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ra1l [238]
3 years ago
13

Which value would complete the last cell?

Physics
1 answer:
GREYUIT [131]3 years ago
5 0

Answer:

4.0

Explanation:

The following data were obtained from the question:

Force (F) = 20 N

Mass (m) = 5 kg

Acceleration (a) =.?

Force is simply defined as the product of mass and acceleration. Mathematically, it is expressed as

Force (F) = mass (m) x acceleration (a)

F = ma

With the above formula, we can obtain th acceleration of the body as follow:

Force (F) = 20 N

Mass (m) = 5 kg

Acceleration (a) =.?

F = ma

20 = 5 x a

Divide both side by 5

a = 20/5

a = 4 m/s²

Therefore, the value that will complete the last cell in the question above is 4.

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Answer:

Density of 127 I = \rm 1.79\times 10^{14}\ g/cm^3.

Also, \rm Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.

Explanation:

Given, the radius of a nucleus is given as

\rm r=kA^{1/3}.

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  • \rm k = 1.3\times 10^{-13} cm.
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The density of the nucleus is defined as the mass of the nucleus M per unit volume V.

\rm \rho = \dfrac{M}{V}=\dfrac{M}{\dfrac 43 \pi r^3}=\dfrac{M}{\dfrac 43 \pi (kA^{1/3})^3}=\dfrac{M}{\dfrac 43 \pi k^3A}.

For the nucleus 127 I,

Mass, M = \rm 2.1\times 10^{-22}\ g.

Mass number, A = 127.

Therefore, the density of the 127 I nucleus is given by

\rm \rho = \dfrac{2.1\times 10^{-22}\ g}{\dfrac 43 \times \pi \times (1.3\times 10^{-13})^3\times 127}=1.79\times 10^{14}\ g/cm^3.

On comparing with the density of the solid iodine,

\rm \dfrac{Density\ of\ ^{127}I}{Density\ of\ the\ solid\ iodine}=\dfrac{1.79\times 10^{14}\ g/cm^3}{4.93\ g/cm^3}=3.63\times 10^{13}.\\\\\Rightarrow Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.

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Answer:

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Explanation:

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