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2 years ago

I need help me with my question

Physics

1 answer:

lisov135 [29]2 years ago

4 0

The tilt of the moon's axis does not allow for monthly alignment, so the lunar and solar eclipse do not happen every month.

<h3>How do the lunar and solar eclipse occur?</h3>

For the occurrence of lunar and solar eclipse, the sun, moon and the earth must remain in a plan and along a straight line.
When the earth appears in between the sun and the moon, lunar eclipse occurs.
When the moon appears in between the sun and the earth, solar eclipse occurs.
The moon and earth are rotating not only around the sun, but also around the black hole of Milky way galaxy.
So they are not present in a plan as well as in a straight line in every full moon and new moon time.

Thus, we can conclude that the option D is correct.

Learn more about the lunar eclipse and solar eclipse here:

brainly.com/question/8643

#SPJ1

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In the four terrestrial planets, the densest, heaviest materials are at the center and not evenly distributed throughout the pla

Gala2k [10]

<em>The question is incomplete. Please read below the missing content.</em>

B) The 4 terrestrial planets have to as soon as were hot sufficient to be molten (like a liquid).

Terrestrial planets may be described as any planet of the sun machine or any exoplanet that is "Earth-like" in the experience that it is composed primarily of metals and rock, in evaluation to a planet that's a fuel large.

In the sun machine, the terrestrial planets are the inner planets closest to the solar, i.e. Mercury, Venus, Earth, and Mars.

Scientists believe that the densest, heaviest substances are at the center and not evenly allotted can be due to the truth that they need to once be warm enough to be molten. In this example, density will take region, making the heavier below and the lighter on the pinnacle.

In the four terrestrial planets, the densest, heaviest materials are at the center and not evenly distributed throughout the planet. Scientists interpret this observation to mean that:

a.the four terrestrial planets must once have been inside the Sun

b.the four terrestrial planets must once have been hot enough to be molten (like a liquid)

c.the four terrestrial planets must have formed where Jupiter and Saturn now are

d.the four terrestrial planets must have collided with each other many times

e.none of the above

Learn more about terrestrial planets here brainly.com/question/11862047

#SPJ4

3 0

2 years ago

You read in a science magazine that on the Moon, the speed of a shell leaving the barrel of a modern tank is enough to put the s

olga_2 [115]

To solve this problem we will use the definition of the kinematic equations of centrifugal motion, using the constants of the gravitational acceleration of the moon and the radius of this star.

Centrifugal acceleration is determined by

$a_c = \frac{v^2}{r}$

Where,

v = Velocity

r = Radius

From the given data of the moon we know that gravity there is equivalent to

$a = 1.62m/s$

While the radius of the moon is given by

$r = 1.74*10^6m$

If we rearrange the function to find the speed we will have to

$v = \sqrt{ar}$

$v = \sqrt{1.6(1.74*10^6)}$

$v = 1.7km/s$

The speed for this to happen is 1.7km/s

3 0

3 years ago

Describe how the chemical bonds changed during the reaction.

k0ka [10]

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Answer:

In a chemical reaction, the atoms and molecules that interact with each other are called reactants. ... No new atoms are created, and no atoms are destroyed. In a chemical reaction, reactants contact each other, bonds between atoms in the reactants are broken, and atoms rearrange and form new bonds to make the products.

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6 0

3 years ago

Read 2 more answers

Two objects carry initial charges that are q1 and q2, respectively, where |q2| > |q1|. They are located 0.160 m apart and beh

mart [117]

Answer:

$\rm |q_1|=8.0\times 10^{-7}\ C,\ \ \ |q_2| = 4.6\times 10^{-6}\ C.$

Explanation:

According to the Coulomb's law, the magnitude of the electrostatic force between two static point charges $\rm q_1$ and $\rm q_1$ , separated by a distance $\rm r$ , is given by

$\rm F = \dfrac{kq_1q_2}{r^2}.$

where k is the Coulomb's constant.

Initially,

$\rm r = 0.160\ m\\F_i = -1.30\ N.\\\\and \ \ |q_2|>|q_1|.$

The negative sign is taken with force F because the force is attractive.

Therefore, the initial electrostatic force between the charges is given by

$\rm F_i = \dfrac{kq_1q_2}{r^2}.\\-1.30=\dfrac{kq_1q_2}{0.160^2}\\\rm\Rightarrow q_2 = \dfrac{-1.30\times 0.160^2}{q_1k}\ \ \ ..............\ (1).$

Now, the objects are then brought into contact, so the net charge is shared equally, and then they are returned to their initial positions.

The force is now repulsive, therefore, $\rm F_f = +1.30\ N.$

The new charges on the two objects are

$\rm q_1'=q_2' = \dfrac{q_1+q_2}{2}.$

The new force is given by

$\rm F_f = \dfrac{kq_1'q_2'}{r^2}\\+1.30=\dfrac{k\left (\dfrac{q_1+q_2}{2}\right )\left (\dfrac{q_1+q_2}{2}\right )}{0.160^2}\\\Rightarrow \left (\dfrac{q_1+q_2}{2}\right )^2=\dfrac{+1.30\times 0.160^2}{k}\\(q_1+q_2)^2=\dfrac{4\times 1.30\times 0.160^2}{k}\\q_1^2+q_2^2+2q_1q_2=\dfrac{4\times 1.30\times 0.160^2}{k}\\\\$

Using (1),

$\rm q_1^2+\left ( \dfrac{-1.30\times 0.160^2}{q_1k}\right )^2+2\left (\dfrac{-1.30\times 0.160^2}{k} \right )=\dfrac{4\times 1.30\times 0.160^2}{k}\\q_1^2+\dfrac 1{q_1^2}\left ( \dfrac{-1.30\times 0.160^2}{k}\right )^2-\left (\dfrac{6\times 1.30\times 0.160^2}{k} \right )=0\\q_1^4+\left ( \dfrac{-1.30\times 0.160^2}{k}\right )^2-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{k} \right )=0$

$\rm q_1^4+\left ( \dfrac{-1.30\times 0.160^2}{k}\right )^2-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{k} \right )=0\\q_1^4+\left ( \dfrac{-1.30\times 0.160^2}{9\times 10^9}\right )^2-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{9\times 10^9} \right )=0\\q_1^4-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{9\times 10^9} \right )+\left ( \dfrac{-1.30\times 0.160^2}{9\times 10^9}\right )^2=0$

$\rm q_1^4-q_1^2\left (2.22\times 10^{-11} \right )+\left ( 1.37\times 10^{-23}\right ) =0\\\Rightarrow q_1^2 = \dfrac{-(-2.22\times 10^{-11})\pm \sqrt{(-2.22\times 10^{-11})^2-4\cdot (1)\cdot (1.37\times 10^{-23})}}{2}\\=1.11\times 10^{-11}\pm 1.046\times 10^{-11}.\\=6.4\times 10^{-13}\ \ \ or\ \ \ 2.156\times 10^{-11}\\\Rightarrow q_1 = \pm 8.00\times 10^{-7}\ C\ \ \ or\ \ \ \pm 4.64\times 10^{-6}\ C.$

Using (1),

When $\rm q_1 = \pm 8.00\times 10^{-7}\ C$ ,

$\rm q_2=\dfrac{-1.30\times 0.160^2}{\pm 8.00\times 10^{-7}\times 9\times 10^9}=\mp4.6\times 10^{-6}\ C.$

When $\rm q_1=\pm 4.6\times 10^{-6}\ C$ ,

$\rm q_2=\dfrac{-1.30\times 0.160^2}{\pm 4.64\times 10^{-6}\times 9\times 10^9}=\mp7.97\times 10^{-7}\ C\approx 8.0\times 10^{-7}\ C.$

Since, $\rm |q_2|>|q_1|$

Therefore, $\rm |q_1|=8.0\times 10^{-7}\ C,\ \ \ |q_2| = 4.6\times 10^{-6}\ C.$

7 0

3 years ago

What cloud makes hail

MArishka [77]

Answer:

Cumulonimbus

Hail development. Hail is a type of strong precipitation, which is shaped in rainstorms mists (Cumulonimbus). Tempests mists comprises of beads of fluid water (at temperatures lower than 0°с, the beads can be in a thermodynamically instable supercooled condition) and ice gems

Explanation:

3 0

3 years ago