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3 years ago

9

The rate at which light energy is radiated from a source is measured in which of the following units? A. Candela B. Angstrom C.

Lumen D. Lux

2 answers:

svet-max [94.6K]3 years ago

6 0

The correct answer is Lumen

Bumek [7]3 years ago

5 0

A. Candela

Tell me if I got it wright

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What affect does distance have on gravitation force

tankabanditka [31]

As distance increases gravitational force decreases
and as distance decreases gravitational force increases
Therefore, it is inversely proportion.

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Silver has a density of 10.5 g/cm³, and gold has a density of 19.3 g/cm³. Which would have a greater mass, 5 cm³ of silver or 5

Vedmedyk [2.9K]

Gold have a greater mass

<h3>Further explanation </h3>

Density is a quantity derived from the mass and volume

Density is the ratio of mass per unit volume

The unit of density can be expressed in g/cm³ or kg/m³

Density formula:

$\large {\boxed {\bold {\rho ~ = ~ \frac {m} {V}}}}$

ρ = density

m = mass

v = volume

mass of Silver :

$\tt mass=\rho\times V\\\\mass=10.5\times 5=52.5~g$

mass of Gold :

$\tt mass=\rho\times V\\\\mass=19.3\times 5=96.5~g$

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3 years ago

3. A sample of carbon atoms were chemically combined with a sample of oxygen atoms to yield the

NISA [10]

Your mom is the answer

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3 years ago

What is the energy released in the fission reaction 1 0 n + 235 92 U → 131 53 I + 89 39 Y + 16 1 0 n 0 1 n + 92 235 U → 53 131 I

mihalych1998 [28]

<u>Answer:</u> The energy released in the given nuclear reaction is 94.99 MeV.

<u>Explanation:</u>

For the given nuclear reaction:

$_{92}^{235}\textrm{U}+_0^1\textrm{n}\rightarrow _{53}^{131}\textrm{I}+_{39}^{89}\textrm{Y}+16_{0}^1\textrm{n}$

We are given:

Mass of $_{92}^{235}\textrm{U}$ = 235.043924 u

Mass of $_{0}^{1}\textrm{n}$ = 1.008665 u

Mass of $_{53}^{131}\textrm{I}$ = 130.9061246 u

Mass of $_{39}^{89}\textrm{Y}$ = 88.9058483 u u

To calculate the mass defect, we use the equation:

$\Delta m=\text{Mass of reactants}-\text{Mass of products}$

Putting values in above equation, we get:

$\Delta m=(m_U+m_{n})-(m_{I}+m_{Y}+16m_{n})\\\\\Delta m=(235.043924+1.008665)-(130.9061246+88.9058483+16(1.008665))=0.1019761u$

To calculate the energy released, we use the equation:

$E=\Delta mc^2\\E=(0.1019761u)\times c^2$

$E=(0.1019761u)\times (931.5MeV)$ (Conversion factor: $1u=931.5MeV/c^2$ )

$E=94.99MeV$

Hence, the energy released in the given nuclear reaction is 94.99 MeV.

3 0

3 years ago