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solniwko [45]
3 years ago
8

The distance versus this plot for a particular object shows a quadratic relationship. Which column of distance data is possible

for this solution?​

Physics
1 answer:
SOVA2 [1]3 years ago
5 0

Answer:

A

Explanation:

In Column A, the pattern is 1², 2², 3², etc.  So x(t) = t², which is quadratic.

In Column B, the pattern is 2*1, 2*2, 2*3, etc.  So x(t) = 2t, which is linear.

In Column C, the pattern is 9*1, 9*2, 9*3, etc.  So x(t) = 9t, which is linear.

In Column D, the pattern is 1/1, 1/2, 1/3, etc.  So x(t) = 1/t, which is reciprocal.

In Column E, the pattern is 1/1², 1/2², 1/3², etc.  So x(t) = 1/t², which is reciprocal.

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The buoyant force won't change.
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8. Three grams of Bismuth-218 decay to 0.375 grams in one hour. What is the half-
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Explanation:

This problem can be solved using the <u>Radioactive Half Life Formula</u>:  

A=A_{o}.2^{\frac{-t}{H}} (1)  

Where:  

A=0.375 g is the final amount of the material  

A_{o}=3 g is the initial amount of the material  

t=1 h is the time elapsed  

H is the half life of the material (the quantity we are asked to find)  

Knowing this, let's substitute the values and find h from (1):

0.375 g=(3 g)2^{\frac{-1h}{H}} (2)  

\frac{0.375 g}{3 g}=2^{\frac{-1h}{H}} (3)  

Applying natural logarithm in both sides:

ln(\frac{0.375 g}{3 g})=ln(2^{\frac{-1 h}{H}}) (4)  

-2.079=-\frac{1 h}{H}ln(2) (5)  

Clearing H:

H=\frac{-1h}{-2.079}(0.693) (6)  

Finally:

h=0.333 h This is the half-life of the Bismuth-218 isotope

4 0
3 years ago
Which would fall with a greater acceleration in a vacuum-a leaf or a stone?
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Given the following data;

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Substituting into the equation, we have;

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