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Tatiana [17]
3 years ago
14

What is the significance of Nucleotides in Chromosomes?​

Physics
2 answers:
fgiga [73]3 years ago
3 0

Answer:

it comprises of the DNA/RNA bipolymer molecules

yuradex [85]3 years ago
3 0
The nucleotides attach to each other (A with T, and G with C) to form chemical bonds called base pairs, which connect the two DNA strands. Genes are short pieces of DNA that carry specific genetic information.
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What does the equation indicate about the chemical reaction that forms water,
mixer [17]
We won't have a clue until we see the equation.
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3 years ago
two charges attract each other with a force of 1.2 * 10⁻⁶N. Calculate the force that would act between the two charges when dist
Likurg_2 [28]

1. When the distance is doubled: 3\cdot 10^{-7}N

The electrostatic force between two charges is given by:

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

q1 and q2 are the two charges

r is the distance between the two charges

The initial force between the two charges is F=1.2\cdot 10^{-6}N. In this part of the problem, the distance between the two charges is doubled, so we can write

r'=2r

And substituting into the formula, we find the new force:

F'=k\frac{q_1 q_2}{r'^2}=k\frac{q_1 q_2}{(2r)^2}=\frac{1}{4}k\frac{q_1 q_2}{r^2}=\frac{F}{4}

So, the force is reduced to 1/4 of its original value. Therefore, it is

F'=\frac{1.2\cdot 10^{-6} N}{4}=3\cdot 10^{-7}N

2. When the distance is halved: 4.8\cdot 10^{-6}N

The initial force between the two charges is F=1.2\cdot 10^{-6}N. In this part of the problem, the distance between the two charges is halved, so we can write

r'=r/2

And substituting into the formula, we find the new force:

F'=k\frac{q_1 q_2}{r'^2}=k\frac{q_1 q_2}{(r/2)^2}=4k\frac{q_1 q_2}{r^2}=4F

So, the force is quadrupled. Therefore, it is

F'=4(1.2\cdot 10^{-6} N)=4.8\cdot 10^{-6}N

3. When the distance is tripled: 1.33\cdot 10^{-7}N

The initial force between the two charges is F=1.2\cdot 10^{-6}N. In this part of the problem, the distance between the two charges is tripled, so we can write

r'=3r

And substituting into the formula, we find the new force:

F'=k\frac{q_1 q_2}{r'^2}=k\frac{q_1 q_2}{(3r)^2}=\frac{1}{9}k\frac{q_1 q_2}{r^2}=\frac{F}{9}

So, the force is reduced to 1/9 of its original value. Therefore, it is

F'=\frac{1.2\cdot 10^{-6} N}{9}=1.33\cdot 10^{-7}N

4 0
4 years ago
If your horse got bit in the mouth by a rattle snake how long dose she have to live
Gelneren [198K]

Answer: Probably at-least 30 mins to an hour

5 0
3 years ago
Read 2 more answers
Look at the circuit diagram. Which of these components is part of the circuit?
tangare [24]
I think the answer is c AC power source
Hope this help you?
4 0
3 years ago
Viewers of Star Trek hear of an antimatter drive on the Starship Enterprise. One possibility for such a futuristic energy source
Alex_Xolod [135]

Answer:

The magnetic field strength  required to hold anti-protons, moving at 5.70 ✕ 10⁷ m/s in a circular path of 3.20 m in radius is 0.186 T.

Explanation:

To solve the question we note that the magnetic force on a moving charge is given by

F = q·v·B

Where

q = Charge

v = Velocity of the charge =5.70 ✕ 10⁷ m/s

B = Magnetic field

Based on Newton's second law,

Force = Mass, m × Acceleration, a = m × a

Where:

a = Acceleration

m = Mass of anti-proton = Mass of proton = 1.6726219 × 10⁻²⁷ kg

We note that for circular motion, acceleration a is given by

\alpha = \frac{v^{2} }{r} .

Where:

r = Radius = 3.20 m

Therefore, for the circular motion, force, F = \frac{m\cdot v^{2} }{r}

Equating the magnetic force equation to the circular force equation, we have

\frac{m\cdot v^{2} }{r} = q·v·B So that, we find B by making the subject of the formula as follows

B= \frac{m\times v^{2} }{r\times q \times v} . Which gives

B= \frac{m\times v }{r\times q} =  \frac{(1.6726219 \times 10^{-27}) \times (5.7\times 10^{7}) }{(3.20)\times (1.602\times 10^{-19}) } =  0.186 T

The magnetic field strength is

B = 0.186 T

4 0
3 years ago
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