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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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To test the performance of its tires, a car
VLD [36.1K]

Answer:

0.45

Explanation:

Sum of forces in the y direction:

∑F = ma

N − mg = 0

N = mg

There are friction forces in two directions: centripetal and tangential.  The centripetal acceleration is:

ac = v² / r

ac = (31 m/s)² / 333 m

ac = 2.89 m/s²

The total acceleration is:

a = √(ac² + at²)

a = √((2.89 m/s²)² + (3.32 m/s²)²)

a = 4.40 m/s²

Sum of forces:

∑F = ma

Nμ = ma

mgμ = ma

μ = a / g

μ = 4.40 m/s² / 9.8 m/s²

μ = 0.45

5 0
4 years ago
A sled plus passenger with total mass m = 52.3 kg is pulled a distance d = 21.8 m across a horizontal, snow-packed surface for w
Vikki [24]

Answer:

Explanation:

Since the sled plus passenger moves with constant velocity , force applied will be equal to frictional force. Let the force applied be F

a ) Frictional force = μ R = F cosφ

R = mg - F sinφ

μ(mg - F sinφ)  = F cosφ

μmg = F (μsinφ+cosφ)

F = μmg / (μsinφ+cosφ)

Work done

= F cosφ x d

= μmg x cosφ x d / (μsinφ+cosφ)

b )Work done

= 0.13 x 52.3 x 9.8 cos36.7 x 21.8 / ( 0.13 sin36.7 +cos36.7)

= 1164.61 / .87946

1324.23 J

c ) work done on the sled by friction

= - (work done by force)

= - μmg x cosφ x d / (μsinφ+cosφ)

d ) work done on the sled by friction

= - 1324.23 J

3 0
3 years ago
In each of the parts of this question, a nucleus undergoes a nuclear decay. Determine the resulting nucleus in each case.
MA_775_DIABLO [31]

A) Francium-223

In an alpha decay, a nucleus decay emitting an alpha particle, which corresponds to a nucleus of helium: so, it consists of 2 protons and 2 neutrons.

X \rightarrow X' + \alpha

This means that in the decay:

- The original nucleus loses 2 protons --> so its atomic number Z decreases by 2 units

- The original nucleus loses 2 nucleons (2 protons and 2 neutrons) --> so its mass number A decreases by 4 units

In this example, the original nucleus is Ac (Actinium), with

Z = 89

A = 227

After the decay, it must be

Z - 2 = 89 - 2 = 87

A - 4 = 227 - 4 = 223

We see from the periodict table, Z=87 corresponds to Francium (Fr), so the final nucleus will be francium-223 (the isotope of francium with 223 nucleons).

B) Polonium-211

In a beta-minus decay, a neutron in the nucleus turns into a proton, emitting a fast-moving electron (the beta particle) and an anti-neutrino.

n \rightarrow p + e^- + \bar{\nu}

Therefore, in this process:

- The original nucleus gains 1 protons, so its atomic number Z increases by 1 unit

- The original nucleus does not lose/gain nucleons, so its mass number A remains the same

In this example, the original nucleus is Bi (bismuth)-211, with

Z = 83

A = 211

So After the decay, it will be

Z + 1 = 83 + 1 = 84

A = 211

So, the nucleus will be Polonium (Z=84), isotope with 211 nucleons.

C) Neon-22

In a beta-plus decay, a proton in the nucleus turns into a neutron, emitting a fast-moving positron (the beta particle) and a neutrino.

p \rightarrow n + e^+ +\nu

Therefore, in this process:

- The original nucleus loses 1 protons, so its atomic number Z decreases by 1 unit

- The original nucleus does not lose/gain nucleons, so its mass number A remains the same

In this example, the original nucleus is Na (sodium)-22, with

Z = 11

A = 22

So After the decay, it will be

Z - 1 = 11 - 1 = 10

A = 22

So, the nucleus will be Neon (Z=10), isotope with 22 nucleons.

D) Technetium-98

In a gamma decay, an unstable nucleus emits a gamma ray:

X' \rightarrow X + \gamma

In this process, only energy is released (in the form of gamma ray), so there is no gain/loss of protons/neutrons in the process. This means that:

- The atomic number Z remains constant

- The mass number A remains constant

In this example, we have a nucleus of Tc (Technetium)-98, with

Z = 43

A = 98

These numbers will not change during the decay: this means that after the decay, we will still have a nucleus of Technetium-98.

8 0
3 years ago
What color is a carrot?
Mashutka [201]

Answer:

reddish-orrange

Explanation:

please mark me as brainliest

7 0
3 years ago
An object with an initial speed of 4.0 meters per second accelerates uniformly at 2.0 meters per second squared in the direction
leva [86]

Answer:

6 m/s.

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 4 m/s

Acceleration (a) = 2 m/s²

Distance (s) = 5 m

Final velocity (v) =?

The final velocity of the object can obtained as shown below:

v² = u² + 2as

v² = 4² + (2 × 2 × 5)

v² = 16 + 20

v² = 36

Take the square root of both side.

v = √36

v = 6 m/s

Therefore, the final speed of the object is 6 m/s.

8 0
3 years ago
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