1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
dangina [55]
3 years ago
7

What is the weight of a 90kg man standing on the moon, where gmoon = 1.64 m/s2?

Chemistry
1 answer:
Doss [256]3 years ago
3 0

Answer:

The weigth of a 90kg man standing on the moon is <u><em>147.6 N (option C)</em></u>

Explanation:

Weight is called the action exerted by the force of gravity on the body.

The mass (amount of matter that a body contains) of an object will always be the same, regardless of where it is located. Instead, the weight of the object will vary according to the force of gravity acting on it.

The formula that allows you to calculate the weight of any body is:

W = m*g

where:

  • W = weight measured in N.
  • m = mass measured in kg.
  • g = acceleration of gravity measured in m/s². The acceleration of gravity g is the same for all objects that fall due to gravitational attraction, whatever their size or composition. For example, as an approximate value on Earth, g = 9.8 m/s².

In this case,  the mass m has a value of 90 kg and the gravity g has a value of 1.64 m/s², which is the value of the acceleration of gravity of the moon. Then:

W=90 kg* 1.64 m/s²

<u><em>W= 147.6 N</em></u>

Finally, <u><em>the weigth of a 90kg man standing on the moon is 147.6 N (option C)</em></u>

You might be interested in
32. List three examples of substances. Explain why each<br> is a substance. (3.1)
Alex

Answer: Chemical X H3 and f1

Explanation:

8 0
3 years ago
How many molecules are in 1 mol of the chemical equation shown above
aivan3 [116]
There's 6.022×10^23 particles in 1 mole of anything

like there is 1000 grams in 1 kilogram of anything
7 0
3 years ago
Convert 123 in scientific notation
dalvyx [7]
✡ Answer: 1.23*10^2 ✡


- - Add a decimal at the end (to the right) and count till you get to the first number.
So now you have 1.23

- - Now you always want to times it by 10 to the power of how many times you moved it over, in this case, 2

Final answer: 1.23*10^2

✡Hope this helps✡


4 0
3 years ago
Read 2 more answers
Can you tell me about atoms
trasher [3.6K]

Answer:

is the smallest unit of ordinary matter that forms a chemical element.

Explanation:

and were created after the Big Bang 13.7 billion years ago.

4 0
3 years ago
Calculate the concentration in % (m/m) of a solution containing 30.0g of mgcl2 dissolved in 270.0g of h20
qaws [65]
As the question tells you, you need to use the formula

% mass= mass of solute/ mass of solution x 100

mass solute= 30.0 g
mass of solution= 30.0 + 270.0= 300.0 g

% mass= 30.0/ 300.0 x 100= 10%

answer is B
4 0
3 years ago
Other questions:
  • What evidence of a chemical reaction might you see in bleaching a stain
    8·2 answers
  • For the following reaction, 5.05 grams of copper are mixed with excess silver nitrate. The reaction yields 11.0 grams of copper(
    9·1 answer
  • The pH of a Ba(OH) 2 solution is 10.00. What is the H+ ion concentration of this solution?
    9·1 answer
  • A 2.832 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 6.439 grams of CO2 and 2
    6·1 answer
  • What is the primary source of energy in a food chain? A) consumers B) plants C) producers Eliminate D) sun
    10·2 answers
  • Suppose that a certain fortunate person has a net worth of $79.0 billion ($7.90×1010). If her stock has a good year and gains $3
    9·1 answer
  • An element consists of 1.40% of an isotope
    14·2 answers
  • Why does ammonia gas diffuse faster than hydrogen chloride gas?
    14·1 answer
  • How many oxygens are used per glucose molecule in glycolysis?
    8·1 answer
  • Explain why one molecule of nabh4 will reduce only two molecules of m acetylbenzaldehyde to form the corresponding product
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!