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bekas [8.4K]
3 years ago
13

A swimming pool is 1.4 m deep and 12 m long. Is it possible for you to dive to the very bottom of the pool so people standing on

the deck at the end of the pool do not see you?
Physics
1 answer:
icang [17]3 years ago
5 0

Answer:

r = 1.59 m

So this is smaller than the length of the swimming pool and hence we can say that diver will not be visible to the person sitting at deck

Explanation:

If the diver is at the bottom of the swimming pool then we will say that the nobody on the deck will able to see if light coming from the diver will have total internal reflection

so we will have

n_1 sin\theta_1 = n_2 sin\theta_2

here we will have

n_1 = \frac{4}{3} for water

n_2 = 1

\theta_2 = 90

now we have

\frac{4}{3}sin\theta_1 = 1

now we have

\theta_1 = 48.6

now let say the radial distance from the position of diver is"r"

so we have

tan\theta = \frac{r}{h}

tan48.6 = \frac{r}{1.4}

r = 1.59 m

So this is smaller than the length of the swimming pool and hence we can say that diver will not be visible to the person sitting at deck

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Answer:

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Emf induced is given by

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Two large parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 cm. Part A If the surfac
alukav5142 [94]

Answer:

5308.34 N/C

Explanation:

Given:

Surface density of each plate (σ) = 47.0 nC/m² = 47\times 10^{-9}\ C/m^2

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We know, from Gauss law for a thin sheet of plate that, the electric field at a point near the sheet of surface density 'σ' is given as:

E=\dfrac{\sigma}{2\epsilon_0}

Now, as the plates are oppositely charged, so the electric field in the region between the plates will be in same direction and thus their magnitudes gets added up. Therefore,

E_{between}=E+E=2E=\frac{2\sigma}{2\epsilon_0}=\frac{\sigma}{\epsilon_0}

Now, plug in  47\times 10^{-9}\ C/m^2 for 'σ' and 8.85\times 10^{-12}\ F/m for \epsilon_0 and solve for the electric field. This gives,

E_{between}=\frac{47\times 10^{-9}\ C/m^2}{8.854\times 10^{-12}\ F/m}\\\\E_{between}= 5308.34\ N/C

Therefore, the electric field between the plates has a magnitude of 5308.34 N/C

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3 years ago
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