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lidiya [134]
3 years ago
5

The sun is 1.5 × 108 km from Earth. The index of refraction for water is 1.349. How much longer would it take light from the sun

to reach Earth if the space between them were filled with water rather than a vacuum?
Physics
1 answer:
tankabanditka [31]3 years ago
4 0

Answer:

175s

Explanation:

time it takes sunlight to reach the earth in  vacuum

C=light speed=299792458m/s

X=1.5x10^8km=1.5x10^11m

c=X/t

T1=X/c

T1=1.5X10^11/299792458=500.34s

time it takes sunlight to reach the earth in  water:

First we calculate the speed of light in water taking into account the refractive index

Cw=299792458m/s/1.349=222233104.5m/s

T2=1.5x10^11/222233104.5m/s=675s

additional time it would take for the light to reach the earth

ΔT=T2-T1=675-500=175s

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If F(theta)=tan theta=3, find F(theta+pi)
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The period of the tan function is π so (∅ + π) would yield the same value as ∅
F(∅ + π) = 3
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2 years ago
For variables control, a circuit voltage will be measured using a sample of five circuits. The past average voltage for samples
astra-53 [7]

Answer:

Average :

UCL = 4.15

LCL = 2.65

Range :

UCL = 2.75

LCL = 0

Explanation:

Given :

Sample size, n = 5

Average, X = 3.4

Range, R = 1.3

A2 for n = 5 ; equals 0.577 ( X chart table)

For the average :

Upper Control Limit (UCL) :

X + A2*R

3.4 + 0.577(1.3) = 4.1501

Lower Control Limit (LCL) :

X - A2*R

3.4 - 0.577(1.3) = 2.6499

FOR the range :

Upper Control Limit (UCL) :

UCL = D4*R

D4 for n = 5 ; equals = 2.114

UCL = 2.114*1.3 = 2.7482

Lower Control Limit (LCL) :

LCL = D3*R

D3 for n = 5 ; equals = 0

LCL = 0 * 1.3 = 0

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3 years ago
A space explorer is moving through space far from any planet or star. He notices a large rock, taken as a specimen from an alien
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He should push it gently.
This is because the forces of resistance this situation are minimal, so the rock will not slow as it would on Earth. Kicking the rock may result in it travelling too fast and hitting something else, causing damage. Moreover, the rock could start rebounding off of surfaces and create havoc. 
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3 years ago
The body is subjected to a force of 0,4 N m with a shoulder of 5 cm. What is the magnitude of this force?​
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Answer:

8N

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5 0
2 years ago
A glass rod and a steel rod are of equal length at 0C. At 100C they differ in length by
NeX [460]

The given lengths at 0 °C are 2.5 m

Let l₀ be the given lengths of the glass and steel rods at 0 °C. Let l and l' be the lengths of the glass and steel rods at 100 °C respectively.

From our expression for linear expansivity,

l = l₀ + l₀αΔθ where α = linear expansivity of glass = 0.000008/°C and Δθ = temperature change = θ - θ' where θ = 100 °C and θ' = 0 °C. So, Δθ = 100 °C - 0 °C = 100 °C.

Also,

l' = l₀ + l₀α'Δθ where α' = linear expansivity of steel = 0.000012/°C and Δθ = temperature change = θ - θ' where θ = 100 °C and θ' = 0 °C. So, Δθ = 100 °C - 0 °C = 100 °C.

Since the difference in their lengths at 100 °C = 0.001 m, we have that

l - l' = l₀ + l₀αΔθ - (l₀ + l₀α'Δθ)

l - l' = l₀ + l₀αΔθ - l₀ - l₀α'Δθ)

l - l' = l₀αΔθ - l₀α'Δθ

l - l' = l₀(α- α')Δθ

Making l₀ subject of the formula, we have

l₀ = (l - l')/[(α- α')Δθ]

Substituting the values of the variables into the equation, we have

l₀ = (l - l')/[(α- α')Δθ]

l₀ = 0.001 m/[(0.000008/°C - 0.000012/°C)100 °C.]

l₀ = 0.001 m/[(-0.000004/°C)100 °C.]

l₀ = 0.001 m/-0.0004

l₀ = -2.5 m

Neglecting the negative sign,

l₀ = 2.5 m

So, the given lengths at 0 °C are 2.5 m

Learn more about linear expansion here:

brainly.com/question/14089545

6 0
2 years ago
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