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2 years ago

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A piece of charcoal used for cooking is found at the remains of an ancient campsite. A 0.96-kg sample of carbon from the wood ha

s an activity equal to 5.76 ✕ 10^2 decays per minute. Find the age of the charcoal. Hint: Living material has an activity of 15.0 decays/minute per gram of carbon present. yr

1 answer:

Nutka1998 [239]2 years ago

5 0

Answer:

t = 26541.55 years

Explanation:

let the age of charcoal is t

Activity of living material = 15 decays / min /g

Activity of living material = 15 * 1000 decays /min /kg

Acitivity of living material per 0.94 kg A = 0.94 *15 * 1000 decays / min

Activity of after time t is A ' = 5.76*10^2 decay per minute

half life of carbon = 5730 years

designtegration constant $\lambda = \frac{0.693}{5730}$

we know $A ' = A e^{-\lambda t}$

$5.76*10^2 = 0.94 * 15 * 1000 e^{-\frac{0.693}{5730}t}$

$0.040 = e^{-\frac{0.693}{5730}t}$

taking log on both side

$-3.21 = -\frac{0.693}{5730}t$

t = 26541.55 years

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A baseball player leads off the game and hits a long home run. The ball leaves the bat at an angle of 70.0 from the horizontal w

professor190 [17]

Answer: 211.059 m

Explanation:

We have the following data:

$\theta=70\°$ The angle at which the ball leaves the bat

$V_{o}=55 m/s$ The initial velocity of the ball

$g=-9.8 m/s^{2}$ The acceleration due gravity

We need to find how far (horizontally) the ball travels in the air: $x$

Firstly we need to know this velocity has two components:

<u>Horizontally:</u>

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

$V_{ox}=55 m/s cos(70\°)=18.811 m/s$ (2)

<u>Vertically:</u>

$V_{oy}=V_{o}sin \theta$ (3)

$V_{oy}=55 m/s sin(70\°)=51.683 m/s$ (4)

On the other hand, when we talk about parabolic movement (as in this situation) the ball reaches its maximum height just in the middle of this parabola, when $V=0$ and the time $t$ is half the time it takes the complete parabolic path.

So, if we use the following equation, we will find $t$ :

$V=V_{o}+gt=0$ (5)

Isolating $t$ :

$t=\frac{-V_{o}}{g}$ (6)

$t=\frac{-55 m/s}{-9.8 m/s^{2}}$ (7)

$t=5.61 s$ (8)

Now that we have the time it takes to the ball to travel half of is path, we can find the total time $T$ it takes the complete parabolic path, which is twice $t$ :

$T=2t=2(5.61 s)=11.22 s$ (9)

With this result in mind, we can finally calculate how far the ball travels in the air:

$x=V_{ox}T$ (10)

Substituting (2) and (9) in (10):

$x=(18.811 m/s)(11.22 s)$ (11)

Finally:

$x=211.059 m$

8 0

2 years ago

At what displacement of a sho is the energy half kinetic and half potential? what fraction of the total energy of a sho is kinet

expeople1 [14]

As we know that KE and PE is same at a given position

so we will have as a function of position given as

$KE = \frac{1}{2}m\omega^2(A^2 - x^2)$

also the PE is given as function of position as

$PE = \frac{1}{2}m\omega^2x^2$

now it is given that

KE = PE

now we will have

$\frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2x^2$

$A^2 - x^2 = x^2$

$2x^2 = A^2$

$x = \frac{A}{\sqrt2}$

so the position is 0.707 times of amplitude when KE and PE will be same

Part b)

KE of SHO at x = A/3

we can use the formula

$KE = \frac{1}{2}m\omega^2(A^2 - x^2)$

now to find the fraction of kinetic energy

$f = \frac{KE}{TE} = \frac{A^2 - x^2}{A^2}$

$f = \frac{A^2 - (\frac{A}{3})^2}{A^2}$

$f_k = \frac{8}{9}$

now since total energy is sum of KE and PE

so fraction of PE at the same position will be

$f_{PE} = 1 - f_k$

$f_{PE} = 1 - (8/9) = 1/9$

7 0

3 years ago

Which reading strategy helps you to read and understand a passage quickly by making use of your previous knowledge?

Molodets [167]

Taking notes,because you can read back on what you learned

4 0

3 years ago

1. When an individual suffers from anxiety severe enough to ____________, he might be suffering from an anxiety disorder. (Point

zheka24 [161]

The answers to these questions are:

1. <span>When an individual suffers from anxiety severe enough to DISRUPT HIS DAILY LIFE, he might be suffering from an anxiety disorder.

2. 19 MILLION </span><span>people in the United States suffer from anxiety disorders.

3. I</span><span>f a patient complains that whenever he has to get on a plane he begins to sweat, have chest pain, and nausea, he is most likely suffering from PANIC DISORDER.

4. THAT IT RUNS IN FAMILIES. A lot of researchers and medical professionals believe that anxiety may be hereditary.
</span><span>
5. ANXIETY </span><span>is a state of intense apprehension, uncertainty, and fear resulting from the anticipation of a threatening event or situation.

6. </span><span>A twin is MORE likely to suffer from an anxiety disorder if the other twin has some form of anxiety disorder.

7. </span><span>Generalized Anxiety Disorder (GAD) is characterized by PROLONGED AND EXCESSIVE WORRY EVEN WHEN THERE IS NO APPARENT CAUSE FOR CONCERN.

8. </span><span>Specific Phobia is characterized by EXAGGERATED OR IRRATIONAL FEAR OF OBJECTS OR SITUATIONS THAT ARE NOT GENERALLY HARMFUL.

9. </span>Post-Traumatic Stress Disorder (PTSD) is characterized by <span>FLASHBACKS OR NIGHTMARES.

10. </span>Obsessive-Compulsive Disorder is characterized by OBSESSIVE, IRRATIONAL THOUGHTS THAT INCREASE/INTENSIFY FEELINGS OF ANXIETY.

These answers are based from what I understand from various sources. Hope this helps. :)

8 0

3 years ago

Read 2 more answers

A grinding wheel with a moment of inertia of 2 kg-m2 has a 2.50 N-m torque applied to it. What is its final kinetic energy 10 se

grigory [225]

First you do 2.5 n-m / 2 kg-m2
This equals 1.25 rad/s^2

Then multiply rad/s^2 by 10 because of the amount of seconds
This equals 12.5 rad/s^2

Then do this following equation : (1/2)Iwf^2
Do (0.5) 12.5 * 12.5 * 2
This equals 156.25 J (meaning that the answer is C: 156 J)

6 0

3 years ago