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3 years ago

15

Suppose a scientist doing an experiment in nuclear fusion starts out with twelve atoms. How many atoms will most likely result f

rom the experiment

2 answers:

Natasha_Volkova [10]3 years ago

7 0

six, because atoms combine during nuclear fusion

ddd [48]3 years ago

3 0

<span>So we wan't to know what happens during nuclear fusion where we have 12 atoms. Nuclear fusion is a nuclear process where 2 or more nuclei will form 1 or more different nuclei with a release of energy. So if we start with 12 atoms, the most likely result will be 6 atoms. </span>

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State the law of conservation of energy

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Two test tubes are filled with a solution of bromthymol blue. A student exhales through a straw into each tube, and the bromthym

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3 years ago

A tin can has a volume of 1100 cm³ and a mass of 80 g. Approximately how many grams of lead shot can it carry without sinking in

Kruka [31]

Answer:

1020g

Explanation:

Volume of can= $1100cm^3=1100\times 10^{-6}m^3$

$1cm^3=10^{-6}m^3$

Mass of can=80g= $\frac{80}{1000}=0.08kg$

1Kg=1000g

Density of lead= $11.4g/cm^3=11.4\times 10^{3}=11400kg/m^3$

By using $1g/cm^3=10^3kg/m^3$

We have to find the mass of lead which shot can it carry without sinking in water.

Before sinking the can and lead inside it they are floating in the water.

Buoyancy force = $F_b=Weight of can+weight of lead$

$\rho_wV_cg=m_cg+m_lg$

Where $\rho_w=10^3kg/m^3=$ Density of water

$m_c=$ Mass of can

$m_l=$ Mass of lead

$V_c=$ Volume of can

Substitute the values then we get

$1000\times 1100\times 10^{-6}=0.08+m_l$

$1.1-0.08=m_l$

$m_l=1.02 kg=1.02\times 1000=1020g$

$1 kg=1000g$

Hence, 1020 grams of lead shot can it carry without sinking water.

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3 years ago

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3 years ago

Read 2 more answers

A cosmic ray (an electron or nucleus moving ar speeds close to the speed of light) travels across the Milky Way at a speed of 0.

Fiesta28 [93]

Answer:

Cosmic ray's frame of reference: 99,875 years

Stationary frame of reference: 501,891 years

Explanation:

First of all, we convert the distance from parsec into metres:

$d=30,000 pc =9.26\cdot 10^{20} m$

The speed of the cosmic ray is

$v=0.98 c$

where

$c=3.0 \cdot 10^8 m/s$ is the speed of light. Substituting,

$v=(0.98)(3.0\cdot 10^8)=2.94\cdot 10^8 m/s$

And so, the time taken to complete the journey in the cosmic's ray frame of reference (called proper time) is:

$T_0 = \frac{d}{v}=\frac{9.26\cdot 10^{20}}{2.94\cdot 10^8}=3.15\cdot 10^{12} s$

Converting into years,

$T_0 = \frac{3.15\cdot 10^{12}}{(365\cdot 24\cdot 60 \cdot 60}=99,875 years$

Instead, the time elapsed in the stationary frame of reference is given by Lorentz transformation:

$T=\frac{T_0}{\sqrt{1-(\frac{v}{c^2})^2}}$

And substituting v = 0.98c, we find:

$T=\frac{99,875}{\sqrt{1-(\frac{0.98c}{c})^2}}=501,891 years$

3 0

3 years ago