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arsen [322]
3 years ago
6

The LR5 is the specialist submarine for underwater rescue. The average density of seawater is 1028 kg/ m3.

Physics
1 answer:
sladkih [1.3K]3 years ago
5 0

Answer:

P = 7196 [kPa]

Explanation:

We can solve this problem using the expression that defines the pressure depending on the height of water column.

P = dens*g*h

where:

dens = 1028 [kg/m^3]

g = 10 [m/s^2]

h = 700 [m]

Therefore:

P = 1028*10*700

P = 7196000 [Pa]

P = 7196 [kPa]

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If Earth's axis was "straight up and down" instead of tilted, then ...

<span>-- There would be no seasons.

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When you float an ice cube in water, you notice that 90% of it is submerged beneath the surface. Now suppose you put the same ic
Gwar [14]

Answer:

option (c)

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90% of the body is submerged in water.

Now it is immersed in an unknown liquid whose density is less than the density of water.

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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
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)

<em>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.</em>

To find A₀ we can use the following equation:  

A_{0} = N_{0} \lambda   (2)

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

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}}

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

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}}

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\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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