Which quantity does a light-year measure?distancespeedtimevolume

Physics

2 answers:

marissa [1.9K]3 years ago

8 0

Answer:

Distance (i think)

jeyben [28]3 years ago

3 0

Answer: Distance

Explanation:

Distance is the unit of the light year and we can measure the light year by using the distance measurement. In one year, the light can travel at the velocity of 300000 km (Kilometer) in each seconds. Therefore, in every one year, they can travel 10 trillion kilometers.

Light year is also denoted as ly and it travel in one year at the empty space. The distance of the stars are generally measured in the light year as the objects in the space are mostly far away and they are using the small distance unit like astronomical unit.

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Acar accelerates from 4 meters/second to 16 meters/second in 4 seconds. The car's acceleration is

s2008m [1.1K]

To understand this question, you need to understand the concept of acceleration first. Have you ever been in a car and noticed that it was getting faster and faster? That "speeding up" of the car is known as acceleration! Acceleration is essentially the rate at which you speed up.

Okay, so we now know what acceleration is. What are its units? The unit of acceleration is the change in velocity over a period of time: $\frac{∆v}{t}$

If you haven't learned about velocity yet, just think about it as speed for now. The funny-looking triangle, ∆, is a symbol for "the change of." For example, if I started walking at 3 $\frac{feet}{second}$ then sped up to 5 $\frac{feet}{second}$ , then the change in my speed would be 2 $\frac{feet}{second}$ , because I started walking 2 $\frac{feet}{second}$ faster!

Okay, enough with all the explanations. Hopefully, you understand the units now. Let's take a look at the question. A car accelerates from 4 $\frac{meters}{second}$ to 16 $\frac{meters}{second}$ in 4 seconds. What would the acceleration be? Let's set up an equation:

a = $\frac{∆v}{t}$

a is the acceleration, ∆v is the change in velocity, and t is the time elapsed.

Now, let's plug in our values! ∆v is the change in velocity, and to find that we simply have to subtract 16 $\frac{meters}{second}$ by 4 $\frac{meters}{second}$ . That makes sense, right? Back to the equation.

a = $\frac{∆v}{t}$
a = $\frac{16-4}{4}$

(16 - 4 is the change in velocity, and 4 is the number of seconds the car was accelerating)

a = $\frac{12}{4}$

a = 3 ( $\frac{meters}{second^{2}}$ )

We have our answer! The car's acceleration is 3 meters per second^{2}.

(You might be thinking: Wait. Meters per second squared? The reason for that is because acceleration is the rate at which the speed increases! That makes the unit $\frac{\frac{meters}{second}}{second}$ , which can be simplified down to $\frac{meters}{second^{2} }$ )

Let me know if you need clarification on anything I explained here!
- breezyツ