Ask question

Ask question

All categories

Delicious77 [7]

3 years ago

13

A small logo is embedded in a thick block of crown glass (n = 1.52), 4.70 cm beneath the top surface of the glass. The block is

put under water, so there is 1.70 cm of water above the top surface of the block. The logo is viewed from directly above by an observer in air. How far beneath the top surface of the water does the logo appear to be?

1 answer:

harkovskaia [24]3 years ago

3 0

The concept required to solve this problem is the optical relationship that exists between the apparent depth and actual or actual depth. This is mathematically expressed under the equations.

$d'w = d_w (\frac{n_{air}}{n_w})+d_g (\frac{n_{air}}{n_g})$

Where,

$d_g =$ Depth of glass

$n_w =$ Refraction index of water

$n_g =$ Refraction index of glass

$n_{air} =$ Refraction index of air

$d_w =$ Depth of water

I enclose a diagram for a better understanding of the problem, in this way we can determine that the apparent depth in the water of the logo would be subject to

$d'w = d_w (\frac{n_{air}}{n_w})+d_g (\frac{n_{air}}{n_g})$

$d'w = (1.7cm) (\frac{1}{1.33})+(4.2cm)(\frac{1}{1.52})$

$d'w = 4.041cm$

Therefore the distance below the upper surface of the water that appears to be the logo is 4.041cm

You might be interested in

You heat a mug of water to make hot chocolate. Which statement best describes the changes in the water?

PIT_PIT [208]

It is A because the hot chocolate is well, ya know, hot..

8 0

3 years ago

Read 2 more answers

In a thunder and lightning storm there is a rule of thumb that many people follow. After seeing the lightning, count seconds to

lubasha [3.4K]

Answer:

2.837% less than actual value.

Explanation:

Based on given information let's calculate our value.

S = Vxt = 331m/s x 5s = 1655m, that is the total distance that sound would travel in 5 seconds.

1mile = 1609.34meters.

percentage error is.

$\frac{actual-calculated}{actual} *100 = \frac{1609.34-1655}{1609.34} *100 = -2.83%$

negative indicates less than actual value.

3 0

4 years ago

a spacecraft that seems to be motionless in deep space is given some type of quick push. which statement describes what will hap

bogdanovich [222]

An applied force applied causes a body such as the spacecraft to move. The magnitude of the force determines the change in the velocity

What will happen to the spacecraft is given by option C. from among the possible question options.

C. The spacecraft will begin to move and it will continue moving until it is stopped by an equal and opposite force

Reason:

<em>The possible question options obtained from a similar question includes;</em>

<em>A. The spacecraft will move for some time then stop slowly</em>

<em>B. Air resistance will prevent the spacecraft from moving</em>

<em>C. The spacecraft will begin to move and the motion will continue until a force equal and opposite to the applied force stops it</em>

<em>D. The quick push will not cause the spacecraft to move, as a quick push works on Earth only</em>

The state of the space craft = Motionless

Location of the spacecraft = Deep space

Type of force applied = Quick push

The statement that describes what will happen = Required

Solution;

Let <em>F</em>, represent the force applied, and let Δt be the duration of the applied force, we have;

The impulse of the force, F × Δt = m·(v₂ - v₁)

Where;

m = The mass of the spacecraft

v₁ = The initial velocity of the spacecraft = 0

v₂ = Final velocity of the spacecraft

Plugging in v₁ = 0, gives;

F × Δt = m·v₂

The space craft is given a velocity, <em>v₂</em>, and according to Newton's First Law of Motion, it continues moving in a straight line until another force acts on it

Therefore, the correct option is option C. <u>The spacecraft will begin to move and the motion will continue until it is stopped by an equal and opposite force</u>

<u />

Learn more about Newton's First Law of Motion here:

brainly.com/question/20841616

8 0

3 years ago

A ball is thrown straight up into the air. At each of the following instants, is the ball's acceleration ay equal to g, −g, 0, g

Leto [7]

Answer:

a= (-g) from the moment the ball is thrown, until it stops in the air.

a = (0) when the ball stops in the air.

a = (g) since the ball starts to fall.

Explanation:

The acceleration is <em>(-g)</em> <em>from the moment the ball is thrown, until it stops in the air</em> because the movement goes in the opposite direction to the force of gravity. In the instant <em>when the ball stops in the air the acceleration is </em><em>(0)</em> because it temporarily stops moving. Then, <em>since the ball starts to fall, the acceleration is </em><em>(g)</em><em> </em>because the movement goes in the same direction of the force of gravity

6 0

3 years ago

Read 2 more answers

A beam of light strikes a sheet of glass at an angle of 56.6° with the normal in air. You observe that red light makes an angle

Yuri [45]

Answer:

(a). Index of refraction are $n_{red}$ = 1.344 & $n_{violet}$ = 1.406

(b). The velocity of red light in the glass $v_{red} =$ 2.23 × $10^{8} \ \frac{m}{s}$

The velocity of violet light in the glass $v_{violet} =$ 2.13 × $10^{8} \ \frac{m}{s}$

Explanation:

We know that

Law of reflection is

$n_1 \sin\theta_{1} = n_2 \sin\theta_{2}$

Here

$\theta_1$ = angle of incidence

$\theta_2$ = angle of refraction

(a). For red light

1 × $\sin 56.6$ = $n_{red}$ × $\sin 38.4$

$n_{red}$ = 1.344

For violet light

1 × $\sin 56.6$ = $n_{violet}$ × $\sin 36.4$

$n_{violet}$ = 1.406

(b). Index of refraction is given by

$n = \frac{c}{v}$

$n_{red}$ = 1.344

$v_{red} = \frac{c}{n_{red} }$

$v_{red} = \frac{3(10^{8} )}{1.344}$

$v_{red} =$ 2.23 × $10^{8} \ \frac{m}{s}$

This is the velocity of red light in the glass.

The velocity of violet light in the glass is given by

$v_{violet} = \frac{3(10^{8} )}{1.406}$

$v_{violet} =$ 2.13 × $10^{8} \ \frac{m}{s}$

This is the velocity of violet light in the glass.

8 0

3 years ago