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3 years ago

How many minutes does light need to travel from the sun to Jupiter?

Physics

1 answer:

nydimaria [60]3 years ago

3 0

Light travels at a speed of 299,792 kilometers per second; 186,287 miles per second

Mercury it will take 193.0 seconds(3.2 minutes)

Venus it will take 360.0 seconds(6.0 minutes)

Earth it will take 499.0 seconds(8.3 minutes)

Mars It will take 759.9 seconds(12.6 minutes)

Jupiter It will take 2595.0 seconds(43.2 minutes)

Saturn it will take 4759.0 seconds(79.3 minutes)

Uranus it will take 9575.0 seconds(159.6 minutes)

Neptune it will take 14998.0 seconds(4.1 hours)

Pluto it will take 19680.0 seconds(5.5 hours)

You might be interested in

Soap bubbles can display impressive colors, which are the result of the enhanced reflection of light of particular wavelengths f

Bogdan [553]

Answer:

the minimum wall thickness that will enhance the reflection of light is 146.9 nm

Explanation:

Given the data in the question;

At the first interface, a phase shift occurs as the incident light is in air that has less refractive index compare to the thin film of soap bubble.

At the second interface, no shift occurs,

condition for constructive interference;

t = ( m + 1/2) × λ/2n

where m = 0, 1, 2, 3 . . . . . .

now, the condition for the constructive interference;

t = mλ/2n

where t is the thickness of the soap bubble, λ is the wavelength of light and n is the refractive index of soap bubble.

so the minimum thickness of the film which will enhance reflection of light will be;

t $_{min$ = ( m + 1/2) × λ/2n

we substitute

t $_{min$ = ( 0 + 1/2) × 711 /2(1.21)

t $_{min$ = 0.5 × 711/2.42

t $_{min$ = 0.5 × 293.80165

t $_{min$ = 146.9 nm

Therefore, the minimum wall thickness that will enhance the reflection of light is 146.9 nm

8 0

3 years ago

Two charges are located in the x–y plane. If q1 = -2.90 nC and is located at x = 0.00 m, y = 0.840 m and the second charge has m

Lunna [17]

Answer:

Epx= - 21.4N/C

Epy= 19.84N/C

Explanation:

Electric field theory

The electric field at a point P due to a point charge is calculated as follows:

E= k*q/r²

E= Electric field in N/C

q = charge in Newtons (N)

k= electric constant in N*m²/C²

r= distance from load q to point P in meters (m)

Equivalences

1nC= 10⁻⁹C

known data

q₁=-2.9nC=-2.9 *10⁻⁹C

q₂=5nC=5 *10⁻⁹C

r₁=0.840m

$r_{2} =\sqrt{1^{2} +0.8^{2} } =\sqrt{1.64}$

$sin\beta =\frac{0.8}{\sqrt{1.64} } =0.6246$

$cos\beta =\frac{1}{\sqrt{1.64} } =0.7808$

Calculation of the electric field at point P due to q1

Ep₁x=0

$Ep_{1y} =\frac{k*q_{1} }{r_{1}^{2} } =\frac{8.99*10^{9}*2.9*10^{-9} }{0.84^{2} } =36.95\frac{N}{C}$

Calculation of the electric field at point P due to q2

$Ep_{2x} =-\frac{k*q_{2} *cos\beta }{r_{2}^{2} } =-\frac{8.99*10^{9}*5*10^{-9} *0.7808 }{(\sqrt{1.64})^{2} } =-21.4\frac{N}{C}$

$Ep_{2y} =-\frac{k*q_{2} *sin\beta }{r_{2}^{2} } =-\frac{8.99*10^{9}*5*10^{-9} *0.6242 }{(\sqrt{1.64})^{2} } =-17.11\frac{N}{C}$

Calculation of the electric field at point P(0,0) due to q1 and q2

Epx= Ep₁x+ Ep₂x==0 - 21.4N/C =- 21.4N/C

Epy= Ep₁y+ Ep₂y=36.95 N/C-17.11N =19.84N/C

7 0

3 years ago

What speed should a satellite of mass 4,900 kg moving around

Rudiy27

Based on the calculations, the speed required for this satellite to stay in orbit is equal to 1.8 × 10³ m/s.

<u>Given the following data:</u>

Gravitational constant = 6.67 × 10⁻¹¹ m/kg²

Mass of Moon = 7.36 × 10²² kg

Distance, r = 4.2 × 10⁶ m.

<h3>How to determine the speed of this satellite?</h3>

In order to determine the speed of this satellite to stay in orbit, the centripetal force acting on it must be sufficient to change its direction.

This ultimately implies that, the centripetal force must be equal to the gravitational force as shown below:

Fc = Fg

mv²/r = GmM/r²

<u>Where:</u>

m is the mass of the satellite.

M is mass of the Moon.

Making v the subject of formula, we have;

v = √(GM/r)

Substituting the given parameters into the formula, we have;

v = √(6.67 × 10⁻¹¹ × 7.36 × 10²²/4.2 × 10⁶)

v = √(1,168,838.095)

v = 1,081.13 m/s.

Speed, v = 1.8 × 10³ m/s.

Read more on speed here: brainly.com/question/20162935

#SPJ1

8 0

2 years ago

How many minutes would it take a light wave to travel from the planet Venus to Earth? (Average distance from Venus to Earth = 28

Gnesinka [82]

Answer:

$t=2.51min$

Explanation:

The time taken by the light to travel a given distance is defined as:

$t=\frac{d}{c}$

Here c is obviously the speed of light. Now we convert the average distance form Venus to Earth to meters:

$28*10^{6}mi*\frac{1609.34}{1mi}=4.51*10^{10}m$

Finally, we calculate the minutes taken by the light to travel from Venus to Earth:

$t=\frac{4.51*10^{10}m}{3*10^8\frac{m}{s}}\\t=150.33s*\frac{1min}{60s}=2.51min$

6 0

3 years ago

How often is carbon dioxide cycled through the atmosphere

olga55 [171]

Maybe around 350 years, depending on the carbon cycle and the time taken through steps.

5 0

3 years ago