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Georgia [21]
3 years ago
6

A helium-filled balloon occupies a volume of 15 cubic meters at sea level. the balloon is released and raises to a point in the

atmosphere where the pressure is .75 atm. what is its volume if the temperature remains the same?
Physics
1 answer:
Elena L [17]3 years ago
5 0

According to Boyle’s law, For a fixed amount of an ideal gas kept at a fixed temperature, P (pressure) and V (volume) are inversely proportional.

Therefore,

P_{1} V_{1} =P_{2} V_{2}

Given P_{1} = 1 atm , V_{1} = 15 \ cubic \ meter and P_{2} = 0.75\ atm.

Thus,

V_{2} = \frac{P_{1}\times V_{1}  }{ P_{2} } = \frac{1\times 15}{0.75} =20 m^3

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Acid mine drainage is:
Dahasolnce [82]

Answer:

Acid mine drainage is dissolved toxic materials wash from mines into nearby lakes and streams.

Explanation:

Acid mine drainage is the flow of acidic water with pH typically between 2 and 4, and high concentrations of other dissolved toxic materials from mines into nearby lakes and streams. It mainly occurs during metal sulfide mining, when the metal sulfide ore such as pyrite (FeS2) is exposed to water and oxygen from air to produce soluble iron and sulfuric acid.

Microorganisms, especially acidophile bacteria like Acidithiobacillus ferrooxidans grow by pyrite oxidation, i.e., oxidizing the Fe²⁺ in pyrite to Fe³⁺, which again react with pyrite and water to produce sulfuric acid. Then the acidic water flows into nearby water sources and reduces the pH value of water in those sources. As a result, heavy metals such as copper, lead, mercury, etc in other mineral ores also get dissolved into the water. The action of acidophile bacteria also increases the rate and degree of acid-mine drainage process.

The acid mine drainage causes water pollution and adversely affect the aquatic plants and animals. It also results in the contamination of drinking water, corrosion of infrastructures such as bridges, etc.

8 0
3 years ago
You toss a ball straight up with an initial speed of 30m/s. How high does it go, and how long is it in the air (neglecting air r
Brut [27]

Explanation:

Given that,

A ball is tossed straight up with an initial speed of 30 m/s

We need to find the height it will go and the time it takes in the air.

At its maximum height, its final speed, v = 0 and it will move under the action of gravity. Using equation of motion :

v = u +at

Here, a = -g

v = u -gt

i.e. u = gt

t=\dfrac{u}{g}\\\\t=\dfrac{30\ m/s}{9.8\ m/s^2}\\\\t=3.06\ s

So, the time for upward motion is 3.06 seconds. It means that it will in air for 3.06×2 = 6.12 seconds

Let d is the maximum distance covered by it.

d=ut-\dfrac{1}{2}gt^2

Putting all values

d=30(3.06)-\dfrac{1}{2}\times 9.8\times (3.06)^2\\\\d=45.91\ m

Hence, it will go to a height of 45.91 m and it will in the air for 6.12 seconds.

8 0
3 years ago
Priyadarshini means..........​
Blizzard [7]

Answer:

A submission from India says the name Priyadharshini means "Beloved and pleasing to look at" and is of Sanskrit origin. According to a user from India, the name Priyadharshini is of Indian (Sanskrit) origin and means "God gift".

6 0
2 years ago
Read 2 more answers
The lowest note on a grand piano has a frequency of 27.5 Hz. The entire string is 2.00 m long and has a mass of 400 g. The vibra
Norma-Jean [14]

Answer:

1456 N

Explanation:

Given that

Frequency of the piano, f = 27.5 Hz

Entire length of the string, l = 2 m

Mass of the piano, m = 400 g

Length of the vibrating section of the string, L = 1.9 m

Tension needed, T = ?

The formula for the tension is represented as

T = 4mL²f²/ l, where

T = tension

m = mass

L = length of vibrating part

F = frequency

l = length of the whole part

If we substitute and apply the values we have Fri. The question, we would have

T = (4 * 0.4 * 1.9² * 27.5²) / 2

T = 4368.1 / 2

T = 1456 N

Thus, we could conclude that the tension needed to tune the string properly is 1456 N

4 0
3 years ago
Two cars leave towns 680 kilometers apart at the same time and travel toward each other. one car's rate is 10 kilometers per hou
Evgen [1.6K]

D = distance between the cars at the start of time = 680 km

v₁ = speed of one car

v₂ = speed of other car = v₁ - 10

t = time taken to meet = 4 h

distance traveled by one car in time "t" + distance traveled by other car in time "t" = D

v₁ t + v₂ t = D

(v₁ + v₂) t = D

inserting the values

(v₁ + v₁ - 10) (4) = 680

v₁ = 90 km/h

rate of slower car is given as

v₂ = v₁ - 10

v₂ = 90 - 10 = 80 km/h

5 0
3 years ago
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