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3 years ago

12

I need a story about kepler's laws for a child, something a child would understand. And i'd like it to be okay for someone if I

copy anhd paste it or else my parents are gonna kill me and my teacher will be very disappointe, this is for summer school.

2 answers:

andreyandreev [35.5K]3 years ago

8 0

Answer:

(1) The orbits are ellipses, with focal points ƒ1 and ƒ2 for the first planet and ƒ1 and ƒ3 for the second planet. The Sun is placed in focal point ƒ1.

(2) The two shaded sectors A1 and A2 have the same surface area and the time for planet 1 to cover segment A1 is equal to the time to cover segment A2.

(3) The total orbit times for planet 1 and planet 2 have a ratio a13/2 : a23/2

juin [17]3 years ago

8 0

Answer:

g the day off and I will be there at work and I will be there at work and I will

Explanation:

you are not going to be able to make it to the meeting tonight but I can tomorrow

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Please help me ; - ;

Irina18 [472]

Answer:

2 = C

3 = B

1 = A

Explanation:

5 0

2 years ago

Two photographers are competing for business in town. Andrea uses only film photography and Keira uses only digital photography.

____ [38]

There is no right or wrong answer, your teacher wants you to support your own answer with points. As long as the reasons make logical sense you are fine.

I think they both have valid points. Their replies are both true, but from a buyer's perspective who would you purchase services from? You would get different answers depending on who you ask.

If you choose to go old school, obviously you get an actual photo that can be stored physically. This means it is a memory that can be preserved, and it might feel more nostalgic being able to touch the photo.

On the other hand, a digitally stored photo can be altered (photoshop), but it is forever as long as the internet still exists. A physical photo would fade with time, which doesn't happen with a digital photo.

It is definitely easier to argue that digital photography has more advantages (they do, it is why nobody uses film anymore)

Points you can consider:

Can be transferred to the other side of the world instantly

Ability to make copies and print as many photos as you want

Can be stored on cloud/devices and be like that forever

Compare them with film photography to give a more solid response.

3 0

3 years ago

The volume of water in the Pacific Ocean is about 7.0 × 10 8 km 3 . The density of seawater is about 1030 kg/m3. (a) Determine t

Novay_Z [31]

To solve the problem it is necessary to consider the concepts related to Potential Energy and Kinetic Energy.

Potential Energy because of a planet would be given by the equation,

$PE=\frac{GMm}{r}$

Where,

G = Gravitational Universal Constant

M = Mass of Ocean

M = Mass of Moon

r = Radius

From the data given we can calculate the mass of the ocean water through the relationship of density and volume, then,

$m = \rho V$

$m = (1030Kg/m^3)(7*10^8m^3)$

$m = 7.210*10^{11}Kg$

It is necessary to define the two radii, when the ocean is far from the moon and when it is facing.

When it is far away, it will be the total diameter from the center of the earth to the center of the moon.

$r_1 = 3.84*10^8 + 6.4*10^6 = 3.904*10^8m$

When it's near, it will be the distance from the center of the earth to the center of the moon minus the radius,

$r_2 = 3.84*10^8-6.4*10^6 - 3.776*10^8m$

PART A) Potential energy when the ocean is at its furthest point to the moon,

$PE_1 = \frac{GMm}{r_1}$

$PE_1 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.904*10^8}$

$PE_1 = 9.05*10^{15}J$

PART B) Potential energy when the ocean is at its closest point to the moon

$PE_2 = \frac{GMm}{r_2}$

$PE_2 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.776*10^8}$

$PE_2 = 9.361*10^{15}J$

PART C) The maximum speed. This can be calculated through the conservation of energy, where,

$\Delta KE = \Delta PE$

$\frac{1}{2}mv^2 = PE_2-PE_1$

$v=\sqrt{2(PE_2-PE_1)/m}$

$v = \sqrt{\frac{2*(9.361*10^{15}-9.05*10^{15})}{7.210*10^{11}}}$

$v = 29.4m/s$

8 0

3 years ago

The magnitude of the charge of the electron is:

brilliants [131]

Answer:

a. Exactly the same as the magnitude of the charge of the proton.

Explanation:

The elementary charge (e) is the smallest electric charge that can exist in the universe. Any positive or negative electric charge can be expressed as a multiple of the elementary charge, since is the electric charge carried by a single proton or, equivalently, the magnitude of the electric charge carried by a single electron (-1e).

3 0

3 years ago

A number of compounds containing the heavier noble gases, and especially xenon, have been prepared. One of these is xenon hexafl

aalyn [17]

Answer:

The answer is "82.2 torr"

Explanation:

moles of Xe:

$= \frac{0.06}{131.293} \\\\ =0.00045699313 \ \mol$

moles of $F_2$ :

$= \frac{0.0274}{38} \\\\= 0.00072105263\ \ mol$

moles of produced $XeF_2:$

$= 0.00024$

moles of left $Xe$ :

$= 0.00021$

Calculate the Pressure:

$= \frac{(0.08206\times 0.00024 \times 293)}{(.1)} + \frac{(0.08206\times 0.00021 \times 293)}{(.1)} \\\\= 0.10819611 \ \ atm \\\\ = 0.10819611 \times 760 \\\\ = 82.2 \ \ torr$

5 0

2 years ago