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goldenfox [79]
3 years ago
5

The site of proteins is called what

Physics
2 answers:
ValentinkaMS [17]3 years ago
8 0

Answer: Ribosomes are the site of protein synthesis, and they convert the coded information in messenger RNA into an actual protein molecule.

I hope this helped! OvO

Musya8 [376]3 years ago
3 0

Answer: Your question is not clear whether you are referring to sites of protein synthesis or Protein I-sites so I’ve pasted this paragraph from Wikipedia of the later.

Explanation:

I-sites are short sequence-structure motifs that are mined from the Protein Data Bank (PDB) that correlate strongly with three-dimensional structural elements. These sequence-structure motifs are used for the local structure prediction of proteins. Local structure can be expressed as fragments or as backbone angles. Locations in the protein sequence that have high confidence I-sites predictions may be the initiation sites of folding. I-sites have also been identified as discrete models for folding pathways. I-sites consist of about 250 motifs. Each motif has an amino acid profile, a fragment structure (represented by a "paradigm" fragment chosen from a protein in the PDB) and optionally, a 4-dimensional tensor of pairwise sequence covariance.

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an object is 50 cm to the left and an image is 50 cm to the right of a convex lens. What is the focal length?
Andrew [12]

Answer:

An Object son

Explanation:

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9) An electrical appliance has a resistance of 25 N. When this electrical ap-
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(C)

Explanation:

From Ohm's law,

V = IR

Solving for I,

I = V/R

= (230 V)/(25 ohms)

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Is a process that modifies light waves so they vibrate in a single plane
madam [21]

The process you're fishing for is "polarization", but that's a

misleading description.

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8 0
4 years ago
Luis wants to compare the density of three solid objects that are different shapes and sizes. What information does he need to m
tatyana61 [14]

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7 0
3 years ago
In 2017, the company SpaceX became the first private company to send supplies to the International Space Station with a reusable
pav-90 [236]

Answer:

Approximately 3.98\; \rm m \cdot s^{-2}.

Assumption: air resistance on the rocket is negligible. Take g = \rm 9.81\; m \cdot s^{-2}.

Explanation:

By Newton's Second Law of Motion, the acceleration of the rocket is proportional to the net force on it.

\displaystyle \text{Acceleration} = \frac{\text{Net Force}}{\text{Mass}}.

Note that in this case, the uppercase letter \rm M in the units stands for "mega-", which is the same as 10^6 times the unit that follows. For example, \rm 1\; Mg = 10^6\; g, while \rm 1\; MN = 10^6\; N.

Convert the mass of the rocket and the thrust of its engines to SI standard units:

  • The standard unit for mass is kilograms: \displaystyle m = \rm 552\; Mg = 552 \times 10^6\; g \times \frac{1\; \rm kg}{10^3\; g}  = 552 \times 10^3 \; kg.
  • The standard for forces (including thrust) is Newtons: \text{Thrust} = \rm 7.61 \; MN = 7.61 \times 10^6\; N.

At launch, the velocity of the rocket would be pretty low. Hence, compared to thrust and weight, the air resistance on the rocket would be pretty negligible. The two main forces that contribute to the net force of the rocket would be:

  • Thrust (which is supposed to go upwards), and
  • Weight (downwards due to gravity.)

The thrust on the rocket is already known to be \rm 7.61 \times 10^6\; N. Since the rocket is quite close to the ground, the gravitational acceleration on it should be approximately 9.81\; \rm m \cdot s^{-2} = 9.81 \; N \cdot kg^{-1}. Hence, the weight on the rocket would be approximately 9.81\; \rm N \cdot kg^{-1} \times 552 \times 10^3\; kg = 5.41412\times 10^6\; N.

The magnitude of the net force on the rocket would be

\begin{aligned}&\text{Thrust} - \text{Weight} \\ &= 7.61 \times 10^6\; \rm N - 5.41412\times 10^6\; N \\ &\approx 2.19 \times 10^6\; \rm N\end{aligned}.

Apply the formula \displaystyle \text{Acceleration} = \frac{\text{Net Force}}{\text{Mass}} to find the net force on the rocket. To make sure that the output (acceleration) is in SI units (meters-per-second,) make sure that the inputs (net force and mass) are also in SI units (Newtons for net force and kilograms for mass.)

\begin{aligned}\displaystyle &\text{Acceleration} \\ &= \frac{\text{Net Force}}{\text{Mass}} \\ &= \frac{2.19 \times 10^6\; \rm N}{552 \times 10^3\; \rm kg}  \\ &\approx \rm 3.98\; \rm m \cdot s^{-2}\end{aligned}.

6 0
3 years ago
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