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lidiya [134]
3 years ago
6

DNA is coiled into chromosomes in a cell's

Physics
2 answers:
aalyn [17]3 years ago
8 0
In the nucleus of each cell<span>, the </span>DNA<span> molecule is packaged </span>into<span> thread-like structures called </span>chromosomes<span>. Each </span>chromosome<span> is made up of </span>DNA<span> tightly</span>coiled<span> many times around proteins called histones that support its structure</span>
tia_tia [17]3 years ago
6 0

Answer: Cell's nucleus

Explanation :

The full form of DNA is Deoxyribonucleic acid. The genetic information of a cell is organized in the DNA. It is inherited from parents by their children.

DNA is coiled into chromosomes in a cell's nucleus. It has a shape like a double helix. It is twisted in the form of spirals.

Hence, DNA is coiled into chromosomes in a cell's Nucleus.

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A top of rotational inertia 4.0 kg m2 receives a torque of 2.4 nm from a physics professor. the angular acceleration of the body
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Angular acceleration is simply the ratio of the Torque over the rotation inertia, that is:

Angular acceleration = Torque / Rotational inertia

 

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3 years ago
A 0.400-kg ice puck, moving east with a speed of 5.86 m/s , has a head-on collision with a 0.900-kg puck initially at rest.
andreev551 [17]

Answer:

a) The final speed of the 0.400-kg puck after the collision is 2.254 meters per second, b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards, c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.

Explanation:

a) Since collision is perfectly elastic and there are no external forces exerted on pucks system, the phenomenon must be modelled after the Principles of Momentum and Energy Conservation. Changes in gravitational potential energy can be neglected. That is:

Momentum

m_{1}\cdot v_{1,o} + m_{2}\cdot v_{2,o} = m_{1}\cdot v_{1,f} + m_{2}\cdot v_{2,f}

Energy

\frac{1}{2}\cdot (m_{1}\cdot v_{1,o}^{2}+ m_{2}\cdot v_{2,o}^{2})=\frac{1}{2}\cdot (m_{1}\cdot v_{1,f}^{2}+ m_{2}\cdot v_{2,f}^{2})

m_{1}\cdot v_{1,o}^{2} + m_{2}\cdot v_{2,o}^{2} = m_{1}\cdot v_{1,f}^{2} + m_{2}\cdot v_{2,f}^{2}

Where:

m_{1}, m_{2} - Masses of the 0.400-kg and 0.900-kg pucks, measured in kilograms.

v_{1,o}, v_{2,o} - Initial speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

v_{1}, v_{2} - Final speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

If m_{1} = 0.400\,kg, m_{2} = 0.900\,kg, v_{1,o} = +5.86\,\frac{m}{s}, v_{2,o} = 0\,\frac{m}{s}, the system of equation is simplified as follows:

2.344\,\frac{kg\cdot m}{s} = 0.4\cdot v_{1,f} + 0.9\cdot v_{2,f}

13.736\,J = 0.4\cdot v_{1,f}^{2}+0.9\cdot v_{2,f}^{2}

Let is clear v_{1,f} in first equation:

0.4\cdot v_{1,f} = 2.344 - 0.9\cdot v_{2,f}

v_{1,f} = 5.86-2.25\cdot v_{2,f}

Now, the same variable is substituted in second equation and resulting expression is simplified and solved afterwards:

13.736 = 0.4\cdot (5.86-2.25\cdot v_{2,f})^{2}+0.9\cdot v_{2,f}^{2}

13.736 = 0.4\cdot (34.340-26.37\cdot v_{2,f}+5.063\cdot v_{2,f}^{2})+0.9\cdot v_{2,f}^{2}

13.736 = 13.736-10.548\cdot v_{2,f} +2.925\cdot v_{2,f}^{2}

2.925\cdot v_{2,f}^{2}-10.548\cdot v_{2,f} = 0

2.925\cdot v_{2,f}\cdot (v_{2,f}-3.606) = 0

There are two solutions:

v_{2,f} = 0\,\frac{m}{s} or v_{2,f} = 3.606\,\frac{m}{s}

The first root coincides with the conditions before collision and the second one represents a physically reasonable solution.

Now, the final speed of the 0.400-kg puck is: (v_{2,f} = 3.606\,\frac{m}{s})

v_{1,f} = 5.86-2.25\cdot (3.606)

v_{1,f} = -2.254\,\frac{m}{s}

The final speed of the 0.400-kg puck after the collision is 2.254 meters per second.

b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards.

c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.

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