1.29eV

Explanation:

In order to calculate the kinetic energy of the proton you first take into account the uncertainty principle, which is given by:

\Delta x \Delta p\geq \frac{h}{4\pi}      (1)

Δx : uncertainty of position = 2.0pm = 2.0*10^-12m

Δp: uncertainty of momentum = ?

h: Planck's constant = 6.626*10^-34 J.s

You calculate the minimum possible value of Δp from the equation (1):

\Delta p=\frac{h}{4\pi \Delta x}=\frac{6.626*10^{-34}J.s}{4\pi(2.0*10^{-12}m)}\\\\\Delta p=2.63*10^{-23}kg.\frac{m}{s}

The minimum kinetic energy is calculated by using the following formula:

k=\frac{(\Delta p)^2}{2m}       (2)

m: mass of the proton = 1.67*10^{-27}kg

k=\frac{(2.63*10^{-23}kgm/s)^2}{2(1.67*10^{-27}kg)}=2.08*10^{-19}J

in eV you have:

2.08*10^{-19}J*\frac{6.242*10^{18}eV}{1J}=1.29eV

The kinetic energy of the proton is 1.29eV

7 0
2 years ago
During this chemical reaction energy is released. In the chemistry lab, this would be indicated by an increase in temperature or
PtichkaEL [24]

Answer:

C) the reactants have more potential energy than the products.

Explanation:

6 0
2 years ago
Read 2 more answers
Consider your moment of inertia about a vertical axis through the center of your body, both when you are standing straight up wi
jeka94

Answer:

     I₁ / I₂ = 1.43

Explanation:

To find the relationship of the two inertial memits, let's calculate each one, let's start at the moment of inertia with the arms extended

Before starting let's reduce all units to the SI system

       d₁ = 42 in (2.54 10⁻² m / 1 in) = 106.68 10⁻² m

       d₂ = 38 in = 96.52 10⁻² m

The moment of inertia is a scalar quantity for which it can be added, the moment of total inertia would be the moment of inertia of the man (cylinder) plus the moment of inertia of each arm

        I₁ = I_man + 2 I_ arm

Man indicates that we can approximate them to a cylinder where the average diameter is

         d = (d₁ + d₂) / 2

         d = (106.68 + 96.52) 10-2 = 101.6 10⁻² m

The average radius is

         r = d / 2 = 50.8 10⁻² m = 0.508 m

The mass of the trunk is the mass of man minus the masses of each arm.

        M = M_man - 0.2 M_man = 80 (1-0.2)

        M = 64 kg

The moments of inertia are:

A cylinder with respect to a vertical axis:         Ic = ½ M r²

A rod that rotates at the end:                            I_arm = 1/3 m L²

Let us note that the arm rotates with respect to man, but this is at a distance from the axis of rotation of the body, so we must use the parallel axes theorem for the moment of inertia of the arm with respect to e = of the body axis.

           I1 = I_arm + m D²

Where D is the distance from the axis of rotation of the arm to the axis of the body

          D = d / 2 = 101.6 10⁻² /2 = 0.508 m

Let's replace

          I₁ = ½ M r² + 2 [(1/3 m L²) + m D²]

Let's calculate

         I₁ = ½ 64 (0.508)² + 2 [1/3 8 1² + 8 0.508²]

         I₁ = 8.258 + 5.33 + 4.129

         I₁ = 17,717 Kg m² / s²

Now let's calculate the moment of inertia with our arms at our sides, in this case the distance L = 0,

          I₂ = ½ M r² + 2 m D²

          I₂ = ½ 64 0.508² + 2 8 0.508²

          I₂ = 8,258 + 4,129

          I₂ = 12,387 kg m² / s²

The relationship between these two magnitudes is

          I₁ / I₂ = 17,717 /12,387

          I₁ / I₂ = 1.43

3 0
2 years ago
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