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UkoKoshka [18]
3 years ago
7

Why is light-year not related to time? (I just need scientific evidence why it isn't related to time)

Physics
1 answer:
valkas [14]3 years ago
6 0

Answer:

A light year is a measure of distance as opposed to a time measurement. This is because a light year is the distance light can travel in the time of a year.

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How did you manage to overcome the difficulties that arose?
Flauer [41]

Answer:

I just toughed it out and talked with friends

Explanation:

4 0
2 years ago
A projectile is launched at ground level with an initial speed of 54.5 m/s at an angle of 35.0° above the horizontal. It strikes
Alchen [17]
<h2>Answer: x=125m, y=48.308m</h2>

Explanation:

This situation is a good example of the projectile motion or parabolic motion, in which we have two components: x-component and y-component. Being their main equations to find the position as follows:

x-component:

x=V_{o}cos\theta t   (1)

Where:

V_{o}=54.5m/s is the projectile's initial speed

\theta=35\° is the angle

t=2.80s is the time since the projectile is launched until it strikes the target

x  is the final horizontal position of the projectile (the value we want to find)

y-component:

y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}   (2)

Where:

y_{o}=0  is the initial height of the projectile (we are told it  was launched at ground level)

y  is the final height of the projectile (the value we want to find)

g=9.8m/s^{2}  is the acceleration due gravity

Having this clear, let's begin with x (1):

x=(54.5m/s)cos(35\°)(2.8s)   (3)

x=125m   (4)  This is the horizontal final position of the projectile

For y (2):

y=0+(54.5m/s)sin(35\°)(2.8s)-\frac{(9.8m/s^{2})(2.8s)^{2}}{2}   (5)

y=48.308m   (6)  This is the vertical final position of the projectile

4 0
3 years ago
what would the mass be of an object that was moving at a velocity of 35 m/s and has a kinetic energy of 500j be?
bearhunter [10]

Answer:

0.816 kg

Explanation:

E_k=\frac{1}{2}mv^{2}\\

so m=\frac{2E_k}{v^2}=\frac{2\times500}{35^2}=0.816 kg

5 0
2 years ago
A body which has surface area 5cm² and temperature of 727°C radiates 300J of energy in one minute. Calculate it's emissivity giv
cestrela7 [59]
<h2>Answer: 0.17</h2>

Explanation:

The Stefan-Boltzmann law establishes that a black body (an ideal body that absorbs or emits all the radiation that incides on it) "emits thermal radiation with a total hemispheric emissive power proportional to the fourth power of its temperature":  

P=\sigma A T^{4} (1)  

Where:  

P=300J/min=5J/s=5W is the energy radiated by a blackbody radiator per second, per unit area (in Watts). Knowing 1W=\frac{1Joule}{second}=1\frac{J}{s}

\sigma=5.6703(10)^{-8}\frac{W}{m^{2} K^{4}} is the Stefan-Boltzmann's constant.  

A=5cm^{2}=0.0005m^{2} is the Surface area of the body  

T=727\°C=1000.15K is the effective temperature of the body (its surface absolute temperature) in Kelvin.

However, there is no ideal black body (ideal radiator) although the radiation of stars like our Sun is quite close.  So, in the case of this body, we will use the Stefan-Boltzmann law for real radiator bodies:

P=\sigma A \epsilon T^{4} (2)  

Where \epsilon is the body's emissivity

(the value we want to find)

Isolating \epsilon from (2):

\epsilon=\frac{P}{\sigma A T^{4}} (3)  

Solving:

\epsilon=\frac{5W}{(5.6703(10)^{-8}\frac{W}{m^{2} K^{4}})(0.0005m^{2})(1000.15K)^{4}} (4)  

Finally:

\epsilon=0.17 (5)  This is the body's emissivity

3 0
3 years ago
How many grams of CO2 are in 10 mol of the compound?
Zepler [3.9K]

For our problem, this means that one mole of CO2 has a mass of 44.01 grams. So 22 grams divided by 44.01 grams is roughly 0.5 moles of CO2.


hope it helps

4 0
3 years ago
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