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Gemiola [76]
2 years ago
13

Consider again the objects you ranked by distance. Suppose each object emitted a burst of light right now. Rank the objects from

left to right based on the amount of time it would take this light to reach Earth, from longest time to shortest time.a. star on far side of Andromeda Galaxyb. star on near side of Andromeda Galaxyc. star on far side of Milky Way Galaxyd. star near center of Milky Way Galaxy1. Orion Nebula2. Alpha Centauri3. Pluto4. The Sun
Physics
1 answer:
Vika [28.1K]2 years ago
5 0

Answer:

Following are the solution to this question:

Explanation:

That light takes a very long time to hit the planet, and the object is far off the earth. The light of such an item near to the planet takes less time to enter it. The star is 2,5 million light-years from the Planet on the far side of the Andromeda Galaxy. But on the other hand, the moon is 15 crore miles from the earth, so sunlight is quickly reached on the ground as the other thing.  

That milky way away from the earth is 66,500 light-years far, that distance between Earth and Orion nebula is 1,344 light-years, with such a distance of 4,367 light-years. The earth is 5.2261 trillion km apart from Pluto.

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A maroon plane has a takeoff speed of 88.3 m/s and requires 1635 m to reach that speed. Determine the acceleration of the plane
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18.5164213 if you divide them both you get that number so the volceity is the number shown above.
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3 years ago
Find the length of an arc with a radius of 6.0m swept across 2.5 radians.
soldier1979 [14.2K]

Hello!

In terms of an arc length, radians is the measurement of an arc length in terms of the original radius. This arc length is 2.5 radians, so we multiply it by our original radius.

2.5(6)=15

Therefore, the arc length is 15 meters.

I hope this helps!


7 0
3 years ago
Set the initial bead height to 3.00 m. Click Play. Notice that the ball makes an entire loop. What is the minimum height require
strojnjashka [21]

Answer:

h> 2R

Explanation:

For this exercise let's use the conservation of energy relations

starting point. Before releasing the ball

       Em₀ = U = m g h

Final point. In the highest part of the loop

       Em_f = K + U = ½ m v² + ½ I w² + m g (2R)

where R is the radius of the curl, we are considering the ball as a point body.

      I = m R²

      v = w R

we substitute

       Em_f = ½ m v² + ½ m R² (v/R) ² + 2 m g R

       em_f = m v² + 2 m g R

Energy is conserved

       Emo = Em_f

       mgh = m v² + 2m g R

       h = v² / g + 2R

 

The lowest velocity that the ball can have at the top of the loop is v> 0

      h> 2R

3 0
2 years ago
You throw a tennis ball straight up (neglect air resistance). It takes 7.0 seconds to go up and then return to your hand. How fa
sattari [20]

Answer:

Velocity of throwing = 34.335 m/s

Explanation:

Time taken by the tennis ball to reach maximum height, t = 0.5 x 7 = 3.5 seconds.

Let the initial velocity be u, we have acceleration due to gravity, a = -9.81 m/s² and final velocity = 0 m/s

Equation of motion result we have v = u + at

Substituting

             0 = u - 9.81 x 3.5

             u = 34.335 m/s

Velocity of throwing = 34.335 m/s

6 0
3 years ago
A transparent oil with index of refraction 1.28 spills on the surface of water (index of refraction 1.33), producing a maximum o
Ad libitum [116K]

Answer:

The thickness of the oil slick is 1.95\times10^{-7}\ m

Explanation:

Given that,

Index of refraction = 1.28

Wave length = 500 nm

Order m = 1

We need to calculate the thickness of oil slick

Using formula of thickness

2nt= m\lambda

Where, n = Index of refraction

t = thickness

\lambda = wavelength

Put the value into the formula

2\times1.28 \times t=1\times\times500\times10^{-9}

t = \dfrac{1\times\times500\times10^{-9}}{2\times1.28 }

t=1.95\times10^{-7}\ m

Hence, The thickness of the oil slick is 1.95\times10^{-7}\ m

4 0
2 years ago
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