M)³ / 6 = 4.2e9 m³
<span>so its mass is </span>
<span>M = 3300kg/m³ * 4.2e9m³ = 1.4e13 kg </span>
<span>and so its KE at 16 km/s = 16000 m/s is </span>
<span>KE = ½ * 1.4e13kg * (16000m/s)² = 1.8e21 J
</span># of bombs N = 1.8e21J / 4.0e16J/bomb = 44 234 bombs
<span>give or take.
</span>
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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.
Answer:
The value is 
Explanation:
From the question we are told that
The weight of the block is 
The dimension of the block is 
Generally two atmosphere is equivalent to

Generally 1 atm = 
The area of the block would be evaluated using width and height because we need for the smaller surface to be in contact with the ground in order to maximize the pressure and minimize number of blocks
So

=> 
Generally the force due to this blocks is mathematically represented as

Here N is the number of blocks
So

=> 