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shusha [124]
4 years ago
10

What is the difference between dna and rna?

Physics
2 answers:
Romashka-Z-Leto [24]4 years ago
6 0
The answer is : DNA<span> is a long polymer with deoxyriboses and phosphate backbone. Having  four </span>different<span> nitrogenous bases: adenine, guanine, cytosine and thymine. </span>RNA<span> is a polymer with a ribose and phosphate backbone. Four </span>different<span> nitrogenous bases: adenine, guanine, cytosine, and uracil. Hope this Helps! Good Luck!</span>
Gekata [30.6K]4 years ago
6 0

Answer:

DNA:

Deoxyribonucleic acid acts as the genetic material of almost all the living organisms. The deoxyribose pentose sugar is present in its DNA structure. The nitrogenous bases present in DNA are adenine, guanine, thymine and cytosine. DNA is highly stable in nature and cannot acts as the catalyst.

RNA:

Ribonucleic acid acts as the genetic material in case of few viruses only. The oxyribose pentose sugar is present in RNA structure. The nitrogenous bases present in RNA are adenine, guanine, cytosine and uracil. RNA is highly unstable in nature and acts as a catalyst.

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What system will break the chewed food down to a form your cell can use
horsena [70]
If I'm not mistaken it should be the digestive system due to the fact that our mouths and stomachs break down food and our intestines absorb any water and nutrients
4 0
4 years ago
The half life of uranium-235 is 4. 5 billion years. If 0. 5 half-lives have elapsed, how many years have gone by?
igor_vitrenko [27]

Answer:

2.25 billion years

Explanation:

i am smort

5 0
2 years ago
Please help Math Phys
Lorico [155]

Answer:

269 m

45 m/s

-58.6 m/s

Explanation:

Part 1

First, find the time it takes for the package to land.  Take the upward direction to be positive.

Given (in the y direction):

Δy = -175 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

(-175 m) = (0 m/s) t + ½ (-9.8 m/s²) t²

t = 5.98 s

Next, find the horizontal distance traveled in that time:

Given (in the x direction):

v₀ = 45 m/s

a = 0 m/s²

t = 5.98 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (45 m/s) (5.98 s) + ½ (0 m/s²) (5.98 s)²

Δx = 269 m

Part 2

Given (in the x direction):

v₀ = 45 m/s

a = 0 m/s²

t = 5.98 s

Find: v

v = at + v₀

v = (0 m/s²) (5.98 s) + (45 m/s

v = 45 m/s

Part 3

Given (in the y direction):

Δy = -175 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: v

v² = v₀² + 2aΔy

v² = (0 m/s)² + 2 (-9.8 m/s²) (-175 m)

v = -58.6 m/s

6 0
3 years ago
(a) A load of coal is dropped (straight down) from a bunker into a railroad hopper car of inertia 3.0 × 104 kg coasting at 0.50
Firlakuza [10]

Answer:

a) m=20000Kg

b) v=0.214m/s

Explanation:

We will separate the problem in 3 parts, part A when there were no coals on the car, part B when there is 1 coal on the car and part C when there are 2 coals on the car. Inertia is the mass in this case.

For each part, and since the coals are thrown vertically, the horizontal linear momentum p=mv must be conserved, that is, p=m_Av_A=m_Bv_B=m_Cv_C, were each velocity refers to the one of the car (with the eventual coals on it) for each part, and each mass the mass of the car (with the eventual coals on it) also for each part. We will write the mass of the hopper car as m_h, and the mass of the first and second coals as m_1 and m_2 respectively

We start with the transition between parts A and B, so we have:

m_Av_A=m_Bv_B

Which means

m_hv_A=(m_h+m_1)v_B

And since we want the mass of the first coal thrown (m_1) we do:

m_hv_A=m_hv_B+m_1v_B

m_hv_A-m_hv_B=m_1v_B

m_1=\frac{m_hv_A-m_hv_B}{v_B}=\frac{m_h(v_A-v_B)}{v_B}

Substituting values we obtain

m_1=\frac{(3\times10^4Kg)(0.5m/s-0.3m/s)}{0.3m/s}=20000Kg=2\times10^4Kg

For the transition between parts B and C, we can write:

m_Bv_B=m_Cv_C

Which means

(m_h+m_1)v_B=(m_h+m_1+m_2)v_C

Since we want the new final speed of the car (v_C) we do:

v_C=\frac{(m_h+m_1)v_B}{(m_h+m_1+m_2)}

Substituting values we obtain

v_C=\frac{(3\times10^4Kg+2\times10^4Kg)(0.3m/s)}{(3\times10^4Kg+2\times10^4Kg+2\times10^4Kg)}=0.214m/s

5 0
3 years ago
Which is an example of an abiotic factor in a pound environment
zvonat [6]
A. The water

Biotic factors mean living this such as frogs, grasshoppers, snakes, etc.

Aboriginal factors means non-living such as water, temperature, sunlight, etc.
Abiotic factors are the non-living parts of the environment that have a major influence on living organisms which makes water an abiotic factor.

Hope this helped, if it did please like (if you want to :)
4 0
4 years ago
Read 2 more answers
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