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irga5000 [103]
3 years ago
12

What is the weight of a rock with a mass of 3.6 kilograms

Physics
1 answer:
Marysya12 [62]3 years ago
4 0
3.85 pounds is the answer
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At a certain elevation, the ________ , the air becomes saturated and water-vapor molecules ________.
andriy [413]

At certain altitude, the temperature of air decrease, The air becomes saturated and water vapour molecules starts condensing.


As the altitude of air increase, the atmospheric pressure decrease due to which the temperature of the air decrease. The water molecules in the atmosphere start condensing, which saturate the air (that is air can no hold water molecules), due to which the water vapour molecules starts condensing and falls on the earth in the form of rain.


4 0
3 years ago
With the water trap being used, all the possible sources of error would be eliminated.
Phoenix [80]

Answer:

True

Explanation:

7 0
3 years ago
Which of these help create radio waves?
inysia [295]
The one that help create radio waves is :
Changing electric and magnetic fields applied at right angles

Radio waves are transverse wave, which means  that the oscillations occurring perpendicular to the direction of energy transfer

hope this helps
3 0
3 years ago
HELP PLEASE!! GIVING BRAINLIEST!! If you answer this correctly ill answer some of your questions you have posted! (20pts)
anyanavicka [17]

Explanation:

potential energy =360800J

mass(m)=?

height (h)=25m

g=9.8m/s²

we have

potential energy =360800J

mgh=360800J

m×9.8×25=360800

m=360800/(9.8×25)=1472.653061kg

8 0
3 years ago
Read 2 more answers
An object whose specific gravity is 0.850 is placed in water. What fraction of the object is below the surface of the water?
Fynjy0 [20]

Answer:

The fraction of the object that is below the surface of the water is ¹⁷/₂₀

Explanation:

Given;

specific gravity of the object, γ = 0.850

Specific gravity is given as;

specific \ gravity = \frac{density \ of the \ object}{density \ of \ water}\\\\0.85= \frac{density \ of the \ object}{1000 \ kg/m^3} \\\\density \ of the \ object = 850 \ kg/m^3

Fraction of the object's weight below the surface of water is calculated as;

= \frac{850}{1000} \ \times\ 100\%\\\\= 85 \% \\\\= \frac{17}{20}

Therefore, the fraction of the object that is below the surface of the water is ¹⁷/₂₀

8 0
3 years ago
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