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Elena-2011 [213]
3 years ago
11

The concentration of ozone in los angeles is 0.67 ppm on a summer day. what is the partial pressure of o3 if the total pressure

is 735 torr?
Physics
1 answer:
Katarina [22]3 years ago
5 0
We use the Raoult's Law to answer this question.

Partial Pressure = Mole Fraction*Total Pressure

When concentration is expressed in ppm, it means 'parts per million'. So, there can be 0.67 mol of ozone per 10⁶ mol of mixture.

Mole fraction = 0.67/10⁶ = 0.67×10⁻⁶

Partial pressure = 0.67×10⁻⁶(735 torr) = <em>4.92×10⁻⁴ torr</em>
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A 1.05 kg block slides with a speed of 0.865 m/s on a frictionless horizontal surface until it encounters a spring with a force
djyliett [7]

Answer:

a) U = 0 J    

k = 0.393 J

E = 0.393 J

b) U = 0.0229J

k = 0.370 J

E = 0.393 J

c) U = 0.0914 J

k = 0.302 J

E = 0.393 J

d) U = 0.206 J

k = 0.187 J

E = 0.393 J

e) U = 0.366 J

k = 0.027 J

E = 0.393 J

Explanation:

Hi there!

The equations of kinetic energy and elastic potential energy are as follows:

k = 1/2 · m · v²

U = 1/2 · ks · x²

Where:

m = mass of the block.

v = velocity.

ks = spring constant.

x = displacement of the string.

a) When the spring is not compressed, the spring potential energy will be zero:

U = 1/2 · ks · x²

U = 1/2 · 457 N/m · (0 cm)²

U = 0 J

The kinetic energy of the block will be:

k = 1/2 · m · v²

k = 1/2 · 1.05 kg · (0.865 m/s)²

k = 0.393 J

The mechanical energy will be:

E = k + U = 0.393 J + 0 J = 0.393 J

This energy will be conserved, i.e., it will remain constant because there is no work done by friction nor by any other dissipative force (like air resistance). This means that the kinetic energy will be converted only into spring potential energy (there is no thermal energy due to friction, for example).

b) The spring potential energy will be:

U = 1/2 · 457 N/m · (0.01 m)²

U = 0.0229 J

Since the mechanical energy has to remain constant, we can use the equation of mechanical energy to obtain the kinetic energy:

E = k + U

0.393 J = k + 0.0229 J

0.393 J - 0.0229 J = k

k = 0.370 J

c) The procedure is now the same. Let´s calculate the spring potential energy with x = 0.02 m.

U = 1/2 · 457 N/m · (0.02 m)²

U = 0.0914 J

Using the equation of mechanical energy:

E = k + U

0.393 J = k + 0.0914 J

k = 0.393 J - 0.0914 J = 0.302 J

d) U = 1/2 · 457 N/m · (0.03 m)²

U = 0.206 J

E = 0.393 J

k = E - U = 0.393 J - 0.206 J

k = 0.187 J

e) U = 1/2 · 457 N/m · (0.04 m)²

U = 0.366 J

E = 0.393 J

k = E - U = 0.393 J - 0.366 J = 0.027 J.

4 0
3 years ago
A jet airplane is in level flight. The mass of the airplane is m=9010kg. The airplane travels at a constant speed around a circu
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Answer:

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ω = 0.01353 rad/sec

The direction \theta = tan^{-1} ( \dfrac{\omega ^2 R}{g})

\theta = tan^{-1} ( \dfrac{0.01353 ^2 \times (9.77\times 1609)}{9.81})

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L = 88388.1  ÷ 0.9596

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Explanation:

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What is the relationship between the valence electrons of an atom and the chemical bonds the atom can form?​
stellarik [79]

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