Without counting wind resistance, They will both reach the ground at the same time. If we apply the concept of kinematics, such as the equation vf^2=vi^2 + 2ad. This equation doesn't count how big or how heavy the mass is, it only focuses on how fast where they in the start and how far are both of them from the ground. So if they both have the same distance and same initial veloctity, then they will reach the ground at the same time.
For example, Try dropping a pen and a paper(Vertically) at the same height, you'll see they'll reach the ground at the same time.
If you count wind resistance, the heavier ball will hit the ground faster, because the air molecules will resist the lighter ball compared to the heavier ball.
Explanation:
1-How many moles of NazCOs are in 10.0 ml of a 2.0 M solution?
2-How many moles of NaCl are contained in 100.0 ml of a 0.20 M solution?
3- What weight (in grams) of H2SO4 would be needed to make 750.0 ml of
2.00 M solution?
4-What volume (in ml) of 18.0 M H2SO4 is needed to contain 2.45 g H2S04?
It is customary to work in SI units.
Calculate the volume of the concrete.
V = 3.7*2.1*5.8 cm³ = 45.066 cm³ = 45.066 x 10 ⁻⁶ m³
The mass is 43.8 g = 43.8 x 10⁻³ kg
The density is mass/volume.
Density = (43.8 x 10⁻³ kg)/(45.066 x 10⁻⁶ m³) = 971.9 kg/m³
Answer: 971.9 kg/m³
The heat remains constant because there’s nothing to cool it down
