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Andrej [43]
3 years ago
12

The most common purpose for fracking in the u.s is to get

Physics
1 answer:
sveta [45]3 years ago
5 0
Natural gas is the answer.
You might be interested in
If continuous spectrum light from a star passes through a cool, low-density gas on its way to your telescope and spectroscope, _
natima [27]

Answer: a dark (absorption) line

Explanation:

This is as a result of absorption of electromagnetic a specific wavelength. The pattern followed by such lines is characteristic of specific atoms in the path of radiation.

6 0
3 years ago
A cannonball is tired with an initial speed of 40m/s and a launch angle of 30degrees from a cliff that is 25m tall. a. What is t
ella [17]

Answer:

Part a)

t = 5.1 s

Part b)

x = 176.7 m

Explanation:

Initial speed of the launch is given as

v_i = 40 m/s

angle = 30 degree

Now the two components of the velocity

v_x = v_i cos30

v_x = 34.6 m/s

similarly we have

v_y = v_i sin30

v_y = 20 m/s

Part a)

Now to find the time of flight we will have

\Delta y = v_y t + \frac{1}{2}at^2

-25 = 20 t -\frac{1}{2}(9.81)t^2

t = 5.1 s

Part b)

Distance at which the ball will land is given as

x = v_x t

x = 34.6 (5.1)

x = 176.7 m

Part c)

Since x direction there is no acceleration so it will move with uniform speed while in y direction it will accelerate due to gravity

7 0
3 years ago
Springfield's "classic rock" radio station broadcasts at a frequency of 102.1 mhz. what is the length of the radio wave in meter
Mila [183]
The frequency of the radio wave is:
f=102.1 MHz = 102.1 \cdot 10^6 Hz

The wavelength of an electromagnetic wave is related to its frequency by the relationship
\lambda= \frac{c}{f}
where c is the speed of light and f the frequency. Plugging numbers into the equation, we find
\lambda= \frac{3 \cdot 10^8 m/s}{102.1 \cdot 10^6 Hz}= 2.94 m
and this is the wavelength of the radio waves in the problem.
7 0
3 years ago
PHYSICS CIRCUIT QUESTION PLEASE HELP!! 20 Points!
dimulka [17.4K]
This really calls for a blackboard and a hunk of chalk, but
I'm going to try and do without.

If you want to understand what's going on, then PLEASE
keep drawing visible as you go through this answer, either
on the paper or else on a separate screen.

The energy dissipated by the circuit is the energy delivered by
the battery.  We'd know what that is if we knew  I₁ .  Everything that
flows in this circuit has to go through  R₁ , so let's find  I₁  first.

-- R₃ and R₄ in series make 6Ω.
-- That 6Ω in parallel with R₂ makes 3Ω.
-- That 3Ω in series with R₁ makes 10Ω across the battery.
--  I₁ is  10volts/10Ω  =  1 Ampere.

-- R1:  1 ampere through 7Ω ... V₁ = I₁ · R₁ = 7 volts .

-- The battery is 10 volts. 
    7 of the 10 appear across R₁ .
   So the other 3 volts appear across all the business at the bottom.

-- R₂:  3 volts across it = V₂. 
           Current through it is  I₂ = V₂/R₂ = 3volts/6Ω = 1/2 Amp.

-- R3 + R4:  6Ω in the series combination
                     3 volts across it
                     Current through it is I = V₂/R = 3volts/6Ω = 1/2 Ampere

--  Remember that the current is the same at every point in
a series circuit.  I₃  and  I₄  must be the same 1/2 Ampere,
because there's no place in the branch where electrons can
be temporarily stored, no place for them to leak out, and no
supply of additional electrons.

-- R₃:  1/2 Ampere through it = I₃ .
           1/2 Ampere through 2Ω ... V₃ = I₃ · R₃ = 1 volt

-- R₄:  1/2 Ampere through it = I₄
           1/2 Ampere through 4Ω ... V₄ = I₄ · R₄ = 2 volts

Notice that  I₂  is 1/2 Amp, and (I₃ , I₄) is also 1/2 Amp.
So the sum of currents through the two horizontal branches is 1 Amp,
which exactly matches  I₁  coming down the side, just as it should.
That means that at the left side, at the point where R₁, R₂, and R₃ all
meet, the amount of current flowing into that point is the same as the
amount flowing out ... electrons are not piling up there.

Concerning energy, we could go through and calculate the energy
dissipated by each resistor and then addum up.  But why bother ?
The energy dissipated by the resistors has to come from the battery,
so we only need to calculate how much the battery is supplying, and
we'll have it.

The power supplied by the battery  = (voltage) · (current)

                                                         =  (10 volts) · (1 Amp) = 10 watts .

"Watt" means "joule per second".
The resistors are dissipating 10 joules per second,
and the joules are coming from the battery.

             (30 minutes) · (60 sec/minute)  =  1,800 seconds

             (10 joules/second) · (1,800 seconds)  =  18,000 joules  in 30 min

The power (joules per second) dissipated by each individual resistor is

                       P  =  V² / R
             or
                       P  =  I² · R ,

whichever one you prefer.  They're both true.

If you go through the 4 resistors, calculate each one, and addum up, you'll
come out with the same 10 watts / 18,000 joules total. 

They're not asking for that.  But if you did it and you actually got the same
numbers as the battery is supplying, that would be a really nice confirmation
that all of your voltages and currents are correct.
7 0
3 years ago
The pressure in a container will stay the same if you ——?
elena55 [62]
I think it’s D...not sure tho !
8 0
3 years ago
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