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3 years ago

A 59.3 kg diver jumps off a board

Physics

1 answer:

ser-zykov [4K]3 years ago

6 0

The board is 2.50m high.

Why?

We can calculate how high was the board applying the Law of Conservation of Mechanical Energy. This Law states that the mechanical energy (kinematic and potential) will be conserved during the motion.

It can be described with the following formula:

$E_{M_{1}}=E_{M_{2}}\\\\PE_{1}+KE_{1}=PE_{2}+KE_{2}$

$PE=m*g*h\\KE=\frac{1}{2}m*v^{2}$

At the top of the boar, the kinetic energy is equal to 0.

At the water, the potential energy is equal to 0.

So,

$PE_{1}=KE_{2}\\\\m*g*h=1450J\\\\59.3kg*9.8\frac{m}{s^{2}}*h=1450J\\ \\h=\frac{1450J}{59.3kg*9.8\frac{m}{s^{2}}}=2.50m$

Hence, we have that the board is 2.50m high.

Have a nice day!

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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

Illusion [34]

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):

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

I hope it helps you!

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Answer:

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