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puteri
23 days ago
8

Does a fish appear closer or farther from a person wearing swim goggles with an air pocket in front of their eyes than the fish

really is? Does the fish see the person's face closer or farther than it really is? Explain your answer.
Physics
1 answer:
Sav [3.1K]23 days ago
4 0

Response:

In this scenario, the refractive index of seawater is 1.33, while the index for air is 1. Because of this, the refraction angle is smaller than the angle of incidence, making the fish appear closer

Conversely, when viewed from the fish's perspective looking at a person's face, the angle has increased, making the person appear further away

Clarification:

This discussion is analyzed by applying the refraction law, which states that when a light ray crosses from one medium to another, its path bends due to differing indices of refraction,[ [TAG_20]]

n₁ sin θ₁ = n₂ sin θ₂

where n₁ and n₂ are the refractive indices and θ represents the angles for each medium.

Here, with seawater being 1.33 and air at 1, the refraction angle remains lesser than the angle of incidence, leading to the fish appearing nearer

1 sin θ₁ = 1.33 sin θ₂

θ₂ = sin⁻¹ ( 1/1.33 sin θ₁)

When the fish gazes at the human face, the angle's reason increases, hence making the face seem more distant

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Keith_Richards [3271]
Based on my findings, within a period of 2 hours, there are certain atoms remaining. N = N0 * 2^(-t/6.020) = N = N0 * 2^-0.33223 = 0.7943 N0 Thus, the quantity of atoms that undergo disintegration is N0 - N = N0 * (1 - 0.79430) = 0.2057 N0 This must equate to 15 mCi = 15 * 3.7 * 10^7 = 5.55 * 10^8 atoms N0 = 5.55 * 10^8 / 0.2057 = 2.698 * 10^9 atoms Consequently, 2.698 * 10^9 atoms represents the value of N0.
4 0
1 month ago
A mouse runs along a baseboard in your house. The mouse's position as a function of time is given by x(t)=pt2+qt, with p = 0.36
ValentinkaMS [3465]

Response:

0.60 m/s

Details:

The average speed between times t = a and t = b can be expressed as:

v_avg = (x(b) − x(a)) / (b − a)

Given the function x(t) = 0.36t² − 1.20t, and considering the interval from 1.0 to 4.0:

v_avg = (x(4.0) − x(1.0)) / (4.0 − 1.0)

v_avg = [(0.36(4.0)² − 1.20(4.0)) − (0.36(1.0)² − 1.20(1.0))] / 3.0

v_avg = [(5.76 − 4.8) − (0.36 − 1.20)] / 3.0

v_avg = [0.96 − (-0.84)] / 3.0

v_avg = 0.60

The average speed calculated is 0.60 m/s.

5 0
2 months ago
A force is applied to a block sliding along a surface (Figure 2). The magnitude of the force is 15 N, and the horizontal compone
Softa [3030]

If my calculations are accurate, the angle is 67.5 degrees.

4 0
2 months ago
Read 2 more answers
the flow energy of 124 L/min of a fluid passing a boundary to a system is 108.5 kJ/min. Determine the pressure at this point
serg [3582]

Answer:

The pressure measured at this moment is 0.875 mPa

Explanation:

Given that,

Flow energy = 124 L/min

Boundary to system P= 108.5 kJ/min

P=1.81\ kW

We are tasked with finding the pressure here

Applying the pressure formula

P=F\times v

P=A_{1}P_{1}\times v_{1}

Here, A_{1}v_{1}=Q_{1}

Where, v refers to velocity

Insert the values into the equation

1.81 =P_{1}\times0.124\times\dfrac{1}{60}

P_{1}=\dfrac{1.81\times60}{0.124}

P_{1}=875.80\ kPa

P_{1}=0.875\ mPa

Therefore, the pressure at this moment is 0.875 mPa

5 0
1 month ago
A circular surface with a radius of 0.057 m is exposed to a uniform external electric field of magnitude 1.44 × 104 N/C. The mag
Softa [3030]

Answer:

57.94°

Explanation:

We understand that the formula for flux is

\Phi =E\times S\times COS\Theta

where Ф represents flux

           E indicates electric field

           S denotes surface area

        θ signifies the angle between the electric field direction and the surface normal.

It is given that Ф= 78 \frac{Nm^{2}}{sec}

                          E=1.44\times 10^{4}\frac{Nm}{C}

                          S=\pi \times 0.057^{2}

                         COS\Theta =\frac{\Phi }{S\times E}

 =   \frac{78}{1.44\times 10^{4}\times \pi \times 0.057^{2}}

 =0.5306

 θ=57.94°

4 0
1 month ago
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