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Irina18 [472]
3 years ago
11

Q1. After three half-lives of an isotope, 1 billion of the original isotope's atoms still remain in a certain amount of this ele

ment. How many atoms of the daughter product would you expect to be present?
Q2. By measuring the amounts of parent isotope and daughter product in the minerals contained in a rock, and by knowing the half-life of the parent isotope, a geologist can calculate the absolute age of the rock. A rock contains 125 g of a radioisotope with a half-life of 150 000 years and 875 g of its daughter product. How old is the rock according to the radiometric dating method?
Physics
1 answer:
77julia77 [94]3 years ago
5 0
 <span>If 1 eighth equals 1 billion 7 eighth equals 7 billion. 

The asker of the second question needs a tutorial in radiometric dating. There is little likelihood that the daughter isotope has the same atomic weight as the parent isotope. To measure the mass isotopes doesn't tell us how many atoms of each exist. To get around that let's pretend — which will likely serve the purpose ineptly intended — that the values give an the particle ratio, 125:875. 

The original parent isotope count was 125 + 875 = 1000. The remaining parent isotope is 125/1000 or 1/8. 1/8 = (1/2)^h, where h is the number of half-lives. 

h = log (1/8) ÷ log(1/2) = 3 

And 3 half-lives • 150,000 years/half-life = 450,000 years.</span>
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3 years ago
A car traveling at 30 m/s drives off a cliff that is 50 meters high? How far away does it land?
Semenov [28]

Answer:

The maximum range R_{max}= 132. 72 m

Explanation:

Given,

The initial velocity of the car, u = 30 m/s

The height of the cliff, h = 50 m

Let the car drives off the cliff with a horizontal velocity of 30 m/s.

The formula for a projectile that is projected from a height h from the ground is given by the relation

                                R_{max}= \frac{u}{g}\sqrt{u^{2} + 2gh }  m

Where,

                          g - acceleration due to gravity

Substituting the values in the above equation

                   R_{max}= \frac{30}{9.8}\sqrt{30^{2} + 2X9.8X50 }  

                                          = 132.72  m

Hence, the car lands at a distance, R_{max}= 132. 72 m            

3 0
3 years ago
A 3.63.kgkg chihuahua charges at a speed of 3.3m/s3.3m/s. What is the magnitude of the average force needed to bring the chihuah
sergiy2304 [10]

Answer:

23.96 N

Explanation:

From the question given above, the following data were obtained:

Mass of Chihuahua (m) = 3.63 kg

Velocity (v) = 3.3m/s

Time (t) = 0.50 s

Force (F) =?

Next, we shall determine the acceleration of the Chihuahua. This can be obtained as follow:

Velocity (v) = 3.3m/s

Time (t) = 0.50 s

Acceleration (a) =?

a = v/t

a = 3.3/0.5

a = 6.6 m/s²

Thus, the acceleration of the Chihuahua is 6.6 m/s².

Finally, we shall determine the force need to stop the Chihuahua as shown below:

Mass of Chihuahua (m) = 3.63 kg

Acceleration (a) = 6.6 m/s².

Force (F) =?

F = ma

F = 3.63 × 6.6

F = 23.96 N

Therefore, a force of 23.96 N is needed to stop the Chihuahua.

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3 years ago
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In the diagram, the arrow shows the movement of electric charges through a wire connected to a battery.
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Due to difference in electric potential.

<h3>What is Potential difference?</h3>

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Here , the potential difference is the difference in electric potential between two charged substances.

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