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tankabanditka [31]
3 years ago
14

Why are light-years more convenient than miles, kilometers, or astronomical units (AU) for measuring the distances to stars and

galaxies?
Physics
1 answer:
lana66690 [7]3 years ago
8 0

Answer:

Stars and galaxies are very far from the Earth. If their distance were denoted by km or AU then the numbers would be very large. Light-year makes the representation of distance to be concise.

Explanation:

Stars and galaxies are very far from the Earth. If their distance were denoted by km or AU then the numbers would be very large. Light-year makes the representation of distance to be concise.

The distance light travels in one year is called a light-year.

1\ lightyear = 3\times 10^{8}\times 365.25\times 24\times 60\times 60 = 9.46728\times 10^{15}\ m

As it can be seen here that 1 light year = 9.46728×10¹⁵ m expressing distances of stars and galaxies which are more than 1 light-year becomes easier. Converting it to km will make it 9.46728×10¹² km does not make much difference.

1 Astronomical unit (almost the distance between the Earth and Sun) = 149597870700 m is also not enough to represent the distances of stars and galaxies.

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It has been suggested that a heat engine could be developed that made use of the fact that the temperature several hundred meter
lesantik [10]

Answer:

efficiency = 5.4%

Explanation:

Efficiency of heat engine is given as

\eta = \frac{W}{Q_{in}}

now we will have

W = Q_1 - Q_2

so we will have

\eta = 1 - \frac{Q_2}{Q_1}

now we know that

\frac{Q_2}{Q_1} = \frac{T_2}{T_1}

so we have

\eta = 1 - \frac{T_2}{T_1}

\eta = 1 - \frac{273+6}{273+22}

\eta = 0.054

so efficiency is 5.4%

8 0
3 years ago
Where are variables represented in a graph?
m_a_m_a [10]

Answer:variable or dependent values are represented on y axis of graph

Explanation:

4 0
3 years ago
While walking between gates at an airport, you notice a child running along a moving walkway. Estimating that the child runs at
Mashcka [7]

Answer:

The speed of the moving walkway is 1.50 m/s

Explanation:

The position of the child can be calculated using the following equation:

x = x0 + v · t

Where :

x = position of the child at time t.

v = velocity of the child.

t = time.

When the child runs in the same direction as the walkway, the velocity of the child will be its  velocity relative to the walkway plus the velocity of the walkway. Then, if we place the origin of the frame of reference at the start of the walkway:

x = x0 + v · t

25 m = 0 m + (2.8 m/s + v) · t₁

Where v is the velocity of the walkway

On its way back, the velocity of the child relative to the walkway is in the opposite direction to the velocity of the walkway. Then:

x = x0 + v · t

0 m = 25 m + (-2.8 m/s + v) · t₂

We also know that t₁ + t₂ = 25 s

Then: t₁ = 25 - t₂

So, we can write the following system of equations:

25 m = (2.8 m/s + v) · (25 s - t₂)

-25 m = (-2.8 m/s + v) · t₂

Let´s take the second equation and solve it for t₂

-25 m / (-2.8 m/s + v) = t₂

Now, let´s replace t₂ in the first equation:

25 m = (2.8 m/s + v) · (25 s + 25 m / (-2.8 m/s + v))

Let´s sum the fraction: 25 s + 25 m / (-2.8 m/s + v)

25 m = (2.8 m/s + v) · (25 s ·(-2.8 m/s + v) + 25 m) / (-2.8 m/s + v)

multiply by (-2.8 m/s + v) both sides of the equation:

25 m(-2.8 m/s + v) = (2.8 m/s + v) · (-70 m + 25 s · v + 25 m)

Apply distributive property:

-70 m²/s +25 m·v = -196 m²/s +70 m·v +70 m²/s -70 m·v +25 s ·v² + 25 m v

56 m²/s = 25 s · v²

56 m²/s / 25 s = v²

v = 1.50 m/s

The speed of the moving walkway is 1.50 m/s

7 0
3 years ago
How do you find displacement when all you have is time with no velocity, and then in order to find velocity you have to have dis
saw5 [17]
If you really have nothing else but time, then you can't. There must be some other shred of information. Search around. Look under rocks.
3 0
4 years ago
In addition, many predictions based on the idea have led to additional observations that support it. Which best describes this i
Law Incorporation [45]
The answer is: Theory
6 0
2 years ago
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