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3 years ago

13

The actual depth of a shallow pool 1.00 m deep is not the same as the apparent depth seen when you look straight down at the poo

l from above. How deep (in cm) will it appear to be

1 answer:

DedPeter [7]3 years ago

4 0

Answer:

d' = 75.1 cm

Explanation:

It is given that,

The actual depth of a shallow pool is, d = 1 m

We need to find the apparent depth of the water in the pool. Let it is equal to d'.

We know that the refractive index is also defined as the ratio of real depth to the apparent depth. Let the refractive index of water is 1.33. So,

$n=\dfrac{d}{d'}\\\\d'=\dfrac{d}{n}\\\\d'=\dfrac{1\ m}{1.33}\\\\d'=0.751\ m$

or

d' = 75.1 cm

So, the apparent depth is 75.1 cm.

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Points A (-5,6), B (2,-2), and C (-6,-3) are placed in three different quadrants of a Cartesian coordinate system. Convert each

AURORKA [14]

Answer: A ( $\sqrt{61}$ ,309.8°)

B (2 $\sqrt{2}$ , 315°)

C ( $3\sqrt{5}$ , 26.56°)

Explanation: To transform rectangular coordinates into polar coordinates use:

$r=\sqrt{x^{2}+y^{2}}$ and $\theta=tan^{-1}(\frac{y}{x})$

For point A:

$r=\sqrt{(-5)^{2}+6^{2}}$

$r=\sqrt{61}$

$\theta=tan^{-1}(\frac{6}{-5})$

$\theta=tan^{-1}(-1.2)$

$\theta=-50.2$ °

Point A is in the II quadrant, so we substract the angle for 360° since it is in degrees:

$\theta=360-50.2$

$\theta=$ 309.8°

Polar coordinates for point A is ( $\sqrt{61}$ , 309.8°)

For point B:

$r=\sqrt{2^{2}+(-2)^{2}}$

$r=\sqrt{8}$

$r=2\sqrt{2}$

$\theta=tan^{-1}(\frac{-2}{2} )$

$\theta=tan^{-1}(1)$

$\theta=-45$ °

Point B is in IV quadrant, so:

$\theta=360-45$

$\theta=$ 315°

Polar coordinates for point B is ( $2\sqrt{2}$ , 315°)

For point C:

$r=\sqrt{(-6)^{2}+(-3)^{2}}$

$r=\sqrt{45}$

$r=3\sqrt{5}$

$\theta=tan^{-1}(\frac{-3}{-6} )$

$\theta=tan^{-1}(0.5)$

$\theta=$ 26.56°

Polar coordinates for point C is ( $3\sqrt{5}$ , 26.56°)

3 0

3 years ago

The accompanying table shows measurements of the Hall voltage and corresponding magnetic field for a probe used to measure magne

aalyn [17]

0.125 mm . is the thickness of the sample.

<h3>What do you mean by hall voltage ?</h3>

The Hall effect is the creation of a voltage difference (the Hall voltage) across an electrical conductor, which is transverse to an applied magnetic field perpendicular to the current and an electric current in the conductor. Edwin Hall made the discovery in 1879.

We need to know the material's current, magnetic field, length, number of charge carriers, and area in order to calculate the Hall voltage. The Hall voltage is computed using the formula: v=IBlneA=(100A)(1.5T)(1.0102m)(5.91028/m3)(1.61019C)(2.0105m2)=7.9106V.

lof4

First we have to plot those point Then we can use some computer program to fit those point linearly to get slope

of that graph a and interception b. We already know, from theory, that Hall's voltage AVH and magnitude of

magnetic field B are connected as

Δ $V_{H} =\frac{I}{nqt} B$

where I is current trough probe, n is concentration of charge carriers, q = 1.6 • 10¯19 C is charge of charge

carries and t is thickness of the material. We have put the data from the problem on a graph and fitted linearly and

got

a = 100 μ $\frac{V}{T}$

b = —0.02 μV.

As we can see, our result are in agreement with theoretical assumptions because interception b is almost O, and a

is asked relation between Hall's voltage A VH and magnitude of magnetic field B. Then we can write

ΔVH = $100X10^{-6}$ V/TB

(4) Then we can use result (4) and numbers from the textbook to calculate the thickness of the sample as

$a=\frac{I}{nqt} \\t=\frac{I}{anq} \\t=\frac{.200A}{100X10^{-6}X 1.6 X10^{-19}X10^{26} } \\t=0.125mm$

To learn more about the hall voltage , Visit: brainly.com/question/19130911

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8 0

2 years ago

A non-ideal 12.2 V battery is connected across a resistor R. The internal resistance of the battery is 1.9Ohm. Calculate the pot

Brums [2.3K]

Answer:

R=100 Ohm, V=11.97 volts and I=0.12 amperes

R=10 Ohm, V=10.25 volts and I=1.20 amperes

R=2 Ohm, V=6.26 volts

Explanation:

The potential difference (voltage) of a battery with internal resistance is:

$V=\xi-Ir$ (1)

with $\xi$ the electromotive force (the voltage the batteries say to has) , I the current and r the internal resistance. By Ohm's law the current that passes through the resistor is:

$I=\frac{V}{R}$ (2)

using (2) on (1):

$V=\xi-\frac{V*r}{R}$

solving for V:

$V+\frac{V*r}{R}=\xi$

$V=\frac{\xi}{1+\frac{r}{R}}$ (3)

R=100 Ohm

$V=\frac{12.2}{1+\frac{1.9}{100}}=11.97 V$

R=10 Ohm

$V=\frac{12.2}{1+\frac{1.9}{10}}=10.25 V$

R=2 Ohm

$V=\frac{12.2}{1+\frac{1.9}{2}}=6.26 V$

Because we have now the values of I on the circuit (is the same through all the components because is a series circuit)

We use back substitution on (1) to find the current:

R=100 Ohm

$I=\frac{V}{R}=\frac{11.97}{100}=0.12 A$

R=10 Ohm

$I=\frac{V}{R}=\frac{11.97}{10}=1.20 A$

7 0

4 years ago

A car is driven 125.0 km due west then 65.0 km due south. What is the magnitude of its displacement?

kodGreya [7K]

pythagoras' theorem on right angled triangle. sides 125, 65

sqrt (125^2 +65^2)

5 0

3 years ago

what will happen to the pressure in a gas if you compressed it into a volume that was one third the size? why?

Annette [7]

Answer:

Pressure will triple (Boyle's Law)

Explanation:

Assuming a constant temperature: (compressing gases usually raises the temp significantly):

P1V1 = P2V2

P1V1 / V2 = P2

Now change V2 to 1/3 V2:

P1V1 / (1/3 V2 ) = P2 / (1/3 )

P1V1/ (1/3 v2 ) = 3 P2 <======= THE PRESSURE WILL TRIPLE

This is Boyle's Law

5 0

2 years ago