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Scorpion4ik [409]
3 years ago
15

Which of these is an example of physical change

Chemistry
2 answers:
Yuki888 [10]3 years ago
6 0
I would say d sorry if it is wrong
Julli [10]3 years ago
5 0

Answer:

it is A mark me big brain pls

Explanation:

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Consider two concentric spheres whose radii differ by t = 1 nm, the "thickness" of the interface. At what radius (in nm) of sphe
Gala2k [10]

Answer:

Radius of the interior sphere = 3.847 nm

Explanation:

The volume of the shell (Vs) is equal to the difference of the volume of the outer sphere (Vo) and the volume of the inner sphere (Vi). Then:

V_s=V_o-V_i=V_i\\V_o=2*V_i\\

If we express the radius of the outer sphere (ro) in function of the radius of the inner sphere (ri), we have (e being the shell thickness):

r_o=r_i+e

The first equation becomes

\frac{4 \pi}{3}*r_o^{3}  = 2 * \frac{4 \pi}{3}*r_i^{3}  \\\\\frac{4 \pi}{3}*(r_i+e)^{3}  = 2 * \frac{4 \pi}{3}*r_i^{3}  \\\\(r_i+e)^{3}  = 2 *r_i^{3}  \\\\(r_i^{3}+3r_i^{2}e+3r_ie^{2}+e^{3})=2r_i^{3}\\\\-r_i^{3}+3r_i^{2}e+3r_ie^{2}+e^{3}=0\\\\-r_i^{3}+3r_i^{2}+3r_i+1=0

To find ri that satisfies this equation we have to find the roots of the polynomial.

Numerically, it could be calculated that ri=3.847 nm satisfies the equation.

So if the radius of the interior sphere is 3.847 nm, the volume of the interior sphere is equal to the volume of the shell of 1nm.

8 0
3 years ago
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