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Semmy [17]
3 years ago
13

What is a structure that organizes motion of chromosomes

Physics
2 answers:
Damm [24]3 years ago
5 0
I think centrioles organize the motion of chromosomes. Hope this helps.
strojnjashka [21]3 years ago
4 0
<span>Cytoplasm: <span>the entire contents of the cell, exclusive of the nucleus and bounded by the plasma membrane.

Hope this helped. :)</span></span>
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Now, using your mass (in kg), and the figures for g (in the table below), you can calculate your weight on other planets.
Licemer1 [7]

Answer:

1) Weight on Mercury

F =W=mg=68.11 \times 3.61 m.s^{-2}

Explanation:

do the same to the rest and use your calculator to find the weight in N.

3 0
3 years ago
What happens when an electron moving from the 3rd energy level to the 1st energy level?
Sholpan [36]

Answer:

A photon of wavelength 103 nm is released

Explanation:

When an electron in an atom jumps from a higher energy level to a lower energy level, it releases a photon whose energy is equal to the difference in energy between the two levels.

For example, if we are talking about a hydrogen atom, the energy of the levels are:

E_1 = -13.6 eV\\E_2 = -3.4 eV\\E_3 = -1.5 eV

So, the energy of the photon released when the electron jumps from the level n=3 to n=1 is

\Delta E = E_3 - E_1 = -1.5 -(-13.6)=12.1 eV

In Joules,

\Delta E =12.1\cdot 1.6\cdot 10^{-19} = 1.94\cdot 10^{-18} J

We can also find the wavelength of this photon, using the equation:

\Delta E = \frac{hc}{\lambda}\rightarrow  \lambda=\frac{hc}{\Delta E}=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{1.94\cdot 10^{-18}}=1.03\cdot 10^{-7} m = 103 nm

7 0
4 years ago
As a person squeezes and applies pressure to the outside of a balloon, air particles inside
mixas84 [53]

The air particle inside the balloon will collide more with each other and the temperature inside the balloon will increase.

As a person squeezed and applies the pressure to the outside of a balloon, the air particle inside the balloon gains energy and collide with each other, the particle of the air also try leave the balloon surface will implies equal pressure on the wall of the balloon, as the pressure outside the balloon increase, the inside pressure will also increase.

7 0
3 years ago
We wish to determine the electric field at a point near a positively charged metal sphere (a good conductor). We do so by bringi
quester [9]

Answer:

Less than.

Explanation:

We have the positive charged metal sphere and we have to determine the electric field at a point near to it. In order to find that if we bring the positive test charge at that point then as we know that "like charges repel" so their electric field lines will repel each other resulting in a weaker electric field.    

    However if we bring the negative test charge at that point then of course there will be attraction and also the the electric field lines will direct from the positive to negative resulting in a stronger electric field between them. So there will be larger electric field then before.

"In this case, It can be concluded that electric field will be less than it was at this point before the test charge was present."

5 0
3 years ago
A cannon of mass 5.71 x 103 kg is rigidly bolted to the earth so it can recoil only by a negligible amount. The cannon fires a 7
zheka24 [161]

Answer:

541.14 m/s

Explanation:

We are given that

Mass of cannon=m_1=5.71\times 10^3 kg

Mass of shell,m_2=73.5 kg

Initial velocity of shell,v=547 m/s

We have to find the velocity of shell fired from this loose cannon.

According to law of conservation of momentum

m_1v_1+m_2v_2=m_1u_1+m_2u_2

Initial momentum of system=0

m_1v_1=-m_2v_2

v_1=-\frac{m_2v_2}{m_1}

When the cannon is bolted to the ground then only shell moves and kinetic energy of system equals to kinetic energy of shell

Kinetic energy of shell,K.E=\frac{1}{2}mv^2=\frac{1}{2}\times 73.5(547)^2}=1.09\times 10^7 J

K.E of shell=\frac{1}{2}m_1v^2_1+\frac{1}{2}m_2v^2_2

K.E of shell=\frac{1}{2}m_1(-\frac{m_2v_2}{m_1})^2+\frac{1}{2}m_2v^2_2

K.E of shell=\frac{1}{2}(\frac{m^2_2v^2_2}{m_1}+\frac{1}{2}m_2v^2_2)

2K.E of shell=m_2v^2_2(\frac{m_2}{m_1}+1)

Velocity of shell fired from this loose cannon,v_2=\sqrt{\frac{2k.E}{m_2(\frac{m_2}{m_1}+1)}

v_2=\sqrt{\frac{2\times 1.09\times 10^7}{73.5(\frac{73.5}{5.71\times 10^3}+1)}

v_2=541.14m/s

Hence, the velocity of shell fired from this loose cannon would be 541.14 m/s

5 0
3 years ago
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