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NeTakaya
3 years ago
15

 A container of gas is at a pressure of 3.7 × 105 Pa How much work is done by the gas if its volume expands by 1.6 m3?

Physics
1 answer:
cupoosta [38]3 years ago
5 0

Answer:

Work is done by the gas  = 5.92 x 10⁵ J = 592 kJ

Explanation:

Work done at fixed pressure, W = PΔV

Pressure, P = 3.7 x 10⁵ Pa

Change in volume, ΔV = 1.6 m³

Substituting the values of pressure and change in volume we will get

Work done at fixed pressure, W = PΔV =  3.7 x 10⁵ x 1.6 = 5.92 x 10⁵ J

Work is done by the gas  = 5.92 x 10⁵ J = 592 kJ

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How do geologist determine the age of a fault line? What types of evidence is specifically used?
Dmitriy789 [7]
To determine the relative age of different rocks, geologists start with the assumption that unless something has happened, in a sequence of sedimentary rock layers, the newer rock layers will be on top of older ones. This is called the Rule of Superposition. ... With absolute age dating, you get a real age in actual
6 0
3 years ago
Which type of wave is a sound wave ?
mestny [16]

Technically, both B and D are correct when transmitted through solids, but your answer (and the answer I got from taking the test) will be

D) Longitudinal


Hope this helps!

4 0
3 years ago
The beautiful Multnomah Falls in Oregon are approximately 206m high. If the Columbia river is flowing horizontally at 2.90m/s ju
zhenek [66]

Answer:

The overall velocity of the water when it hits the bottom is:

v_f=63.61\ \frac{m}{s}

Explanation:

Use the law of conservation of energy.

Call it instant [1] to the moment when the water is just before reaching the falls.

At this moment its height h is 206 meters and its velocity  horizontally v_i is v_i = 2.90m/s.

At the instant [1] the water has gravitational power energy E_g

E_g = mgh

The water also has kinetic energy Ek.

E_k = 0.5mv_i ^ 2

Then the Total E1 energy is:

E_1 = mgh + 0.5mv_i ^ 2

In the instant [2] the water is within an instant of touching the ground. At this point it only has kinetic energy, since the height h = 0. However at time [2] the water has maximum final velocity v_f

So:

E_2 = 0.5mv_f ^ 2

As the energy is conserved then E_1 = E_2

mgh + 0.5mv_i ^ 2 = 0.5mv_f ^ 2

Now we solve for v_f.

gh + 0.5v_i ^ 2 = 0.5v_f ^ 2\\\\9.8(206) + 0.5(2.90) ^ 2 = 0.5v_f 2\\\\v_f^2 = \frac{9.8(206) + 0.5(2.90) ^ 2}{0.5}\\\\v_f = \sqrt{\frac{9.8(206) + 0.5(2.90) ^ 2}{0.5}}\\\\v_f=63.61\ \frac{m}{s}

7 0
3 years ago
What does Δx stand for? What SI units are used to measure it?
pav-90 [236]

Answer:

Displacement. Units of measure metre (m)

Explanation:

Delta x Δx stands for displacement, or change in position.

4 0
3 years ago
Light of wavelength 630 nm is incident on a long, narrow slit. Determine the angular deflection of the first diffraction minimum
asambeis [7]

Answer:

a) 1.8°

b) 0.18°

c) 0.018°

Explanation:

Wavelength (λ) = 630nm = 630 *10^-9m

The equation that describes the angular deflection of a dark band is

Wsin(βm) = mλ

w = width of the single slit

λ = wavelength of the light

βm = angular deflection of the mth dark band.

a) In order to get the angular deflection of the first dark band for a slit with 0.02mm width, substitute w = 0.02mm = 0.02*10^-3 , m = 1 , λ= 630*10^-9

0.02*10^-3 sin(β1) = 1 * 630*10^-9

Sin(β1) = 630 * 10^-9 / 0.02*10^-3

Sin(β1) = 0.0315

β1 = Sin^-1(0.0315)

= 1.8°

b) substitute w = 0.2mm = 0.2*10^-3 , m = 1 , λ= 630*10^-9

0.2*10^-3 sin(β1) = 1 * 630*10^-9

Sin(β1) = 630 * 10^-9 / 0.2*10^-3

Sin(β1) = 0.00315

β1 = Sin^-1(0.00315)

= 0.18°

c) substitute w = 2mm = 2*10^-3 , m = 1 , λ= 630*10^-9

2*10^-3 sin(β1) = 1 * 630*10^-9

Sin(β1) = 630 * 10^-9 / 2*10^-3

Sin(β1) = 3.15*10^-4

β1 = Sin^-1(3.15*10^-4)

= 0.018°

6 0
3 years ago
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