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IceJOKER [234]
3 years ago
15

An aminoacyl-trna that enters the a site of the ribosome will next occupy which site?

Physics
1 answer:
kkurt [141]3 years ago
5 0

Answer:

The P site

Explanation:

After the formation of peptide linkage, the trna of P site becomes free, more to the E site and then slip away. The A site trna has dipeptide instead of a single aminoacid.

In the prescence of translocase and energy from GTP, the ribosome move in such a way the peptidyl bearing trna of A site come to lie with P sit. This exposes next codon to A site. A new aminoacid trna complex reaches the fresh codon exposed at A site.

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How many grams are in 6.53 moles of Mn?
Westkost [7]

Answer:

359 g Mn

General Formulas and Concepts:

  • Dimensional Analysis
  • Reading the Periodic Table of Elements

Explanation:

<u>Step 1: Define</u>

6.53 mol Mn

<u>Step 2: Find conversion</u>

1 mol Mn = 54.94 g Mn

<u>Step 3: Dimensional Analysis</u>

<u />6.53 \hspace{3} mol \hspace{3} Mn(\frac{54.94 \hspace{3} g \hspace{3} Mn}{1 \hspace{3} mol \hspace{3} Mn} ) = 358.758 g Mn

<u>Step 4: Simplify</u>

<em>We are given 3 sig figs.</em>

358.758 g Mn ≈ 359 g Mn

3 0
2 years ago
A ski jumper has a mass of 59.6 kg. She is moving with a speed of 23.4 m/s at a height of 44.6 meters above the ground. Determin
Daniel [21]

Hello!

Use the formula:

M = k * p

Data:

M = Mechanic energy

k = Kinetic energy

p = Potencial energy

Descomposing:

M = (0,5*mv²) + (mgh)

Replacing:

M = (0,5 * 59,6 kg * (23,4 m/s)²) + (59,6 kg * 9,81 m/s² * 44,6 m)

M = 16317,28 J + 26076,54 J

M = 42393,82 J

The mechanic energy is <u>42393,82 Joules.</u>

7 0
3 years ago
during the course of songraphic exam, you notice lateral splaying of echoes in the far field. what can you do to improve the ima
horrorfan [7]

Answer:

lateral splaying of echoes in the far field can be improved by Increasing the maximum number of transmit focal zones and optimize their location.

8 0
2 years ago
Hooke's law describes a certain light spring of unstretched length 34.8 cm. When one end is attached to the top of a doorframe a
DaniilM [7]

Answer:

a) k = 1343.6\,\frac{N}{m}, b) l = 0.501\,m\,(50.1\,cm)

Explanation:

a) The Hooke's law states that spring force is directly proportional to change in length. That is to say:

F \propto \Delta l

In this case, the force is equal to the weight of the object:

F = m\cdot g

F = (8.22\,kg)\cdot (9.807\,\frac{m}{s^{2}} )

F = 80.614\,N

The spring constant is:

k = \frac{F}{\Delta l}

k = \frac{80.614\,N}{0.408\,m-0.348\,m}

k = 1343.6\,\frac{N}{m}

b) The length of the spring is:

F = k\cdot (l-l_{o})

l = l_{o} + \frac{F}{k}

l=0.348\,m+\frac{205\,N}{1343.6\,\frac{N}{m} }

l = 0.501\,m\,(50.1\,cm)

6 0
3 years ago
according to isaac newtons theory of gravitation the force exerted between two objects is dependent on- A- an objects weight. B-
puteri [66]
A an objects weight hope this helps! :D 
3 0
2 years ago
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