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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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How many different values of ml are possible in the 5d sublevel?
m_a_m_a [10]
<span>There are 5 different values of ml in the 5d sublevel (-2, -1, 0, 1, and 2).</span>
4 0
3 years ago
A 3.0-kg block starts at rest at the top of a 37° incline, which is 5.0 m long. Its speed when it reaches the bottom is 2.0 m/s.
Mama L [17]

Answer: f_{r} = 16.49N

Explanation: The object is placed on an inclined plane at an angle of 37° thus making it weight have two component,

W_{x} = horizontal component of the weight = mgsinФ

W_{y} = vertical component of weight = mgcosФ

Due to the way the object is positioned, the horizontal component of force will accelerate the object thus acting as an applied force.

by using newton's law of motion, we have that

mgsinФ - f_{r} = ma

where m = mass of object=5kg

a = acceleration= unknown

Ф = angle of inclination = 37°

g = acceleration due to gravity = 9.8m/s^{2}

f_{r} = frictional force = unknown

we need to first get the acceleration before the frictional force which is gotten by using the equation below

v^{2} = u^{2} + 2aS

where v = final velocity = 2m/s

u = initial velocity = 0m/s (because the object started from rest)

a= unknown

S= distance covered = length of plane = 5m

2^{2} = 0^{2} + 2*a*5\\\\4= 10 *a\\\\a = \frac{4}{10} \\a = 0.4m/s^{2}

we slot in a into the equation below to get frictional force

mgsinФ - f_{r} = ma

3 * 9.8 * sin 37 - f_{r} = 3* 0.4

17.9633 - f_{r} =  1.2

f_{r} = 17.9633 - 1.2

f_{r} = 16.49N

4 0
3 years ago
Un cubo de madera de densidad 0.780 g/cm³ mide 11.2 cm en un lado. Cuando se coloca en agua, ¿qué altura del bloque flotará sobr
Stolb23 [73]

Answer:

2.464 cm above the water surface

Explanation:

Recall that for the cube to float, means that the volume of water displaced weights the same as the weight of the block.

We calculate the weight of the block multiplying its density (0.78 gr/cm^3) times its volume (11.2^3  cm^3):

weight of the block = 0.78 * 11.2^3  gr

Now the displaced water will have a volume equal to the base of the cube (11.2 cm^2) times the part of the cube (x) that is under water. Recall as well that the density of water is 1 gr/cm^3.

So the weight of the volume of water displaced is:

weight of water = 1 * 11.2^2 * x

we make both weight expressions equal each other for the floating requirement:

0.78 * 11.2^3 = 11.2^2 * x

then x = 0.78 * 11.2 cm = 8.736 cm

This "x" is the portion of the cube under water. Then to estimate what is left of the cube above water, we subtract it from the cube's height (11.2 cm) as follows:

11.2 cm - 8.736 cm = 2.464 cm

6 0
3 years ago
a car moving with a speed of 10 metre per second is accelerator at the rate of 2 metre per second square find its velocity after
STatiana [176]

a = ( v(2) - v(1) ) ÷ ( t(2) - t(1) )

2 = ( v(2) - 10 ) ÷ ( 6 - 0 )

2 × 6 = v(2) - 10

v(2) = 12 + 10

v(2) = 22 m/s

7 0
3 years ago
How and why will my weight change if i took a trip to the moon
MatroZZZ [7]

weight less on moon than on earth.

high on lift off - G force

low in orbit.

zero at a point between earth and moon

7 0
3 years ago
Read 2 more answers
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