Aspherical object has a density ρ. if it is compressed under high pressure to quarter of its original diameter, its density will now be

its new density would be 0.016 times that of its original.

Explanation:

The density of a spherical object can be calculated using following formula

ρ = 4/3 x x r³

Here r is the radius of the spherical object

and is a mathematical constant having value 22/7

Let's assume the original diameter of the object is d.

Therefore its radius will be = d/2 = 0.5d

Let us plug in this value and find density ρ₁

ρ₁ = 4/3 x x (r)³

ρ₁ = 4/3 x 22/7 x (0.5d)³

Now we are compressing the object so that its diameter becomes one quarter of the original.

The new diameter of the object would be d/4 = 0.25d

Therefore the new radius would be 0.25d/2 = 0.125d

Let us find the new density ρ₂

ρ₂ = 4/3 x 22/7 x (0.125d)³

Let us compare ρ₁ and ρ₂ now

ρ₂/ρ₁ = ( 4/3 x 22/7 x (0.125d)³) / ( 4/3 x 22/7 x (0.5d)³ )

We can cancel out 4/3 and 22/7 here

ρ₂/ρ₁ = (0.125d)³ / (0.5d)³

d gets cancelled.

ρ₂/ρ₁ = 0.01562

ρ₂ = 0.01562 ρ₁

ρ₂ is new density and ρ₁ is original density

Therefore we can say that its new density would be 0.016 times that of its original.

answer: a tank of height h is filled with water and sits on the ground. water squirts from a hole at a height y above the ground and has a range r. for two y values, 0 and h, r is zero. determine the value of y for which the range will be a maximum.

explanation:

it means there are 3 atoms of oxygen