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2 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?

1 answer:

barxatty [35]2 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 hollow sphere of radius 0.200 m, with rotational inertia I = 0.0484 kg·m2 about a line through its center of mass, rolls witho

d1i1m1o1n [39]

Answer:

Part a)

$KE_r = 8 J$

Part b)

v = 3.64 m/s

Part c)

$KE_f = 12.7 J$

Part d)

$v = 2.9 m/s$

Explanation:

As we know that moment of inertia of hollow sphere is given as

$I = \frac{2}{3}mR^2$

here we know that

$I = 0.0484 kg m^2$

R = 0.200 m

now we have

$0.0484 = \frac{2}{3}m(0.200)^2$

$m = 1.815 kg$

now we know that total Kinetic energy is given as

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{R})^2$

$20 = \frac{1}{2}(1.815)v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2$

$20 = 1.5125 v^2$

$v = 3.64 m/s$

Part a)

Now initial rotational kinetic energy is given as

$KE_r = \frac{1}{2}I(\frac{v}{R})^2$

$KE_r = \frac{1}{2}(0.0484)(\frac{3.64}{0.200})^2$

$KE_r = 8 J$

Part b)

speed of the sphere is given as

v = 3.64 m/s

Part c)

By energy conservation of the rolling sphere we can say

$mgh = (KE_i) - KE_f$

$1.815(9.8)(0.900sin27.1) = 20- KE_f$

$7.30 = 20 - KE_f$

$KE_f = 12.7 J$

Part d)

Now we know that

$\frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{r})^2 = 12.7$

$\frac{1}{2}(1.815) v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2 = 12.7$

$1.5125 v^2 = 12.7$

$v = 2.9 m/s$

8 0

2 years ago

How might the distance between the needle and the water droplets have affected the movement of the droplets?

shutvik [7]

Answer:

The behavior of droplets trapped in geometric structures is essential to droplet manipulation applications such as for droplet transport. Here we show that directional droplet movement can be realized by a V-shaped groove with the movement direction controlled by adjusting the surface wettability of the groove inner wall and the cross sectional angle of the groove. Experiments and analyses show that a droplet in a superhydrophobic groove translates from the immersed state to the suspended state as the cross sectional angle of the groove decreases and the suspended droplet departs from the groove bottom as the droplet volume increases. We also demonstrate that this simple grooved structure can be used to separate a water-oil mixture and generate droplets with the desired sizes. The structural effect actuated droplet movements provide a controllable droplet transport method which can be used in a wide range of droplet manipulation applications.

Explanation:

BOOM NOW I WINNNNNNNNNNNn

5 0

2 years ago

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Pls help my science teacher is a horrible teacher...

Anettt [7]

First of all, looks like your teacher is indeed pretty horrible. Secondly, the constraints to consider would be proper weight distribution, methods to minimize excessive motion of the building structure, and quantities such as volume and density, which would help in determining the optimal structure. Keeping the frequency of oscillation for a building low in case of an earthquake or natural disaster would also be a priority.

4 0

2 years ago

A heat engine does 200 j of work per cycle while exhausting 600 j of heat to the cold reservoir. what is the engine's thermal ef

AveGali [126]

The thermal efficiency of an engine is
$\eta= \frac{W}{Q}$
where
W is the work done by the engine
Q is the heat absorbed by the engine to do the work

In this problem, the work done by the engine is W=200 J, while the heat exhausted is Q=600 J, so the efficiency of the machine is
$\eta= \frac{W}{Q}= \frac{200 J}{600 J}=0.33 = 33 \%$

8 0

2 years ago

Which has a greater force: a semi-truck at rest or a moving bicycle?

Wittaler [7]

Although the semi truck certainly has a larger mass, it is not in motion and therefore does not have any momentum. The bicycle however has both mass and velocity and therefore has the larger momentum of the pair.

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2 years ago

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