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3 years ago

What is the simplest relationship between the angular wavenumber k and just one of the other kinematic variables?

Physics

1 answer:

Inessa [10]3 years ago

6 0

Answer : $k=\dfrac{\omega}{k}$

Explanation :

The word kinematics means the study of motion. Kinematic variables gives the description of the motion of the body.

Displacement, Velocity, acceleration and time are associated with the motion of particle.

Wavenumber is defined as the frequency of wave, which is measured in cycles per unit distance.

Wave number is also defines as:

$k=\dfrac{2\pi}{\lambda}$

where, $\lambda$ is wavelength

$k=\dfrac{2\pi}{\lambda}\times \dfrac{T}{T}$

since, $\dfrac{2\pi}{T}=\omega \ and\ \dfrac{1}{V}=\dfrac{T}{\lambda}$

So, $k=\dfrac{\omega}{V}$

Where, $\omega$ is angular frequency and $V$ is velocity of wave.

Hence, the relationship between the angular wavenumber k and and kinamatic variable V is

$k=\dfrac{\omega}{V}$

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3 years ago

A tree falls in a forest. How many years must pass before the 14C activity in 1.03 g of the tree's carbon drops to 1.02 decay pe

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Answer:

t = 5.59x10⁴ y

Explanation:

To calculate the time for the ¹⁴C drops to 1.02 decays/h, we need to use the next equation:

$A_{t} = A_{0}\cdot e^{- \lambda t}$ (1)

where $A_{t}$ : is the number of decays with time, A₀: is the initial activity, λ: is the decay constant and t: is the time.

To find A₀ we can use the following equation:

$A_{0} = N_{0} \lambda$ (2)

where N₀: is the initial number of particles of ¹⁴C in the 1.03g of the trees carbon

From equation (2), the N₀ of the ¹⁴C in the trees carbon can be calculated as follows:

$N_{0} = \frac{m_{T} \cdot N_{A} \cdot abundance}{m_{^{12}C}}$

where $m_{T}$ : is the tree's carbon mass, $N_{A}$ : is the Avogadro's number and $m_{^{12}C}$ : is the ¹²C mass.

$N_{0} = \frac{1.03g \cdot 6.022\cdot 10^{23} \cdot 1.3\cdot 10^{-12}}{12} = 6.72 \cdot 10^{10} atoms ^{14}C$

Similarly, from equation (2) λ is:

$\lambda = \frac{Ln(2)}{t_{1/2}}$

where t 1/2: is the half-life of ¹⁴C= 5700 years

$\lambda = \frac{Ln(2)}{5700y} = 1.22 \cdot 10^{-4} y^{-1}$

So, the initial activity A₀ is:

$A_{0} = 6.72 \cdot 10^{10} \cdot 1.22 \cdot 10^{-4} = 8.20 \cdot 10^{6} decays/y$

Finally, we can calculate the time from equation (1):

$t = - \frac{Ln(A_{t}/A_{0})}{\lambda} = - \frac {Ln(\frac{1.02decays \cdot 24h \cdot 365d}{1h\cdot 1d \cdot 1y \cdot 8.20 \cdot 10^{6} decays/y})}{1.22 \cdot 10^{-4} y^{-1}} = 5.59 \cdot 10^{4} y$