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Naddika [18.5K]
2 years ago
8

Find the radioactivity of a 1 g sample of 226Ra given that

a1" title="t_{1/2}=1620" alt="t_{1/2}=1620" align="absmiddle" class="latex-formula"> years and Avogadro's number = 6.023 × 10^{23}.
Physics
1 answer:
dangina [55]2 years ago
6 0

Answer:

Explanation:

No of atoms of Ra in 1 g of sample = 6.023 x 10²³ / 226

N = 2.66 x 10²¹

disintegration constant λ = .693 / half life

half life = 1620 x 365 x 60 x 60 x 24 = 5.1 x 10¹⁰ s

disintegration constant λ = .693 / 5.1 x 10¹⁰

radioactivity dn / dt = λN

= (.693 / 5.1 x 10¹⁰ )  x 2.66 x 10²¹

= .3614 x 10¹¹ per sec

= 3.614 x 10¹⁰ / s

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a 1/4 watt resistor or a 20 watt amplifier.

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3 years ago
Under certain circumstances, potassium ions (K+) in a cell will move across the cell membrane from the inside to the outside. Th
choli [55]

Answer:

1.368\times 10^{-20}\ J

Explanation:

q = Charge in the potassium ion = 19e-18e

e = Charge of electron = 1.6\times 10^{-19}\ C

V_2-V_1 = Change in potential = 0-(-85.5\times 10^{-3})

Change in electric potential is given by

E=q(V_2-V_1)\\\Rightarrow E=(19e-18e)(0-(-85.5\times 10^{-3})\\\Rightarrow E=1.6\times 10^{-19}\times 85.5\times 10^{-3}\\\Rightarrow E=1.368\times 10^{-20}\ J

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3 0
3 years ago
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A 250 g object is hung onto a spring. It stretches 18 cm. Find the spring's spring constant
Juliette [100K]
<h2>Answer: 13.61 N/m</h2>

Hooke's law establishes that the elongation of a spring is directly proportional to the modulus of the force F applied to it, <u>as long as the spring is not permanently deformed</u>:

F=k (x-x_{o})    (1)

Where:

k is the elastic constant of the spring. The higher its value, the more work it will cost to stretch the spring.

x_{o} is the length of the spring without applying force.

x is the length of the spring with the force applied.

According to this, we have a spring where only the force due gravity is applied.

In other words, the force applied is the weigth W of the block:

W=m.g   (2)

Where m=250g=0.25kg is the mass of the block and g=9.8\frac{m}{s^{2}}  is the gravity acceleration.

W=(0.25kg)(9.8\frac{m}{s^{2}})   (3)

W=2.45N   (4)

Knowing the force applied W and x=18cm=0.18m and x_{o}=0, we can substitute the values in equation (1) and find k:

W=k (x-x_{o})    (5)

2.45N=k (0.18m-0m)    (6)

<u>Finally:</u>

k=13.61\frac{N}{m}  

4 0
3 years ago
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