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vagabundo [1.1K]
2 years ago
8

A. It decreases gradually. B. It increases gradually. C. It falls sharply. D. It remains the same.

Physics
2 answers:
babymother [125]2 years ago
8 0

Answer:

sure, it remains the same by observation

harina [27]2 years ago
3 0

Answer:

It remains the same.

Explanation:

» At 100°C, it is the boiling point of water. And it starts changing from liquid state to gaseous state

{}

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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
4 years ago
Determina el volumen de 450 g de alcohol
Rufina [12.5K]

Luego de utilizar la fórmula del volumen de un líquido encontramos que si el acohol tiene 450 gramos entonces su volumen es de

La fórmula del volumen de la densidad de un líquido es la siguiente:

Densidad= Masa / Volumen

Volumen =Masa / Densidad

Es conocido que la densidad del alcohol es 789 kg/m³, y como sabemos por dato que tiene 450gr el siguiente paso es la sustitución:

Volumen=0.45/ 789

Volumen = 0,00057 m³    

6 0
2 years ago
The wavelength of a light wave is 700 nm in air; this light appears red. If this wave enters a pool of water, its wavelength bec
hodyreva [135]

Answer

given,

wavelength of light in air = 700 nm

Wavelength of light in water = 530 nm

We know that speed of light changes when it moves from one medium to another.

And the frequency of the wavelength does not changes if the medium changes.

we also know that,

                                    v = ν λ

From the above equation we can say that if frequency is constant so, with the change in  velocity changes wavelength will also change.

Hence, wavelength is the property of the wave which determines color.

6 0
3 years ago
In which direction does a bag at rest move when a force of 20 newtons is applied from the right? A. in the direction of the appl
-Dominant- [34]

The correct answer is A. In the direction of applied force. This is because acceleration occurs n the direction of applied force according to Newtons second law of motion which states that the acceleration of a body is directly proportional to the applied force and takes place in the direction of force.

5 0
3 years ago
Read 2 more answers
I need to lift a 2000kg car, 1.798m and the joules required is 35240.8. Converted to watt (W = 35240.8/5 (s)) I got 7048.16 W. I
marusya05 [52]
This is a very interesting problem ... mainly because it's different from
the usual questions in the Physics neighborhood.

I can discuss it with you, but maybe not quite give you a final answer
with the information you've given in the question.

I agree with all of your calculations so far ... the total energy required,
and the power implied if the lift has to happen in 5 seconds.

First of all, let's talk about power.  I'm assuming that your battery is
a "car" battery, and I'm guessing you measured the battery voltage
while the car was running.  Turn off the car, and you're likely to read
something more like 13 to 13.8 volts.
But that's not important right now.  What I'm looking for is the CURRENT
that your application would require, and then to look around and see whether
a car battery would be capable of delivering it.

   Power = (volts) x (current)

   7,050 W  =  (14 volts) x (current)

   Current = (7,050 watts / 14 volts) =  503 Amperes. 

That kind of current knocks the wind out of me.  I've never seen
that kind of number outside of a power distribution yard.
BUT ... I also know that the current demand from a car battery during
starting is enormous, so I'd better look around online and try to find out
what a car battery is actually capable of.

I picked a manufacturer's name that I'd heard of, then picked their
recommended battery for a monster 2003-model car, and looked at
the specs for the battery.

The spec I looked at was the 'CCA' ... cold cranking Amps.
That's the current the battery is guaranteed to deliver for 30 seconds,
at a temperature of 0°F, without dropping below 12 volts.

This battery that I saw is rated  803 Amps  CCA !

OK.  Let's back up a little bit.  I'm pretty sure the battery you have
is a nominal "12-volt" battery.  Let's say you use to start lifting the lift. 
As the lift lifts, the battery voltage sags.  What is the required current
if the battery immediately droops to 12V and stays there, while delivering
7,050 watts continuously ?

          Power = (volts) x (current)

          7,050 W = (12 V) x (current)

            Current = (7,050 W / 12 V)  =  588 Amps . 

Amazingly, we may be in the ball park.
If the battery you have is rated by the manufacturer for 600 Amps
CCA (0°F) or CA (32°F), then the battery can deliver the current
you need.
BUT ... you can't conduct that kind of current through ear-bud wire,
or house wiring wire.  I'm not even so sure of jumper-cables. 
You need thick, no-nonsense cable, AND connections with a lot of
area ... No alligator clips.  Shiny nuts and bolts with no crud on them.

Now ... I still want to check the matter of the total energy.
I'm sure you're OK, because the CCA and CA specifications talk about
30 seconds of cranking, and you're only talking about 5 seconds of lifting.
But I still want to see the total energy requirement compared to the typical
battery specification ... 'AH' ... ampere-hours.

You're talking about 35,000 joules

                          = 35,000 watt-seconds

                         =  35,000 volt-amp-seconds.

               (35,000 volt-amp-sec) x (1 hour/3600 sec) / (12 volt)               

           =  (35,000 x 1) / (3600 x 12)  volt-amp-sec-hour / sec-volt

           =    0.81 Amp-Hour  .

That's an absurdly small depletion from your car battery.
But just because it's only  810 mAh, don't get the idea that you can
do it with a few rechargeable AA batteries out of your camera.
You still need those 600 cranking amps.  That would be a dead short
for a stack of camera batteries, and they would shrivel up and die.

Have I helped you at all ?
5 0
3 years ago
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