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4 years ago

Examine the graph showing the half-life of the radioactive isotope substance x.

Physics

1 answer:

kow [346]4 years ago

8 0

Answer:

T_{1/2} = 5,776 10³ years

We see this life time as it is very close to the life time of carbon 14, so it could be used for dating ancient objects

Explanation:

The radioactive decay of described by an equation of the form

N = N₀ $e^{-\lambda t}$

where N is the number of atoms present, N₀ is the number of initial atoms λ is the activity of the material.

The average life time is defined as the time for which the number of remainng atoms is N = N₀ / 2

$T_{1/2}$ = ln 2 /λ

With these expressions, the best method to determine the average life time is to find the activity of the first equation.

For this we look for a point on the graph as accurate as possible,

N₀ = 100, N = 30 and t = 10,000 years

we substitute in the equation

30 = 100 e^{-\lambda 10000}

ln 0.3 = - λ 10000

λ = - (ln 0.3) / 10000

λ = 1.2 10-4

now we can find the average life time

$T_{1/2}$ = ln 2 / 1,2 10-4

T_{1/2} = 5,776 10³ years

We see this life time as it is very close to the life time of carbon 14, so it could be used for dating ancient objects

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A common method to measure thermal conductivity of a biomaterial is to insert a long metallic probe axially into the center of a

tia_tia [17]

Answer:

The thermal conductivity of the biomaterial is approximately 1.571 watts per meter-Celsius.

Explanation:

Let suppose that thermal conduction is uniform and one-dimensional, the conduction heat transfer ( $\dot Q$ ), measured in watts, in the hollow cylinder is:

$\dot Q = \frac{2\cdot k\cdot L}{\ln \left(\frac{D_{o}}{D_{i}} \right)}\cdot (T_{i}-T_{o})$

Where:

$k$ - Thermal conductivity, measured in watts per meter-Celsius.

$L$ - Length of the cylinder, measured in meters.

$D_{i}$ - Inner diameter, measured in meters.

$D_{o}$ - Outer diameter, measured in meters.

$T_{i}$ - Temperature at inner surface, measured in Celsius.

$T_{o}$ - Temperature at outer surface, measured in Celsius.

Now we clear the thermal conductivity in the equation:

$k = \frac{\dot Q}{2\cdot L\cdot (T_{i}-T_{o})}\cdot \ln\left(\frac{D_{o}}{D_{i}} \right)$

If we know that $\dot Q = 40.8\,W$ , $L = 0.6\,m$ , $T_{i} = 50\,^{\circ}C$ , $T_{o} = 20\,^{\circ}C$ , $D_{i} = 0.01\,m$ and $D_{o} = 0.04\,m$ , the thermal conductivity of the biomaterial is:

$k = \left[\frac{40.8\,W}{2\cdot (0.6\,m)\cdot (50\,^{\circ}C-20\,^{\circ}C)}\right]\cdot \ln \left(\frac{0.04\,m}{0.01\,m} \right)$

$k \approx 1.571\,\frac{W}{m\cdot ^{\circ}C}$

The thermal conductivity of the biomaterial is approximately 1.571 watts per meter-Celsius.

8 0

3 years ago

If it takes Ashley 3 seconds to run from the batters box to first base at an average speed of 6.5 meters per second, what is the

Sidana [21]

19.5 meters because 6.5*3 is 19.5

5 0

3 years ago

Read 2 more answers

Which of the following statements about electric field lines due to static charges are true? (Select all that apply.)

anastassius [24]

Answer:

a. Electric field lines can never cross each other.

d. Wider spacing between electric field lines indicates a lower magnitude of electric field.

Explanation:

a. The electric field lines cannot be crossed, since this would mean that there would be more than one electric field vector for the same point at the place where the crossing occurs.

d. The space between the field lines is inversely proportional to the intensity of the electric field.

7 0

3 years ago

An alien spaceship streaks past Spanos football stadium along the direction of play at 0.8c. A football field is 120 yards long

Viktor [21]

Answer:

(i) Length of stadium, L₀ = 72 yards; width of stadium W = 55 yards (ii) 5 hours (iii) The people on earth. This is because they are not moving at relativistic speed relative to the game. (iv) The people on earth. This is because they are not moving at relativistic speed relative to the game.

Explanation:

(i) Given Length of stadium, L₀ = 120 yards, width of stadium, W₀ = 55 yards, speed of alien spaceship in direction of play, v = 0.8c

We use the equation for length contraction to determine the length of stadium L measured by the alien spaceship, since it is moving in the direction of game play and length contraction only occurs in the direction of motion. So, L = $\sqrt{1-\beta^{2} }$ L₀ where β = $\frac{v}{c}$ v= speed of alien spaceship and c= speed of light. So, β = $\frac{v}{c}$ = $\frac{0.8c}{c}$ = 0.8. Therefore, L = $\sqrt{1-0.8^{2} }$ L₀= $\sqrt{1-0.64}$ L₀= $\sqrt{0.36}$ L₀ = 0.6L₀ = 0.6 × 120 = 72 yards. So, the alien spaceship measures a length of 72 yards. The alien spaceship measures a width,W of 55 yards because there is no length contraction in the direction perpendicular to that of its motion. So, W = W₀ = 55 yards.

(ii) Since the game begins at 1:30 PM Pacific Standard Time (PST) and ends 4:30 PM Pacific Standard Time (PST), the proper time t₀, which is the duration of the event is 4:30 - 1:30 = 3 hours. The time measured by the alien spaceship t (since the time dilates)is given by t = t₀/ $\sqrt{1 - \beta ^{2} }$ . From (i) above β = 0.8 and t₀ = 3 hours. So, t = 3/ $\sqrt{1 - 0.8^{2} }$ = 3/0.6 = 5 hours.

(iii) The people in Spanos football stadium. This is because they are not moving at relativistic speed relative to the game. Since $\frac{v}{c}$ ≈ 0 then β ≈ 0 So, L = $\sqrt{1-\beta^{2} }$ L₀= So, L = $\sqrt{1-0}$ L₀thus L = L₀ for the spectators in spanos football field (iv) The people in Spanos football stadium. This is because they are not moving at relativistic speed relative to the game. Since $\frac{v}{c}$ ≈ 0 then β ≈ 0. So, t = t₀/ $\sqrt{1 - \beta ^{2} }$ = t₀/ $\sqrt{1-0}$ = t₀.Thus t = t₀ for the spectators in spanos football field.

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3 years ago

Multiply the following three numbers and report your answer to the correct number of significant figures: 0.020cm x 50cm x 11.1c

Neko [114]

Answer:11.1

Explanation:

Three significant figures

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3 years ago

Examine the graph showing the half-life of the radioactive isotope substance x￼.

Examine the graph showing the half-life of the radioactive isotope substance x.