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

Answers

Answer 1

Answer:

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,

$n=(d)/(d')\n\nd'=(d)/(n)\n\nd'=(1\ m)/(1.33)\n\nd'=0.751\ m$

d' = 75.1 cm

So, the apparent depth is 75.1 cm.

Answer 2

Answer:

The apparent depth of a 1.00-meter-deep pool, when viewed from above, is around 75.2 centimeters. This difference is due to light refraction in water, causing optical distortion.

When observing a shallow pool of 1.00 meter depth from above, the apparent depth is altered by the phenomenon of light refraction in water. Light bends as it passes from air into water, affecting the way objects are perceived underwater.

The apparent depth is less than the actual depth due to this bending of light. To calculate the apparent depth, one can use the Snell's Law formula, which relates the angles of incidence and refraction to the refractive indices of the two media.

However, a simplified formula for the apparent depth (d') in terms of the actual depth (d) is given by d' = d/n, where 'n' is the refractive index of water (approximately 1.33). Therefore, in this case, the pool's apparent depth, when viewed from above, will be approximately 75.2 centimeters, making it shallower than it appears at first glance due to the optical effects caused by light traveling through water.

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Complete question below:

"What is the apparent depth, in centimeters, when looking straight down at a shallow pool that is 1.00 meter deep? Note that the apparent depth is different from the actual depth due to the refraction of light in water."

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An object essentially at infinity is moved to a distance of 90 cm in front of a thin positive lens. In the process its image distance triples. Determine the focal length of the lens.

Answers

Answer:

67.5 cm

Explanation:

u = - 90 cm, v = 3 x u = 3 x 90 = 270 cm

let f be the focal length

Use lens equation

1 / f = 1 / v - 1 / u

1 / f = 1 / 270 + 1 / 90

1 / f = 4 / 270

f = 67.5 cm

Final answer:

To determine the focal length of the lens, we use the lens formula and set up an equation based on the given information. Solving for the image distance, we find that it is zero, indicating the image is formed at infinity. Therefore, the focal length of the lens is 90 cm.

Explanation:

To determine the focal length of the lens, we can use the lens formula:

1/f = 1/v - 1/u

Where f is the focal length, v is the image distance, and u is the object distance.

Given that the image distance triples when the object is moved from infinity to 90 cm in front of the lens, we can set up the following equation:

1/f = 1/(3v) - 1/(90)

Multiplying through by 90*3v, we get:

90*3v/f = 270v - 90*3v

90*3v/f = 270v - 270v

90*3v/f = 0

Simplifying further, we find that: v = 0

When the image distance is zero, it means the image is formed at infinity, so the lens is focused at the focal point. Therefore, the focal length of the lens is 90 cm.

A test charge of 13 mC is at a point P where an external electric field is directed to the right and has a magnitude of 4 3 106 N/C. If the test charge is replaced with another test charge of 23 mC, what happens to the external electric field at P

Answers

Answer:

The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C, but the direction is still to the right.

Explanation:

From coulomb's law, F = Eq

Thus,

F = E₁q₁

F = E₂q₂

Then

E₂q₂ = E₁q₁

$E_2 = (E_1q_1)/(q_2)$

where;

E₂ is the external electric field due to second test charge = ?

E₁ is the external electric field due to first test charge = 4 x 10⁶ N/C

q₁ is the first test charge = 13 mC

q₂ is the second test charge = 23 mC

Substitute in these values in the equation above and calculate E₂.

$E_2 = (4*10^6*13)/(23) = 2.26 *10^6 \ N/C$

The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C when 13 mC test charge is replaced with another test charge of 23 mC.

However, the direction of the external field is still to the right.

In a Venn diagram, the separate circles contain characteristics unique to each item being compared and the intersection contains characteristics that are common to both items being compared. Ernie is working on the Venn diagram below to compare the career pathways of Biotechnology Research and Development and Diagnostic Services.What else could Ernie put in the common section?

Collecting data and analyzing results
Designing and implementing systems
Maintaining and using diagnostic equipment
Designing and using laboratory equipment
Mark this and return

Answers

Another thing that Ernie put in the common section is collecting data and analyzing results.

What is a Venn diagram?

A Venn diagram is used to show a representation of data. The center of the Venn diagram is often used to indicate the data set that is the same.

Looking at the Venn diagram, another thing that Ernie put in the common section is collecting data and analyzing results.

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I think its- Collecting data and analyzing results
Thats what I put.

What is the magnitude of a point charge that would create an electric field of 1.18 N/C at points 0.822 m away?

Answers

Answer:

q = 8.85 x 10⁻¹¹ C

Explanation:

given,

Electric field, E = 1.18 N/C

distance, r = 0.822 m

Charge magnitude = ?

using formula of electric field.

$E = (kq)/(r^2)$

k is the coulomb constant

$q= (Er^2)/(k)$

$q= (1.18* 0.822^2)/(9* 10^9)$

q = 8.85 x 10⁻¹¹ C

The magnitude of charge is equal to q = 8.85 x 10⁻¹¹ C

If the solution described in the introduction is cooled to 0 ∘c, what mass of kno3 should crystallize? enter your answer numerically in grams.

Answers

14 g is the solubility per 100 g water, since it is difficult to read the graph.
Then, in 130 g H20 the solubility would be 14 g KNO3/100 g H2O x 130 g H2O = 18 g KNO3
The question asks how much crystallizes.
Initial 34.0 g minus 18.0 g still dissolved = 16.0 g crystallizes.

KNO3 of 10g will undergo crystallization at 0 °

Because the heavier the KNO3 mass will require a higher temperature in the dissolution process.

Further explanation

Potassium nitrate is a nitrate salt compound from potassium with the molecular formula KNO3. Potassium nitrate salt can be made by reacting potassium chloride with sodium nitrate. If the saturated solution each of the solution is mixed with each other, then it will form sodium chloride salt because NaCl in water is small, the salt will settle. By cooling the filtered filtrate KNO3 will undergo crystallization

This compound decomposes with oxygen evolution at 500 ° C according to the reaction equation:

2 NaNO3 (s) -> 2NaNO 2 (s) + O2 (g)

Crystallization is separation by forming crystals so that the mixture can be separated. A gaseous or liquid substance can cool or condense and form crystals because it undergoes a crystallization process. Crystals will also form from a solution that will be saturated with a certain solvent. The more the number of crystals, the better, because the less likely to be polluted by dirt.

Potassium Nitrate has a physical white powder that is easily soluble in water and odorless. Meanwhile, to analyze the structure and characteristics of Potassium Nitrate MM2 data processing is used in the Chemoffice 15.0 application. This data processing is used to determine the shape of compounds, types of bonds in molecular movement compounds and other parts that can not be observed directly by the eye without the aid of tools. And for the form of compounds in 2 dimensions and 3 dimensions used Chemdraw 15.0 and Chem3D 15.0 applications

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Details

Grade: High School

Subject: Chemistry

Keyword: kno3, nitrate, crystallization

Assume: The bullet penetrates into the block and stops due to its friction with the block.

The compound system of the block plus the

bullet rises to a height of 0.13 m along a

circular arc with a 0.23 m radius.

Assume: The entire track is frictionless.

A bullet with a m1 = 30 g mass is fired

horizontally into a block of wood with m2 =

4.2 kg mass.

The acceleration of gravity is 9.8 m/s2 .

Calculate the total energy of the composite

system at any time after the collision.

Answer in units of J.

Taking the same parameter values as those in

Part 1, determine the initial velocity of the

bullet.

Answer in units of m/s.

Answers

To solve this problem we will start considering the total energy of the system, which is given by gravitational potential energy of the total of the masses. So after the collision the system will have an energy equivalent to,

$E_T = (m_1+m_2)gh$

Here,

$m_1$ = mass of bullet

$m_2$ = Mass of Block of wood

The ascended height is 0.13m, so then we will have to

PART A)

$E_T = (m_1+m_2)gh$

$E_T = (0.03+4.2)(9.8)(0.13)$

$E_T = 5.389J$

PART B) At the same time the speed can be calculated through the concept provided by the conservation of momentum.

$m_1v_1+m_2v_2 = (m_1+m_2)v_f$

Since the mass at the end of the impact becomes only one in the system, and the mass of the block has no initial velocity, the equation can be written as

$m_1v_1 =(m_1+m_2)v_f$