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Alex787 [66]
3 years ago
8

Protein synthesis is vital for cell growth and metabolism. Transcription and translation are known as the central dogma of biolo

gy in which proteins are synthesized from DNA. How is transcription similar to translation in terms of base pairing? A) in transcription, RNA is formed; in translation, a polypeptide is formed B) transcription starts at the promoter; translation begins at the start codon C) in transcription, DNA binds to mRNA; in translation, codons bind to anti-codons D) in transcription, C pairs with G and A pairs with T to make mRNA; in translation, A pairs with U and G pairs with C to make tRNA
Physics
2 answers:
uranmaximum [27]3 years ago
8 0

Transcription and translation are very similar in terms of pairing. In transcription, DNA must bind with mRNA to make a transcript to send outside of the nucleus. In translation, codons bind with anti-codons to make amino acids which then form proteins.


Liono4ka [1.6K]3 years ago
6 0
I  think the answer is D i'm not for sure tho
 

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Write a question about how changing temperature affects gas
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Pete is driving down 7th Street. He drives 300 meters in 18 seconds. Assuming he does not speed up or slow down, what is his spe
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16.67m/s

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3 years ago
1) A boy drags a wooden crate with a mass of 20 kg, a distance of 12 m, across a rough level floor at a constant speed of 1.5 m/
mojhsa [17]

Answer: a) 49.560 and 21.13 b) i) 50 N, ii) 196 N iii) 196 N iv) 47.685 N

c) i) 594.72 ii) 0 iii) 0 iv) 0

d) 594.72

Explanation: question a)

The force is inclined at an angle of 25° to the horizontal

The horizontal component of force = 50 cos 25° = 49.560 N

The vertical component of force = 50 sin 30°= 21.130N

Question b)

i) according to the question applied force is 50 N

ii) if g = 9.8m/s², w=mg where m = mass of object = 20kg hence weight = 20* 9.8 = 196 N

iii) the normal force is the force the floor exerts on the body as a result of the weight of the object.

Normal reaction R = W = mg, we already deduced that w = mg, hence R = 196 N.

iv) according to newton's laws of motion

F - Fr = ma

F = applied force = horizontal component of force = 49.560 N.

We need to get the acceleration (a) by using Newton laws of motion before we can be able to compute the frictional force..

The body started from rest hence initial velocity u = 0

Final velocity v = 1.5m/s distance covered (s) = 12m

v ² = u² + 2as

But u = 0

v² = 2as

1.5² = 2(a) * 12

2.25 = 24a

a = 2.25/24 = 0.09735m/s²

From F - Fr = ma

49.560 - Fr = 20 * 0.09735

49.560 - Fr = 1.875

Fr = 49.560 - 1.875

Fr = 47.685 N

Question c)

i) The applied force = 49.560 N, distance covered = 12m

Work done = force * distance

Work done = 49.560 * 12

Work done = 594.72 J

ii) the weight of the object does not make the object move a distance, hence work done = 0 ( since distance covered is 0)

iii) the normal force is the same thing as the weight and they did not cover any distance hence work done is zero.

iv) the frictional force does not cover any distance, hence work done is zero.

Question d)

The total work done = work done by applied force + work done by weight + work done by normal reaction + work done by frictional force.

Total work done = 594.72 + 0 + 0 + 0 = 594.72 J

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