3 years ago

Plzz answer this question correctly

Physics

1 answer:

Lelechka [254]3 years ago

6 0

Answer:

I'm not sure sorry.... shshsjsk

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DiKsa [7]

Answer with Explanation:

We are given that

$D_1=0.15mGy$

$D_2=0.22mGy$

$Q_1=6$

$Q_2=10$

a.We have to find the total dose

Total dose= $D=D_1+D_2$

Using the formula then, we get

$D=0.15+0.22$

$D=0.37mGy$

b.We have to find the total dose equivalent

Total dose equivalent=H= $D_1Q_1+D_2Q_2$

Using the formula

$H=0.15(6)+0.22(10)$

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3 years ago

3.00 m^3 of water is at 20.0°C.

romanna [79]

Answer:

$\triangle V = 0.02484m^3$

Explanation:

Given

$V_1 = 3.00m^3$ --- initial volume

$T_1 = 20.0^oC$ --- initial temperature

$T_2 = 60.0^oC$ --- final temperature

$\gamma = 207*10^{-6$ --- coefficient of thermal expansion:

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To do this, we make use of cubic expansivity formula

$\triangle V = \gamma * V_2 * (T_2 - T_1)$

So, we have:

$\triangle V = 207 * 10^{-6} * 3.00 * (60.0 - 20.0)$

$\triangle V = 207 * 10^{-6} * 3.00 * 40.0$

$\triangle V = 0.02484m^3$

The volume will expand by $0.02484m^3$

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