1- A calculation that determines how much greenhouse has an individual generates
2-physical processes in chemical reactions that move carbon through the biosphere and geosphere
3-and overall increase in the worlds temperatures attributed to increased concentrations of greenhouse gases
4-components of air mixture that trap infrared energy
5- rain water that has a pH lower than normal
I think this is right,
Answer:
6.86 × 10²⁴ kg
Explanation:
The mass of the earth m = density of earth, ρ × volume of earth, V
m = ρV
The density of the earth, ρ = 5515 kg/m³ and since the earth is a sphere, its volume is the volume of a sphere V = 4πr³/3 where r = radius of the earth = 6.67 × 10⁶ m
Since m = ρV
m = ρ4πr³/3
So, substituting the values of the variables into the equation for the mass of the earth, m, we have
m = 5515 kg/m³ × 4π(6.67 × 10⁶ m)³/3
m = 5515 kg/m³ × 4π × 296.741 × 10¹⁸ m³/3
m = 5515 kg/m³ × 1189.9639π × 10¹⁸ m³/3
m = 6546105.64378π × 10¹⁸ kg/3
m = 20565197.400122 × 10¹⁸ kg/3
m = 6855065.8 × 10¹⁸ kg
m = 6.8550658 × 10²⁴ kg
m ≅ 6.86 × 10²⁴ kg
Answer:
the answer should be the third statement
Recall that mass is the amount of matter present in a body. That means it's a property that is consistent regardless of the body's current location, gravity's pull on the body, etc.
Let's not confuse mass with weight (which is a force computed as Weight = mass x acceleration). Mass will remain constant and that means that whether the object is on Earth or on Mars, its mass remains the same. Thus, the object will still have 2.00 kg as mass on Mars.
Answers: 2.00 kilograms
