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koban [17]
3 years ago
10

Rowan is walking in a shallow, clear bay, in still water just over her knees. When she looks down at her feet in the sand, she n

otes that they appear closer to her than usual. Normally, when she tips her head forward to look at her feet, her feet are 1.60 m from her eyes. How far away do her feet appear when they are under 0.60 m of water?
Physics
1 answer:
Mamont248 [21]3 years ago
6 0

Answer:

Rowan's feet appears at a distance of 0.451 m under water

Solution:

As per the question:

Actual depth in water, d = 0.60 m

Normal distance of her feet from her eyes, D = 1.60 m

Now,

We know that refractive index of water, n_{w} = 1.33

Thus

To calculate the apparent height of her feet:

n_{w} = \frac{d}{Apparent\ Height, H_{apparent}}

H_{apparent} = \frac{d}{n_{w}} = \frac{0.60}{1.33} = 0.451\ m

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Which of the following units would need to be converted before being used for a calculation
Andreyy89

The units which would need to be converted before being used for a calculation are cm and 0 ks.

<h3>What is Unit?</h3>

This is referred to a standard which is used to make comparisons in the aspect of measurement.

The units cm and 0 ks aren't in their standard form which is m and s respectively.

Read more about Unit here brainly.com/question/4895463

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6 0
1 year ago
What factor is responsible for altering the speed of an electromagnetic wave? A) amplitude B) wavelength Eliminate C) type of me
Nataly [62]

The correct answer is C) type of medium. Electromagnetic waves travel faster in solids than in liquids, and faster in liquids than in gases.

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3 years ago
A helicopter is ascending vertically. a passenger accidentally drops her wallet out the sides of the helicopter when it is 160 m
Advocard [28]

Answer:

(E)56.0 m/s

Explanation:

Height =h=-160 m

Because the wallet moving in downward direction

Time=t=7 s

Final speed of wallet=v=0

We have to find the speed of helicopter ascending  at the moment when the passenger let go of the wallet.

v^2-u^2=2gh

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Substitute the values

0-u^2=2(-160)\times 9.8

u^2=3136

u=\sqrt{3136}=56m/s

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8 0
3 years ago
A 150kg person stands on a compression spring with spring constant 10000n/m and nominal length of 0.50.what is the total length
Ivahew [28]

Answer:

<em>The total length of the spring would be 0.65 m</em>

Explanation:

The Concept

Hooke's law evaluates the increment of  spring in relation to the force acting on the body. Hooke's law states that for a spring undergoing deformation, the  force applied is directly proportional to the deformation experienced by the spring. Hooke's law is represented thus;

F = k x ..................1

where F is the force applied to the spring

k is the spring constant

x is the spring stretch or extension

Step by Step Calculations

We have to obtain x before adding it to the nominal length, We make x the subject formula in equation 1;

x = F/k

but F = m x g

so, x = (m x g)/k

given that, the mass of the person m =150 kg

g is the acceleration due to gravity = 9.81 m/s^{2}

k is the spring constant = 10000 N/m

then x = (9.81 m/s^{2} x 150 kg)/10000 N/m

x = 0.14715 m

the extension experienced by the spring after the compression is 0.14715 m

The total length of the spring would be;

L = 0.14715 m + 0.5 m = 0.64715

L ≈  0.65 m

Therefore the total length of the spring would be 0.65 m

4 0
3 years ago
A softball is thrown from the origin of an x-y coordinate system with an initial speed of 18 m/s at an angle of 35∘ above the ho
podryga [215]
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6 0
3 years ago
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