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Define light yearend explain how and why light years are used to measure distances in the universe

yan [13]

Light years are the distance light can travel in a full year. They are used because there are really big distances in the universe that are easier to measure with a really large form of measurement (light years).

3 0

3 years ago

A _______ is a repeating disturbance or vibration that transfers or moves energy from place to place without transporting mass.

cupoosta [38]

the answer is a wave

7 0

3 years ago

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An energy source forces a constant current of 2A to flow through a light bulbfilament for twenty seconds. If 4.6 kJ is given off

QveST [7]

Answer:

The voltage drop across the bulb is 115 V

Explanation:

The voltage drop equation is given by:

$V=\frac{\Delta W}{\Delta q}$

Where:

ΔW is the total work done (4.6kJ)

Δq is the total charge

We need to use the definition of electric current to find Δq

$I=\frac{\Delta q}{\Delta t}$

Where:

I is the current (2 A)

Δt is the time (20 s)

$2=\frac{\Delta q}{20}$

$q=40 C$

Then, we can put this value of charge in the voltage equation.

$V=\frac{4600}{40}=115 V$

Therefore, the voltage drop across the bulb is 115 V.

I hope it helps you!

6 0

3 years ago

Diferencia entre transferencia y transformación de energía. Da ejemplos de cada caso.

Firdavs [7]

Answer:

La transformación de energía es un proceso en el que la energía se intercambia entre un sistema y el medio ambiente en al menos dos formas de energía diferentes entre sí. Por ejemplo, un panel solar convierte la energía lumínica en energía eléctrica.

En cambio, en la transferencia de energía, esta no cambia su forma sino que es transmitida de un cuerpo a otro. El ejemplo más claro es el de la fogata, que transmite calor al medio ambiente a través de radiación.

6 0

3 years ago

While entering a freeway, a car accelerates from rest at a rate of 2.40 m/s2 for 12.0 s. (a) Draw a sketch of the situation. (b)

ArbitrLikvidat [17]

Answer:

a) See attached picture, b) We know the initial velocity = 0, initial position=0, time=12.0s, acceleration= $2.40m/s^{2}$ , c) the car travels 172.8m in those 12 seconds, d) The car's final velocity is 28.8m/s

Explanation:

a) In order to draw a sketch of the situation, I must include the data I know, the data I would like to know and a drawing of the car including the direction of the movement and its acceleration, just like in the attached picture.

b) From the information given by the problem I know:

initial velocity =0

acceleration = $2.40m/s^{2}$

time = 12.0 s

initial position = 0

c)

unknown:

displacement.

in order to choose the appropriate equation, I must take the knowns and the unknown and look for a formula I can use to solve for the unknown. I know the initial velocity, initial position, time, acceleration and I want to find out the displacement. The formula that contains all this data is the following:

$x=x_{0}+V_{x0}t+\frac{1}{2}a_{x}t^{2}$

Once I got the equation I need to find the displacement, I can plug the known values in, like this:

$x=0+0(12s)+\frac{1}{2}(2.40\frac{m}{s^{2}} )(12s)^{2}$

after cancelling the pertinent units, I get that my answer will be given in meters. So I get:

$x=\frac{1}{2} (2.40\frac{m}{s^{2}} )(12s)^{2}$

which solves to:

$x=172.8m$

So the displacement of the car in 12 seconds is 172.8m, which makes sense taking into account that it will be accelerating for 12 seconds and each second its velocity will increase by 2.4m/s.

d) So, like the previous part of the problem, I know the initial position of the car, the time it travels, the initial velocity and its acceleration. Now I also know what its final position is, so we have more than enough information to find this answer out.

I need to find the final velocity, so I need to use an equation that will use some or all of the known data and the unknown. In order to solve this problem, I can use the following equation:

$a=\frac{V_{f}-V_{0} }{t}$