Respuesta :
Answer: 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 [tex] \pi [/tex] x r³
Here r is the radius of the spherical object
and [tex] \pi [/tex] 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 [tex] \pi [/tex] 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:
The density will increase by a factor of 64: [tex]\rho_2=64\rho_1[/tex]
Explanation:
Hello,
Density at the first state is defined as ratio between mass and volume:
[tex]\rho_1 =\frac{m}{V_1}[/tex]
The volume of a sphere at the initial condition is:
[tex]V_1=\frac{4}{3} \pi r_1^3[/tex]
If the diameter is quartered we compute volume after the compression as:
[tex]V_2=\frac{4}{3} \pi (r_2)^3\\r_2=\frac{D_2}{2}=\frac{\frac{D_1}{4} }{2} =\frac{D_1}{8} =\frac{2r_1}{8}=\frac{r_1}{4}\\V_2=\frac{4}{3} \pi (\frac{r_1}{4})^3\\V_2=\frac{4}{3} \pi\frac{r_1^3}{64}\\V_2=\frac{1}{48} \pi r_1^3[/tex]
In such a way, the volume is reduced therefore the density is increased by a factor of:
[tex]\frac{V_2}{V_1}= \frac{\frac{1}{48} \pi r_1^3}{\frac{4}{3} \pi r_1^3} =\frac{1}{64}\\ V_2=\frac{1}{64}V_1[/tex]
Finally, as the volume is decreased by a factor of 1/64, the density is increased by a factor of 64 by considering:
[tex]\rho_2 =\frac{m}{V_2}=\frac{m}{\frac{1}{64}V_1} =64\frac{m}{V_1}\\ \rho_2=64\rho_1[/tex]
Best regards.
