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3 years ago

How long does it take light to travel between earth and the nearest star?

Physics

2 answers:

erica [24]3 years ago

7 0

Answer:

8.3 minutes

Explanation:

In the solar system the Sun is the nearest Star to the Earth.

The Speed of light is 300,000 kilometers/second.

The distance is given as 147.67×10⁶ km

The time it will take for the light from the star to reach Earth is :

T= distance / speed

147.67×10⁶/300000 = 492.233 seconds

The nearest star to Earth is the Sun and it takes about 8.3 minutes for its light to reach the Earth.

sergejj [24]3 years ago

5 0

Answer:

It takes 8.204 minutes for light to travel from the Sun to Earth

Explanation:

The speed of light is 300000 km/s and the nearest star to the Earth is the Sun which is 149.67 million km. therefore it takes about

147.67×10⁶/300000 = 492.233 seconds for light from the Sun to reach Earth or 8.204 minutes

The other next nearest star is Alpha Centauri which is 4 light years away

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(4A) The mass of Earth is 5.972 * 10^24 kg, and the radius of Earth is 6,371 km.

faltersainse [42]

Answer:

x₁ = 345100 km

Explanation:

The direction of the attraction forces between the earth and the object, and between the moon and the object, are in opposite direction and (along the straight line between the centers of earth and moon) and as gravity is always attractive, the net force will become zero when both forces are equal. According to this:

Let call "x₁" distance between center of the earth and the object, and

"x₂" the distance between center of the moon and the object, Mt mass of the earth, Ml mass of the moon, m₀ mass of the object

we can express:

F₁ ( force between earth and the object )

F₁ = K * Mt * m₀/ ( x₁)² K is a gravitational constant

F₂ (force between mn and the object)

F₂ = K * Ml * m₀ / (x₂)²

Then:

F₁ = F₂ K*Mt*m₀ / x₁² = K*Ml*m₀ /x₂²

Or simplifying the expression

Mt/ x₁² = Ml/ x₂²

We know that x₁ + x₂ = 384000 Km then

x₁ = 384000 - x₂

Mt/( 384000 - x₂)² = Ml / x₂²

Mt * x₂² = Ml *( 384000 - x₂)²

We need to solve for x₂

Mt * x₂² = Ml *[ ( 384000)² + x₂² - 768000*x₂]

By substitution:

5.972*10∧24*x₂² = 7.348*10∧22 * [ 1.47*10∧11 ] + 7.348*10∧22*x₂² -

7.348*10∧22*768000*x₂

Simplifying by 10∧22

5.972*10²*x₂² = 7.348* [ 1.47*10∧11 ] + 7.348*x₂²- 7.348*768000*x₂

Sorting out

5.972*10²*x₂²- 7.348*x₂² = 10.80*10∧11 - 56,43* 10∧5*x₂

(597,2 - 7,348 )* x₂² = 10.80*10∧11 - 56.43*10∧5*x₂

590x₂² + 56.43*10∧5*x₂ - 10.80*10∧11 = 0

Is a second degree equation

x₂ = -56.43*10∧5 ± √3184*10∧10 + 25488*10∧11 / 1160

x₂ ₁ = -56.43*10∧5 + √3184*10∧10 + 25488*10∧11 / 1160

x₂ ₁ = -56.43*10∧5 + √3184*10∧10 + 254880*10∧10 / 1160

x₂ ₁ = -56.43*10∧5 + 10∧5 [ √3184 + 254880 ] /1160

x₂ ₁ = -56.43*10∧5 + 508* 10∧5 / 1160

x₂ ₁ = 451.27*10∧5/1160

x₂ ₁ = 4512.7*10∧4 /1160

x₂ ₁ = 3.89*10∧4 km (distance between the moon and the object)

x₂ ₁ = 38900 km

x₂ = 38900 km

We dismiss the other solution because is negative and there is not a negative distance

Then the distance between the earth and the object is:

x₁ = 384000 - x₂

x₁ = 384000 - 38900

x₁ = 345100 km

5 0

3 years ago

Based on the graph of velocity over time, which could be the initial velocity and the final velocity for this graph?

dolphi86 [110]

The graph is showing a constant velocity as the line is horizontal. The initial speed is equal to the final speed as there is no changing in acceleration. The second statement is the correct statement

Initial = 2.5 m/s
Final = 2.5 m/s

3 0

3 years ago

Read 2 more answers

A car horn emits a frequency of 400 Hz. A car traveling at 20.0 m/s sounds the horn as it approaches a stationary pedestrian. Wh

Temka [501]

Answer:

The observed frequency by the pedestrian is 424 Hz.

Explanation:

Given;

frequency of the source, Fs = 400 Hz

speed of the car as it approaches the stationary observer, Vs = 20 m/s

Based on Doppler effect, as the car the approaches the stationary observer, the observed frequency will be higher than the transmitted (source) frequency because of decrease in distance between the car and the observer.

The observed frequency is calculated as;

$F_s = F_o [\frac{v}{v_s + v} ] \\\\$

where;

F₀ is the observed frequency

v is the speed of sound in air = 340 m/s

$F_s = F_o [\frac{v}{v_s + v} ] \\\\400 = F_o [\frac{340}{20 + 340} ] \\\\400 = F_o (0.9444) \\\\F_o = \frac{400}{0.9444} \\\\F_o = 423.55 \ Hz \\$

F₀ ≅ 424 Hz.

Therefore, the observed frequency by the pedestrian is 424 Hz.

8 0

3 years ago

How much weight is generated by an 80 kg person on planet earth? On the moon?

otez555 [7]

It will be 133 on the moon

8 0

3 years ago

A small cork with an excess charge of +6.0 μC (1 μC = 10 -6 C) is placed 0.12 m from another cork, which carries a charge of -4.

Alex

Answer:

a.16.125 N b. The force is an attractive force. c. 2.68 × 10¹³ electrons d. 3.75 × 10¹³ electrons

Explanation:

a. What is the magnitude of the electric force between the corks?

The electrostatic force of attraction between the two corks is given by

F = kq₁q₂/r² where k = 9 × 10⁹ Nm²/C², q₁ = +6.0 μC = +6.0 × 10⁻⁶ C, q₂ = -4.3 μC = -4.3 × 10⁻⁶ C and r = distance between the corks = 0.12 m

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

F = kq₁q₂/r²

F = 9 × 10⁹ Nm²/C² × +6.0 × 10⁻⁶ C × -4.3 × 10⁻⁶ C/(0.12 m)²

= -232.2 × 10⁻³ Nm²/(0.0144 m)²

= -16125 × 10⁻³ N

= -16.125 N

So, the magnitude of the force is 16.125 N

b. Is this force attractive or repulsive?

Since the direction of the force is negative, it is directed towards the positively charged cork, so the force is an attractive force.

c. How many excess electrons are on the negative cork?

Since Q = ne where Q = charge on negative cork = -4.3 μC = -4.3 × 10⁻⁶ C and n = number of excess electrons and e = electron charge = -1.602 × 10⁻¹⁹ C

So n = Q/e = -4.3 × 10⁻⁶ C/-1.602 × 10⁻¹⁹ C = 2.68 × 10¹³ electrons

d. How many electrons has the positive cork lost?

We need to first find the number of excess positive charge n'

Q' = n'q where Q = charge on positive cork = + 6.0 μC = + 6.0 × 10⁻⁶ C and n = number of excess protons and q = proton charge = +1.602 × 10⁻¹⁹ C

So n' = Q'/q = +6.0 × 10⁻⁶ C/+1.602 × 10⁻¹⁹ C = 3.75 × 10¹³ protons

To maintain a positive charge, the number of excess protons equals the number of electrons lost = 3.75 × 10¹³ electrons

4 0

3 years ago