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uysha [10]
2 years ago
14

Use the worked example above to help you solve this problem. The half-life of the radioactive nucleus _(88)^(226)text(Ra) is 1.6

103 yr. If a sample initially contains 4.00 1016 such nuclei, determine the following:________.
(a) the initial activity in curies µCi
(b) the number of radium nuclei remaining after 4.4 103 yr nuclei
(c) the activity at this later time µCi
Physics
1 answer:
vfiekz [6]2 years ago
8 0

Answer:

Explanation:

From the information given:

The half-life t_{1/2} = 1.6103 years

The no. of the initial nuclei N_o = 4.00 \times 10^6

Using the formula:

N = N_o exp(-\lambda t)

where;

decay constant \lambda = \dfrac{In2}{1.6*10^3} y^{-1}

∴

N = N_o exp ( \dfrac{-In2}{1.6*10^3}\times 4.4 \times 10^3)

N = N_o exp (- 1.906154747)

The number of radium nuclei N = 5.94 × 10¹⁵

The initial activityA_o = \lambda N_o

A_o =(\dfrac{In (2)}{1.61\times 10^3  \times 365 \times 24 \times 3600}\times 4.00 \times 10^{16})

A_o =546075.8487 \ Bq

Since;

1 curie = 3.7 × 10¹⁰ Bq

Then;

A_o =\dfrac{546075.8487 }{3.7\times 10^{10}}

A_o = 1.47588 \times 10^{-5}Ci

A_o = 14.7588 \  \mu Ci

c) The activity at a later time is:

=5.94 \times 10^{15}( \dfrac{In (2)}{1.60 \times 10^3 \times 365\times 24 \times 3600})

= 81599.09018 \ Bq \\ \\ = \dfrac{81599.09018}{3.7\times 10^{10}} \ Ci \\ \\ = 2.20538 \times 10^6 \ Ci  \\ \\  = 2.20538  \ \mu Ci

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Is this right? plzz anwser soon
-Dominant- [34]

Answer:

yes

Explanation:

6 0
2 years ago
Read 2 more answers
A light spring stretches 0.13 m when a 0.35 kg mass is hung from it. The mass is pulled down from this equilibrium position an a
Alenkasestr [34]

Answer:

v = 1.30 m/s

Explanation:

given,

mass hung = 0.35 Kg

spring stretched when load is hanged  (x)= 0.13 m

now,

weight of the mass attached = Kx

             m g = k x

             0.35 x 9.8 = k x 0.13

                k = 26.38 N/m

now, using conservation of energy

 \dfrac{1}{2}mv^2 = \dfrac{1}{2}kx'^2

 v = \sqrt{\dfrac{kx'^2}{m}}

 v = \sqrt{\dfrac{26.38 \times 0.15^2}{0.35}}

 v = \sqrt{1.6958}

    v = 1.30 m/s

6 0
3 years ago
The current in a series circuit is 19.3 A. When an additional 7.40-Ω resistor is inserted in series, the current drops to 13.4 A
Bogdan [553]

Answer:

16.8ohms

Explanation:

According to ohm's law which states that the current passing through a metallic conductor at constant temperature is directly proportional to the potential difference across its ends.

Mathematically, V = IRt where;

V is the voltage across the circuit

I is the current

R is the effective resistance

For a series connected circuit, same current but different voltage flows through the resistors.

If the initial current in a circuit is 19.3A,

V = 19.3R... (1)

When additional resistance of 7.4-Ω is added and current drops to 13.4A, our voltage in the circuit becomes;

V = 13.4(7.4+R)... (2)

Note that the initial resistance is added to the additional resistance because they are connected in series.

Equating the two value of the voltages i.e equation 1 and 2 to get the resistance in the original circuit we will have;

19.3R = 13.4(7.4+R)

19.3R = 99.16+13.4R

19.3R-13.4R = 99.16

5.9R = 99.16

R= 99.16/5.9

R = 16.8ohms

The resistance in the original circuit will be 16.8ohms

5 0
3 years ago
A 60kg bicyclist (including the bicycle) is pedaling to the
Fittoniya [83]

a) 4 forces

b) 186 N

c) 246 N

Explanation:

a)

Let's count the forces acting on the bicylist:

1) Weight (W=mg): this is the gravitational force exerted on the bicyclist by the Earth, which pulls the bicyclist towards the Earth's centre; so, this force acts downward (m = mass of the bicyclist, g = acceleration due to gravity)

2) Normal reaction (N): this is the reaction force exerted by the road on the bicyclist. This force acts vertically upward, and it balances the weight, so its magnitude is equal to the weight of the bicyclist, and its direction is opposite

3) Applied force (F_A): this is the force exerted by the bicylicist to push the bike forward. Its direction is forward

4) Air drag (R): this is the force exerted by the air on the bicyclist and resisting the motion of the bike; its direction is opposite to the motion of the bike, so it is in the backward direction

So, we have 4 forces in total.

b)

Here we can find the net force on the bicyclist by using Newton's second law of motion, which states that the net force acting on a body is equal to the product between the mass of the body and its acceleration:

F_{net}=ma

where

F_{net} is the net force

m is the mass of the body

a is its acceleration

In this problem we have:

m = 60 kg is the mass of the bicyclist

a=3.1 m/s^2 is its acceleration

Substituting, we find the net force on the bicyclist:

F_{net}=(60)(3.1)=186 N

c)

We can write the net force acting on the bicyclist in the horizontal direction as the resultant of the two forces acting along this direction, so:

F_{net}=F_a-R

where:

F_{net} is the net force

F_a is the applied force (forward)

R is the air drag (backward)

In this problem we have:

F_{net}=186 N is the net force (found in part b)

R=60 N is the magnitude of the air drag

Solving for F_a, we find the force produced by the bicyclist while pedaling:

F_a=F_{net}+R=186+60=246 N

3 0
2 years ago
A heat engine with 0.300 mol of a monatomic ideal gas initially fills a 1000 cm3 cylinder at 500 K . The gas goes through the fo
LuckyWell [14K]

Complete Question:

A heat engine with 0.300 mol of a monatomic ideal gas initially fills a 1000 cm3 cylinder at 500 K . The gas goes through the following closed cycle: - Isothermal expansion to 5000 cm3. - Isochoric cooling to 400 K . - Isothermal compression to 1000 cm3. - Isochoric heating to 500 K .

a) what is the work for one cycle

b) what is the thermal efficiency

Answer:

a) Work done for 1 cycle = 402.13

b) Thermal efficiency = 20%

Explanation:

Number of moles, n = 0.300 mol

Initial Volume, V₁ = 1000 cm³

Temperature, T = 500 K

Isothermal expansion to 5000 cm³

Final volume, V₂ = 5000 cm³

R = 8.314 J/ mol.K

Work done, W = nRT ln(V₂/V₁)

W = (0.3 * 8.314 * 500) * ln(5000/1000)

W = 1247.1 * ln5

W₁ = 2007.13 J

Isochoric cooling

In an Isochoric process, volume is constant i.e. V₂ = V₁ = V

W = nRT ln(V/V)

But  ln(V/V) = ln 1 = 0

Work done, W₂ = 0 Joules

Isothermal Compression to 1000 cm³

V₂ = 1000 cm³

V₁ = 5000 cm³

W = nRT ln(V₂/V₁)

W = 0.3 * 8.314 * 400 ln(1000/5000)

W₃ = -1605 J

Isochoric heating to 500 K

Since there is no change in volume, no work is done

W₄ = 0 J

a) Work done for 1 cycle

W = W₁ + W₂ + W₃ + W₄

W = 2007.13 + 0 + 0 -1605+0

W = 402.13 Joules

b) Thermal efficiency

Thermal efficiency = (Net workdone for 1 cycle)/(Heat absorbed)

Heat absorbed = Work done due to thermal expansion = 2007.13 J

Thermal efficiency = 402.13/2007.13

Thermal efficiency = 0.2

Thermal efficiency = 0.2 * 100% = 20 %

3 0
3 years ago
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