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soldi70 [24.7K]
4 years ago
11

With radiocarbon dating scintists compare an object carbon 14 levels with .....

Physics
1 answer:
Ilia_Sergeevich [38]4 years ago
5 0
<h2>Answer:</h2>

<u>With radiocarbon dating scientists compare an object carbon 14 levels with </u><u>the fossil or rock for which the age measurement is required</u>

<h2>Explanation:</h2>

Radiocarbon, or carbon 14, is an isotope of the element carbon that is unstable and weakly radioactive. The carbon-14 method was developed by the American physicist Willard F. Libby about 1946. It can be used to determine the age of a rock or a fossil by comparing the specimen of the required or fossil and compared it with the carbon 14 sample. Carbon 14 decays at constant rate therefore an estimate of the date at which an organism died can be made by measuring the amount of its residual radiocarbon and comparing it with Carbon 14.

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A 1.2 L weather balloon on the ground has a temperature of 25°C and is at atmospheric pressure (1.0 atm). When it rises to an el
Irina-Kira [14]

Answer:

71.19 C

Explanation:

25C = 25 + 273 = 298 K

Applying the ideal gas equation we have

\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

where P, V and T are the pressure, volume and temperature of the gas at 1st and 2nd stage, respectively. We can solve for the temperature and the 2nd stage:

T_2 = T_1\frac{P_2V_2}{P_1V_1} = 298\frac{0.77*1.8}{1.2*1} = 298*1.155 = 344.19 K = 344.19 - 273 = 71.19 C

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4 years ago
A scientist records the motion of an object. She writes down that the
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The answer is Velocity
3 0
3 years ago
Brainliest please help<br><br>tell me if am right <br>if not correct me <br><br><br>​please
REY [17]

Answer:

See the answers below

Explanation:

To solve this problem we must use the following equation of kinematics.

v_{f}=v_{o}+a*t

where:

Vf = final velocity [m/s]

Vo = initial velocity [m/s]

a = acceleration [m/s²]

t = time [s]

<u>First case</u>

Vf = 6 [m/s]

Vo = 2 [m/s]

t = 2 [s]

6=2+a*2\\4=2*a\\a=2[m/s^{2} ]

<u>Second case</u>

Vf = 25 [m/s]

Vo = 5 [m/s]

a = 2 [m/s²]

25=5+2*t\\t = 10 [s]

<u>Third case</u>

Vo =4 [m/s]

a = 10 [m/s²]

t = 2 [s]

v_{f}=4+10*2\\v_{f}=24 [m/s]

<u>Fourth Case</u>

Vf = final velocity [m/s]

Vo = initial velocity [m/s]

a = acceleration [m/s²]

t = time [s]

v_{f}=5+8*10\\v_{f}=85 [m/s]

<u>Fifth case</u>

Vf = final velocity [m/s]

Vo = initial velocity [m/s]

a = acceleration [m/s²]

t = time [s]

8=v_{o}+4*2\\v_{0}=8-8\\v_{o}=0

8 0
3 years ago
A type of light bulb is labeled having an average lifetime of 1000 hours. It’s reasonable to model the probability of failure of
SashulF [63]

Answer:

0.2592 \ or \ 25.92\%

Explanation:

The exponential density function is given as

f(t)=\left \{ {{0} \atop {ce^{ct}}} \right\\0,t

\mu=\frac{1}{c}\\c=\frac{1}{\mu}\\\\=\frac{1}{1000}=0.001\\\\f(t)=0.001e^{-0.001t}

To find probability that bulb fails with the first 300hrs, we integrate from o to 300:

P(0\leq X\leq 300)=\int\limits^{300}_0 {f(t)} \, dt\\\\=\int\limits^{300}_0 {0.001e^{-001t}} \, dt\\ =|-e^{-0.001t}|  \ 0\leq t\leq 300

P(0\leq X\leq 300)=-0.7408+1\\=0.2592

Hence probability of bulb failing within 300hrs is 25.92% or 0.2592

3 0
3 years ago
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