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3 years ago

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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What is tyndall effect

ivann1987 [24]

Answer:

light scattering by particles in a colloid or in a very fine suspension

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3 years ago

Read 2 more answers

if 130N centripetal force is needed to keep a 0.45kg ball that is attached to a string that is 0.7m long to complete 5 full rota

andrezito [222]

Answer:

centripetal acceleration of the ball is 6.9 m/s/s

tangential speed of the ball is 2.2 m/s

Explanation:

As we know that ball complete 5 rotations in 10 seconds

so frequency of rotation of ball is given as

$f = \frac{5}{10} = 0.5 Hz$

now we know that angular frequency is given as

$\omega = 2\pi f$

$\omega = 2(\pi)(0.5)$

$\omega = \pi rad/s$

Now centripetal acceleration is given as

$a_c = \omega^2 R$

$a_c = \pi^2 (0.7)$

$a_c = 6.9 m/s^2$

now the velocity of the ball at this angular frequency is given as

$v = R\omega$

$v = 0.7 (\pi)$

$v = 2.2 m/s$

4 0

3 years ago

A hot-air balloon plus cargo has a mass of 308 kg and a volume of 2910 m3 on a day when the outside air density is 1.22 kg/m3. T

lesya [120]

9514 1404 393

Answer:

1.114 kg/m³

Explanation:

The total mass of the air in the balloon and the balloon + cargo will be the mass of the displaced air. If d is the density of the air in the balloon, then we have ...

2910d +308 = 2910×1.22

Solving for d, we find ...

2910d = 2919(1.22) -308

d = 1.22 -308/2910

d ≈ 1.114 . . . kg/m³

The density of the hot air is about 1.114 kg/m³.

6 0

3 years ago

A 10.0g piece of copper wire, sitting in the sun reaches a temperature of 80.0 C. how many Joules are released when the copper c

Zolol [24]

Answer:

150.8 J

Explanation:

The heat released by the copper wire is given by:

$Q=mC_s \Delta T$

where:

m = 10.0 g is the mass of the wire

Cs = 0.377 j/(g.C) is the specific heat capacity of copper

$\Delta T=40.0 C - 80.0 C=-40.0 C$ is the change in temperature of the wire

Substituting into the equation, we find

$Q=(10.0 g)(0.377 J/gC)(-40.0^{\circ})=-150.8 J$

And the sign is negative because the heat is released by the wire.

6 0

3 years ago

A 7750 kg space probe, moving nose-first toward Jupiter at 179 m/s relative to the Sun, fires its rocket engine, ejecting 72.0 k

Reika [66]

Answer:

179.47m/s

Explanation:

Using the law of conservation of momentum

m1u1 + m2u2 = (m1+m2)v

m1 and m2 are the masses

u1 and u2 are the initial velocities

v is the final velocity

Substitute

7750(179)+72(230) = (7750+72)v

1,387,250+16560 = 7822v

1,403,810 = 7822v

v = 1,403,810/7822

v= 179.47m/s

Hence the final velocity of the probe is 179.47m/s

7 0

2 years ago