Answer:
Concrete
Explanation:
The speed at which sound propagates is medium dependent. As one of the forms of mechanical waves, sound requires a material medium for propagation from place to place.
- Sound travels with the least speed in air because air particles are far apart and they are randomized.
- Sound travels with the greatest speed in solids. Concrete is the only solid material given in the choice.
The speed of sound increases from air to liquid and to solid.
C metal. most metals are a conductor which an electric current is attracted to.
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I believe this is right but double check to make sure :)
It totally depends on what kind of wave you're talking about.
-- a sound wave from a trumpet or clarinet playing a concert-A pitch is about 78 centimeters long ... about 2 and 1/2 feet. This is bigger than atoms.
-- a radio wave from an AM station broadcasting on 550 KHz, at the bottom of your radio dial, is about 166 feet long ... maybe comparable to the height of a 10-to-15-story building. This is bigger than atoms.
-- a radio wave heating the leftover meatloaf inside your "microwave" oven is about 4.8 inches long ... maybe comparable to the length of your middle finger. this is bigger than atoms.
-- a deep rich cherry red light wave ... the longest one your eye can see ... is around 750 nanometers long. About 34,000 of them all lined up will cover an inch. These are pretty small, but still bigger than atoms.
-- the shortest wave that would be called an "X-ray" is 0.01 nanometer long. You'd have to line up 2.5 billion of <u>those</u> babies to cover an inch. Hold on to these for a second ... there's one more kind of wave to mention.
-- This brings us to "gamma rays" ... our name for the shortest of all electromagnetic waves. To be a gamma ray, it has to be shorter than 0.01 nanometer.
Talking very very very very roughly, atoms range in size from about 0.025 nanometers to about 0.26 nanometers.
The short end of the X-rays, and on down through the gamma rays, are in this neighborhood.
Answer:
1.) 274.5v
2.) 206.8v
Explanation:
1.) Given that In one part of the lab activities, students connected a 2.50 µF capacitor to a 746 V power source, whilst connected a second 6.80 µF capacitor to a 562 V source.
The potential difference and charge across EACH capacitor will be
V = Voe
Where Vo = initial voltage
e = natural logarithm = 2.718
For the first capacitor 2.50 µF,
V = Vo × 2.718
746 = Vo × 2.718
Vo = 746/2.718
Vo = 274.5v
To calculate the charge, use the below formula.
Q = CV
Q = 2.5 × 10^-6 × 274.5
Q = 6.86 × 10^-4 C
For the second capacitor 6.80 µF
V = Voe
562 = Vo × 2.718
Vo = 562/2.718
Vo = 206.77v
The charge on it will be
Q = CV
Q = 6.8 × 10^-6 × 206.77
Q = 1.41 × 10^-3 C
B.) Using the formula V = Voe again
165 = Vo × 2.718
Vo = 165 /2.718
Vo = 60.71v
Q = C × 60.71
Q = C
