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3 years ago

Testing shows that a sample of wood from an artifact contains 25% of the

Physics

2 answers:

daser333 [38]3 years ago

5 0

Answer:

The artifact is 11,460 years old.

<u>Explanation</u>:

Given that,

The half life of the carbon-14 is 5730 years and we are left with 255 of the sample of wood from an arti-fact.

So it takes 5730 years for the sample to reduce into half

Initially there will be 100% of the sample so

after first 5730 years, the sample reduces into 50% percent

Now the left 50% sample will take another 5730 years to decay into half of its amount.

after next 5730 years the sample reduces into 25% percent

So totally after 2 half-life the sample reduces into 25%

That is (5730 +5730) years = 11460 years

Umnica [9.8K]3 years ago

3 0

Answer:5730 years

Explanation:

A P E X

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How much work is done on a small car if a 3150 N force is exerted to move it 75.5 m to the side of the road

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3 years ago

An ideal monatomic gas at 275 K expands adiabatically and reversibly to six times its volume. What is its final temperature (in

Gwar [14]

The final temperature is 83 K.

<u>Explanation</u>:

For an adiabatic process,

$T {V}^{\gamma - 1} = \text{constant}$

$\cfrac{{T}_{2}}{{T}_{1}} = {\left( \cfrac{{V}_{1}}{{V}_{2}} \right)}^{\gamma - 1}$

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3 years ago

A cannon, positioned on a hill, shoots a cannonball horizontally at 23 m/s. The cannonball hits the stone wall 1.96 m below the

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Answer: 14. 49 m

Explanation:

We can solve this problem with the following equations:

$x=V_{o} cos \theta t$ (1)

$y-y_{o}=V_{o} sin \theta t-\frac{1}{2}gt^{2}$ (2)

Where:

$x$ is the horizontal distance between the cannon and the ball

$V_{o}=23 m/s$ is the cannonball initial velocity

$\theta=0\°$ since the cannonball was shoot horizontally

$t$ is the time

$y=0$ is the final height of the cannonball

$y_{o}=1.96 m$ is the initial height of the cannonball

$g=9.8 m/s^{2}$ is the acceleration due gravity

Isolating $t$ from (2):

$t=\sqrt{-\frac{2(y-y_{o})}{g}}$ (3)

$t=\sqrt{-\frac{2(0 m-1.96 m)}{9.8 m/s^{2}}}$ (4)

$t=0.63 s$ (5)

Substituting (5) in (1):

$x=(23 m/s) cos(0\°) 0.63 s$ (6)

Finally:

$x=14.49 m$

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