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2 years ago

Examine the spectra of the four unknown substances shown below. What can you conclude?

Physics

2 answers:

NARA [144]2 years ago

7 0

Line spectra are obtained when individual elements are heated using a high-voltage electrical discharge. This heating causes excitation of the element and a subsequent emission of distinct lines of colored light are obtained. Each element has its own unique emission line spectrum; therefore, if any of the tested substances were the same, their spectra would match. However, this is not the case so none of the substances are the same.

leva [86]2 years ago

7 0

Answer:

none of the substances are the same.

Explanation:

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2 years ago

Why is the air drag on a baseball different than it would be for a smooth ball with no stitches? How does this apply to the desi

yKpoI14uk [10]

Answer:

The stitches and dimples around a baseball and a golf ball respectively, disturbs the air drag on the balls once they are in motion, allowing the them to travel more easily.

Explanation:

The stitches on a baseball disturbs the air drag on the ball when the ball is in motion, allowing the ball to travel more easily. Depending on the orientation of the ball in flight, the drag changes as the flow is disturbed by the stitches.

A smooth ball with no stitches or dimples has more air drag that opposes the motion.

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2 years ago

You are standing in a building whose height (40m) you throw a ball downward at a angle of -30 at a speed of (10m/s) acceleration

jeyben [28]

Answer: 3.41 s

Explanation:

Assuming the question is to find the time $t$ the ball is in air, we can use the following equation:

$y=y_{o}+V_{o}sin \theta t-\frac{1}{2}gt^{2}$

Where:

$y=0m$ is the final height of the ball

$y_{o}=40 m$ is the initial height of the ball

$V_{o}=10 m/s$ is the initial velocity of the ball

$t$ is the time the ball is in air

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

$\theta=30\°$

Then:

$0 m=40 m+(10 m/s)(sin(30\°))t-\frac{1}{2}9.8 m/s^{2}t^{2}$

$0 m=40 m+5m/s t-4.9 m/s^{2}t^{2}$

Multiplying both sides of the equation by -1 and rearranging:

$4.9 m/s^{2}t^{2}-5m/s t-40 m=0$

At this point we have a quadratic equation of the form $at^{2}+bt+c=0$ , which can be solved with the following formula:

$t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

Where:

$a=4.9$

$b=-5$

$c=-40$

Substituting the known values:

$t=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(4.9)(-40)}}{2(4.9)}$

Solving the equation and choosing the positive result we have:

$t=3.41 s$ This is the time the ball is in air

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1. What is the arrangement of the outer planets? 2. What effect does their placement have the planets?

Rasek [7]

the arrangement of the outer planets is
1. Mercury
2. Venus 
3. Earth 
4. Mars 
5. Jupiter 
6. Saturn 
7. Uranus 
8. Neptune
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