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Ede4ka [16]
3 years ago
5

A radioactive nucleus has a half-life of 5 x 108 years. Assuming that a sample of rock (say, in an asteroid) solidified right af

ter the solar system formed, approximately what fraction of the radioactive element should be left in the rock today?
Physics
1 answer:
Marina86 [1]3 years ago
5 0

Answer:

So the present mass will be 0.0017002 times of the initial mass of the sample.

Explanation:

Given:

  • half life of the given sample, t'=5\times 10^8\ yrs
  • We know that solar system came in to existence approximately 4.6\times 10^{9} years ago.

<u>Let the initial quantity on the formation of the solar system be, </u>m_o<u />

Then the final quantity at present be:

m=m_o\times (\frac{1}{2})^{(t/t')}

m=m_o\times (\frac{1}{2} )^{4.6\times 10^9\div(5\times 10^8)}

m=m_o\times 0.0017002

So the present mass will be 0.0017002 times of the initial mass of the sample.

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What type of rays would you expect to be used frequently at a hospital to make medical diagnoses?
Bess [88]
X rays because to see your bones
7 0
3 years ago
Given that the concentration of bovine carbonic anhydrase is 3.3 pmol ⋅ L − 1 and R max ( V max ) = 222 μmol ⋅ L − 1 ⋅ s − 1 , d
LuckyWell [14K]

Answer:

The turnover number of the enzyme molecule bovine carbonic anhydrase = 67,272,727.27 s^–1.

Explanation:

Given:

The concentration of bovine carbonic anhydrase = total enzyme concentration = Et = 3.3 pmol⋅L^–1 = 3.3 × 10^–12 mol.L^–1

The maximum rate of reaction = Rmax (Vmax) = 222 μmol⋅L^–1⋅s^–1 = 222 × 10^–6 mol.L^–1⋅s^–1

The formula for the turnover number of an enzyme (kcat, or catalytic rate constant) = Rmax ÷ Et = 222 × 10^–6 mol.L^–1⋅s^–1 ÷ 3.3 × 10^–12 mol.L^–1 = 67,272,727.27 s^–1

Therefore, the turnover number of the enzyme molecule bovine carbonic anhydrase = 67,272,727.27 s^–1

3 0
3 years ago
Lus
patriot [66]

Answer:

hey but the person at the top is right

7 0
3 years ago
A punter drops a 2.0 kg ball from rest vertically 1 meter down onto his foot. The ball leaves the foot with a speed of 18 m/s at
pav-90 [236]

Answer:

Explanation:

Given

mass of drop m=2 kg

height of fall h=1 m

ball leaves the foot with a speed of 18 m/s at an angle of 55^{\circ}

Velocity of ball just before the collision with the floor

u^=2gh

u=\sqrt{2gh}

u=\sqrt{2\times 9.8\times 1}=4.42 m/s

Impulse delivered in Y direction

J_y=m(v\sin (55)-(-u))

J_y=2(18\sin (55)+4.42)

J_y=38.32 kg-m/s

Impulse in x direction

J_x=m\times v\cos (55)

J_x=2\times 4.42\cos (55)=20.646

J_{net}=\sqrt{J_x^2+J_y^2}

J_{net}=\sqrt{(38.32)^2+(20.64)^2}

J_{net}=43.52 kg-m/s

at an angle of \tan \phi =\frac{J_y}{J_x}=\frac{38.32}{20.64}

\phi =tan^{-1}(1.856)

\phi =61.7^{\circ}  

7 0
2 years ago
How long does it take a 22 kW steam engine to do 5.6 × 107 J of work? Answer in units of s.
Lena [83]
<h2>Time needed is 2545.45 seconds.</h2>

Explanation:

We know equation for power

               \texttt{Power = }\frac{\texttt{Work}}{\texttt{Time}}

Here we need to find time when for a 22 kW steam engine to do 5.6 × 10⁷ J of work.

So

        Work = 5.6 × 10⁷ J

        Power = 22 kW = 22 x 10³ W

Substituting

        \texttt{Power = }\frac{\texttt{Work}}{\texttt{Time}}\\\\22\times 10^3=\frac{5.6\times 10^7}{\texttt{Time}}\\\\\texttt{Time}=2545.45s

Time needed is 2545.45 seconds.

5 0
3 years ago
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