Answer:
The answer is "
"
Explanation:
For point a:
Energy balance equation:
From the above equation:
because the rate of air entering the tank that is 
constant.
Since the tank was initially empty and the inlet is constant hence,
Interpolate the enthalpy between 
. The surrounding air
temperature:
Substituting the value from ideal gas:
Follow the ideal gas table.
The 
and between temperature 
Interpolate
Substitute values from the table.
For point b:
Consider the ideal gas equation. therefore, p is pressure, V is the volume, m is mass of gas. 
(M is the molar mass of the gas that is 
and R is gas constant), and T is the temperature.
For point c:
Entropy is given by the following formula:
Electro waves in a vacuum air is deals with this and electricity when the air and the electricity it makes electro magnets.
Answer:
<em>The frequency of of the note = 131 Hz.</em>
Explanation:
<em>Frequency:</em><em> Frequency can be defined as the number of complete oscillation completed by a wave in one seconds. The S.I unit of frequency is Hertz ( Hz)</em>
v = λf ............................ Equation 1
Making f the subject of the equation,
f = v/λ .......................... Equation 2
Where v = Speed, λ = wavelength, f = frequency
<em>Given: v = 343 m/s, λ = 2.62 m.</em>
<em>Substituting these values into equation 2</em>
<em>f = 343/2.62</em>
<em>f = 131 Hz</em>
<em>Thus the frequency of of the note = 131 Hz.</em>
Answer: 
Explanation:
When a number is written in scientific notation (representing the number using powers of base ten) it is expressed so that it contains a digit in the place of the units and all other digits after the decimal point, multiplied by the respective exponent.
Then, the significant figures (or significant digits) will be the digits that are before the power of ten.
Now, in the case of the number 299,792,458 if we want to write it with three significant digits, we have to write it in scientific notation as:
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 ?
