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2 years ago

WILL MARK BRAINLIEST! PLS ANSWER ASAP

Physics

2 answers:

irakobra [83]2 years ago

8 0

Answer:

It has sunlight, oxygen, water, and fertile soil.

Explanation:

Sunlight: provides plants (producers) with energy so that the ecosystem is sustainable. Also provides heat and warmth so the planet does not freeze.

Oxygen: obvious. required for human survival and survival of most carbon-based organisms.

Water: required for plant growth and sustainability of carbon-based life.

Fertile soil: a place where plants can grow and produce energy.

I will not write an entire description because your teacher may accuse you of plagiarism and use this topic as proof.

Montano1993 [528]2 years ago

5 0

Answer

What makes up a perfect planet? It is the right distance from the Sun, it is protected from harmful solar radiation by its magnetic field, it is kept warm by an insulating atmosphere, and it has the right chemical ingredients for life, including water and carbon. Proportionate Ozone Layer and Light amount. According to the panspermia hypothesis, microscopic life—distributed by meteoroids, asteroids and other small Solar System bodies—may exist throughout the Universe. This is the perfect planet. In the end a perfect planet includes SUSTAINBLE DEVELOPEMNT IN EVERY ASPECT OF LIFE!

Explanation:

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A freshly prepared sample of radioactive isotope has an activity of 10 mCi. After 4 hours, its activity is 8 mCi. Find: (a) the

Maurinko [17]

Answer:

(a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

Explanation:

Given that,

Activity $R_{0}=10\ mCi$

Time $t_{1}=4\ hours$

Activity R= 8 mCi

(a). We need to calculate the decay constant

Using formula of activity

$R=R_{0}e^{-\lambda t}$

$\lambda=\dfrac{1}{t}ln(\dfrac{R_{0}}{R})$

Put the value into the formula

$\lambda=\dfrac{1}{4\times3600}ln(\dfrac{10}{8})$

$\lambda=0.0000154\ s^{-1}$

$\lambda=1.55\times10^{-5}\ s^{-1}$

We need to calculate the half life

Using formula of half life

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}$

Put the value into the formula

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{1.55\times10^{-5}}$

$T_{\dfrac{1}{2}}=44.719\times10^{3}\ s$

$T_{\dfrac{1}{2}}=11.3\ hr$

(b). We need to calculate the value of N₀

Using formula of $N_{0}$

$N_{0}=\dfrac{3.70\times10^{6}}{\lambda}$

Put the value into the formula

$N_{0}=\dfrac{3.70\times10^{6}}{1.55\times10^{-5}}$

$N_{0}=2.38\times10^{11}\ nuclei$

(c). We need to calculate the sample's activity

Using formula of activity

$R=R_{0}e^{-\lambda\times t}$

Put the value intyo the formula

$R=10e^{-(1.55\times10^{-5}\times30\times3600)}$

$R=1.87\ mCi$

Hence, (a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

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SIZIF [17.4K]

I'm not sure I completely understand the expression you want evaluated.

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2 years ago

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Darya [45]

Answer: