At what height above the earth surface will the gravitational acceleration be 5m

Chemistry

1 answer:

slamgirl [31]3 years ago

8 0

Answer: -

The acceleration due to gravity at height r = a = GM/r²

Rearranging

r² = GM /a

= (6.674 x 10⁻¹¹ x 5.972 x 10²⁴ ) / 5

r = 8.917 x 10⁶ m

r = 8917 Km

Now Radius of earth = 6371 Km

So height = 8917 - 6371 = 2546 Km

You might be interested in

An instrument that allows you to look closely at objects that are invisible to the human eye, including cells

xxMikexx [17]

The answer is Microscope

4 0

3 years ago

To prevent rust, nails are coated with ________. To prevent rust, nails are coated with ________. salt calcium lithium sodium zi

Arturiano [62]

Answer:

the answer to this question is zinc

7 0

3 years ago

How Do Glow Sticks Glow? Explain in 5-6 sentences. IF NOT, YOUR ANSWER WILL BE DELETED

Readme [11.4K]

Answer:

All liquid glow products depend on a chemical process known as CHEMILUMINESCENCE to produce their light. Chemiluminesence is a chemical reaction that causes a release of energy in the form of light. To produce this light the electrons in the chemicals become excited and rise to a higher energy level.To utilise this process glowsticks contain two liquids; hydrogen peroxide and tert-butyl alcohol. When mixed together it is these liquids that create the glow. Fluorescent dyes are also used in the alcohol to alter the colour of the light emitted. Chemiluminesence is a chemical reaction that causes a release of energy in the form of light.

Explanation:

Hope It Helps!

6 0

3 years ago

3.5g of a Certain compound X, known to be made of carbon, hydrogen, and perhaps oxygen, and to have a molecular molar mass of 15

shutvik [7]

Answer:

C₅H₁₀O₅

Explanation:

1. Calculate the mass of each element in 2.78 mg of X.

(a) Mass of C

$\text{Mass of C} = \text{5.13 g CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{1.400 g C}$

(b) Mass of H

$\text{Mass of H} = \text{2.10 g H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g H$_{2}$O}} = \text{0.2349 g H}$

(c) Mass of O

Mass of O = 3.5 - 1.400 - 0.2349 = 1.87 g

2. Calculate the moles of each element

$\text{Moles of C = 1400 mg C}\times\dfrac{\text{1 mmol C}}{\text{12.01 mg C }} = \text{116.6 mmol C}\\\\\text{Moles of H = 234.9 mg H} \times \dfrac{\text{1 mmol H}}{\text{1.008 mg H}} = \text{233.1 mmol H}\\\\\text{Moles of O = 1870 mg O} \times \dfrac{\text{1 mmol O}}{\text{16.00 mg O}} = \text{116 mmol O}$