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zmey [24]
2 years ago
15

A planet travels in an elliptical path around a star, as shown in the figure. As the planet gets closer to the star, the gravita

tional force that the star exerts on the planet increases. Which statement of reasoning best supports and correctly identifies what happens to the magnitude of the force that the planet exerts on the star as the planet gets closer to the star?
The force remains constant because the mass of the planet remains constant.
A

The force increases because it is part of a Newton’s third law pair of forces with the force that the star exerts on the planet.
B

The force decreases because the planet increases its speed as it gets closer to the star.
C

The force fluctuates such that it increases and decreases because the planet does not travel in a perfectly circular path.

Chemistry
1 answer:
Soloha48 [4]2 years ago
3 0

Answer:

The force increases because it is part of a Newton’s third law pair of forces with the force that the star exerts on the planet.

Explanation:

Force between two objects can be expressed by an equation:

F = G • m1 • m2 / r^2,

where m1 and m2 are objects' masses, r is the distance between them, and G is a gravitational constant.

That means that greater the masses or lesser the distance, the force will be greater, and vice versa.

This force exists between any two objects, but is generally extremely weak, so it's best observed with big and large objects with great mass, such as planets and stars.

This force, whatever its magnitude may be, always works on both objects, following the third Newton's law.

So, whatever the force the stat exerts on the planet is, the planet will exert the same amount of force on the star.

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When 0.40 mol Al is mixed with 0.40 mol Br2, the following reaction occurs: 2Al(s) + 3Br2(l) → 2AlBr3(s) Identify the limiting r
Lera25 [3.4K]

Answer:

d. The limiting reactant is Br2, and 0.13 mol Al should remain unreacted.

Explanation:

From the reaction, 2Al(s) + 3Br2(l) → 2AlBr3(s)

2 moles of Al combines with 3 moles of Br to form 2 moles of AlBr3

Hence 1 mole of Br combines with 2/3 moles of Al

or 0.4 moles of Br combines with 2/3×0.4 moles of Al or 0.267 Moles of Al leaving

0.4 - 0.267 = 0.133 moles of Al remaining unreacted

7 0
2 years ago
The accepted value is 1.43. Which correctly describes this student’s experimental data?
Natalija [7]

Answer:

Neither accurate nor precise

Explanation:

The values were not near or even the same as the accepted value thus making it neither accurate nor precise.

4 0
2 years ago
Read 2 more answers
Gamma rays are often used to kill microorganisms in food, in an attempt to make the food safer. Some people contend that this ir
nikdorinn [45]

Answer:

b . Irradiated food is shown to not be radioactive.

Explanation:

If it can be proven that irradiated food is not radioactive, then it will effective dispute the idea that irradiated food are less safe to eat.

  • An irradiated food is one in which ionizing radiations have been employed to improve food quality.
  • Thus, bacteria and other food spoilers can be exterminated from the food.
  • Most irradiated food do not contain radiation and are fit for consumption.

If it can be proven, that this is true, then it will challenge the idea that irradiated foods are not safe.

4 0
1 year ago
According to the equation below, how many moles of Ca(OH)2 are required to react with 1.36 mol H3PO4 to produce Ca3(PO4)2? 3Ca(O
allsm [11]

<u>Answer:</u> The amount of calcium hydroxide needed to react is 2.04 moles

<u>Explanation:</u>

We are given:

Moles of phosphoric acid = 1.36 moles

For the given chemical equation:

3Ca(OH)_2+2H_3PO_4\rightarrow Ca_3(PO_4)_2+6H_2O

By Stoichiometry of the reaction:

2 moles of phosphoric acid reacts with 3 moles of calcium hydroxide

So, 1.36 moles of phosphoric acid will react with = \frac{3}{2}\times 1.36=2.04mol of calcium hydroxide

Hence, the amount of calcium hydroxide needed to react is 2.04 moles

3 0
2 years ago
If the lead can be extracted with 92.5% efficiency, what mass of ore is required to make a lead sphere with a 3.50 cm radius? le
ElenaW [278]

The volume of sphere can be calculated using the following formula:


V=\frac{4}{3}\pi r^{3}


Here, r is radius of the sphere which is 3.5 cm. Putting the value,


V=\frac{4}{3}\pi r^{3}=\frac{4}{3}(3.14)(3.5cm)^{3}=180 cm^{3}


This is equal to the volume of lead, density of lead is 11.34 g/cm^{3} thus, mass of lead can be calculated as follows:


m=d×V


Putting the values,


m=11.34 g/cm^{3}\times 180 cm^{3}=2041.2 g


Let the mass of ore be 1 g, 68.6% of galena is obtained by mass, thus, mass of galena obtained will be 0.686 g.


Now, 86.6% of lead is obtained from this gram of galena, thus, mass of lead will be:


m=0.686×0.866=0.5940 g


Therefore, 0.5940 g of lead is obtained from 1 g of ore for 100% efficiency, thus, for 92.5% efficiency

m=\frac{92.5}{100}\times 0.5940=0.54945 g

1 g of lead obtain from\frac{1}{0.54945}=1.82 grams of ore.


Thus, 2041.2 g of lead obtain from:


2041.2\times 1.82=3.715\times 10^{3}g or 3.715 kg


Therefore, mass of ore required to make lead sphere is 3.715 kg.


6 0
2 years ago
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