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aev [14]
3 years ago
5

a 12.0L container is filled with a gas to a pressure of 2660 torr at 0°C. At what temp. will the pressure inside the container b

e 3040 torr if volume is held constant?
Physics
1 answer:
Alexxx [7]3 years ago
6 0
\frac{p1}{t1} = \frac{p2}{t2}

Pressure Law: constant volume

Convert all Temperatures to Kelvin

0°C= 273K

\frac{2660}{273} = \frac{3040}{t2} \\ \\ 3040 \times 273 = t2 \times 2660 \\ \\ 829920 \div 2660 = t2 \\ \\ t2 = 312

answer= 312Kelvin
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on a aircraft carrier a jet is slowed from 105 mph to a stop in 2 seconds. What is the rate of deceleration?
Bess [88]
105/2 = 52.5/1

The rate of deceleration is 52.5 mps (miles per second).
6 0
3 years ago
Hook's law describes an ideal spring. Many real springs are better described by the restoring force (FSp)s=−kΔs−q(Δs)^3, where q
kvasek [131]

Answer

given,

k = 250 N/m

q = 900 N/m³

(FSp)s=−kΔs−q(Δs)^3

work done = Force x displacement

W = \int {F. dx}

limits are x = 0 to x = 0.15 m

work done

W = \int_0^{0.15} (k x + q x^3)\ dx

W = [\dfrac{kx^2}{2}+\dfrac{qx^4}{4}+ C]_0^0.15

W = \dfrac{300\times 0.15^2}{2}+\dfrac{900\times 0.15^4}{4}

W = 3.375 + 0.1139

W = 3.3488 J

b) % cubic term =\dfrac{0.1139}{3.3488}

   % cubic term =3.4\ %

7 0
3 years ago
Why should galaxy collisions have been more common in the past than they are today?
Roman55 [17]

Answer:

I think d is the answer haha

7 0
3 years ago
(a) A 70-kg person at rest has an oxygen consumption rate Qhum = 14.5 liter/h, 2% of which is supplied by diffusion through the
nevsk [136]

Answer:

(a) fd=1.7058\times 10^{-5}\ L.hr^{-1}.cm(b) [tex]r=26.008\ cm

Explanation:

(a)

  • Oxygen consumption rate of humans, Q_h=14.5\ L.hr^{-1}

area of human skin, A_h=1.7\ m^2

  • diffusion rate through skin of humans, d=2\%\ of\ Q_h
  • ∴d=\frac{2}{100} \times 14.5

d=0.29\ L.hr^{-1}

<u>Flux of diffusion rate, </u>

fd=\frac{d}{A}

fd=\frac{0.29}{17000}

fd=1.7058\times 10^{-5}\ L.hr^{-1}.cm(b)Surface area for a spherical animal:[tex]A=4.\pi.r^2

Diffusion flux rate for animal:

fd=\frac{14.5}{A}

1.7058\times 10^{-5}=\frac{14.5}{4.\pi.r^2}

r=26.008\ cm

4 0
4 years ago
Explain what happens to the sound waves when a singer hits the high pitched notes during the National Anthem. Be sure to use the
Tasya [4]

Answer: The frequency increases as the pitch increases, and the amplitude increases as the volume increases

Explanation:

Waves have the property of:

v = f*λ

where v is the speed of the wave (which is almost constant for soundwaves, v = 340 m/s)

f is the frequency of the wave, and λ is the wavelength.

Now, we know that when the pitch of a note increases, also does the frequency of the soundwave (so the wave oscillates faster).

Now, we also want to include the amplitude of the soundwave in this.

The amplitude is related to the volume of the soundwave (actually is related to the energy, and as higher is the energy, more "loud" is the sound).

As the high pitch part is usually "louder", we can assume that we have an amplitude increase.

Then the answer would be:

"The frequency increases as the pitch increases, and the amplitude increases as the volume increases"

4 0
3 years ago
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