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Law of Universal Gravitation (what factors affect gravity and how?)

olya-2409 [2.1K]

Explanation:

Hey there!

According to the law of gravitation, the factors that affects the gravity are:

Mass of the heavenly bodies.
Distance from the centre of the body to next body.

Mass of heavenly bodies:The mass of heavenly bodies affect the gravity. We know that gravity is directly proportional to the mass of heavenly bodies. so, when the mass is more there is more gravity and where there is less mass there is less gravity.

Distance from the centre of the body to next body: It is one of the factor to influence gravity. When there is more distance there is less gravity and where there is distance more force is there. Is means gravity is inversely proportional to the distance between the bodies.

Hope it helps...

7 0

2 years ago

Is this correct?? please help

NeTakaya

Yes it is but I have to check so wait

4 0

2 years ago

How many moles of glucose does 1.2 x 10^24 molecules represent

sammy [17]

<h3>Answer:</h3>

2.0 mol C₆H₁₂O₆

<h3>General Formulas and Concepts:</h3>

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Chemistry

Atomic Structure

Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Stoichiometry

Using Dimensional Analysis

<h3>Explanation:</h3>

Step 1: Define

1.2 × 10²⁴ molecules C₆H₁₂O₆ (glucose)

Step 2: Identify Conversions

Avogadro's Number

Step 3: Convert

Set up: $\displaystyle 1.2 \cdot 10^{24} \ molecules \ C_6H_{12}O_6(\frac{1 \ mol \ C_6H_{12}O_6}{6.022 \cdot 10^{23} \ molecules \ C_6H_{12}O_6})$
Divide: $\displaystyle 1.99269 \ mol \ C_6H_{12}O_6$

Step 4: Check

Follow sig fig rules and round. We are given 2 sig figs.

1.99269 mol C₆H₁₂O₆ ≈ 2.0 mol C₆H₁₂O₆

3 0

3 years ago

1. Calculate the energy of a photon with a frequency of 3x1015 Hz. ?????

Afina-wow [57]

Answer:

1.99 x 10⁻¹⁸J

Explanation:

Given parameters:

Frequency of the wave = 3 x 10¹⁵Hz

Unknown:

Energy of the photon = ?

Solution:

To solve this problem, we use the expression below;

E = hf

Where E is the energy, h is the Planck's constant and f is the frequency

Now insert the parameters and solve for E;

E = 6.63 x 10⁻³⁴ x 3 x 10¹⁵ = 19.9 x 10⁻¹⁹J or 1.99 x 10⁻¹⁸J

6 0

3 years ago

A syringe containing 1.63 mL of oxygen gas is cooled from 91.0 ∘C to 0.9 ∘C. What is the final volume Vf of oxygen gas? (Assume

Archy [21]

Answer:

Vf = 1.22 mL

Explanation:

If we assume that the pressure is constant and the number of moles does not change, we can say that the volume of the gas is modified in a directly ratio, to the Absolute Temperature.

Let's convert the values:

91°C + 273 = 364K

0.9°C + 273 = 273.9K

Volume decreases if the temperature is decreases

Volume increases if the T° increases

V₁ / T₁ = V₂ / T₂ → 1.63mL /364K = V₂ / 273.9K

V₂ = (1.63mL /364K) . 273.9K → 1.22 mL

5 0

3 years ago

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