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3 years ago

At a certain elevation, the , the air becomes saturated and water-vapor molecules .

1 answer:

4 0

At certain altitude, the temperature of air decrease, The air becomes saturated and water vapour molecules starts condensing.

As the altitude of air increase, the atmospheric pressure decrease due to which the temperature of the air decrease. The water molecules in the atmosphere start condensing, which saturate the air (that is air can no hold water molecules), due to which the water vapour molecules starts condensing and falls on the earth in the form of rain.

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A city bus travels along its route from Jackson Street 8 km to Aurora Avenue, where it turns and continues 13 km.

Elina [12.6K]

im a bit confused on what the question is:(

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3 years ago

How can a 1kg ball have more kinetic energy than a 100kg ball? Explain both using words and by providing a numerical example

MariettaO [177]

1 kg ball can have more kinetic energy than a 100 kg ball as increase in velocity is having greater impact on K.E than increase in mass.

<u>Explanation</u>:

We know kinetic energy can be judged or calculated by two parameters only which is mass and velocity. As kinetic energy is directly proportional to the $(velocity)^2$ and increase in velocity leads to greater effect on translational Kinetic Energy. Here formula of Kinetic Energy suggests that doubling the mass will double its K.E but doubling velocity will quadruple its velocity:

$\text { Kinetic Energy }=\frac{1}{2} m v^{2}$

Better understood from numerical example as given:

If a man A having weight 50 kg run with speed 5 m/s and another man B having 100 kg weight run with 2.5 m / s. Which man will have more K.E?

This can be solved as follows:

$\text { Kinetic Energy of } \mathrm{A}=\frac{1}{2} 50 \times 5^{2}=625 \mathrm{J}$

$\text { Kinetic Energy bf } \mathrm{B}=\frac{1}{2} 100 \times 2.5^{2}=312.5 \mathrm{J}$

It shows that man A will have more K.E.

Hence 1 kg ball can have more K.E than 100 kg ball by doubling velocity.

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3 years ago

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3 years ago

8. Mark is sitting in a chair, legs out in front of him resting on an ottoman. His knees are.....

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3 years ago

Read 2 more answers

An iron wire has a cross-sectional area equal to 5.00×10⁻⁶ m² . Carry out the following steps to determine the drift speed of th

Doss [256]

In mass, there are 55.85 × 10⁻³ kg/mol in in 1 mole of iron.
The molar density of iron is equal to 1.41 × 10⁵ mol/m³.
The density of iron atoms is equal to 8.49 × 10²⁸ atoms/m³.
The number density of conduction electrons is equal to 1.70 × 10²⁹ conduction electrons/m³.
The drift speed of conduction electrons is equal to 2.21 × 10⁻⁴ m/s.

<h3>How to calculate the drift speed of the conduction electrons?</h3>

Mathematically, the drift speed of the conduction electrons can be calculated by using this formula:

V = (m × σ × V)/ρ × e × f × l)

V = I/(n × A × Q)

Where:

U represents the drift speed of the conduction electrons, in m/s.

m represents the molecular mass of the metal, in kg.

e represents the elementary charge, in C.

f represents the number of free electrons per atom.

σ represents the electric conductivity of the medium at a particular temperature in S/m.

ρ represents the density of the conductor, in kg/m³.

ℓ represents the length of the conductor, in m.

ΔV represents the voltage applied or potential difference across the conductor in V.

<h3>How many kilograms are there in 1 mole of iron? </h3>

Molar mass of iron = 55.85 g/mol.

In Kilograms, we have:

Mass = 55.85 × 1/1000

Mass = 55.85 × 10⁻³ kg/mol.

For the molar density of iron, we have:

Molar density = density/molar mass

Molar density = 7874/0.056

Molar density = 1.41 × 10⁵ mol/m³.

For the density of iron atoms, we have:

Density of iron atoms = Avogadro's constant × molar density

Density of iron atoms = 6.023 × 10²³ × 1.406 × 10⁵

Density of iron atoms = 8.49 × 10²⁸ atoms/m³.

For the number density of conduction electrons, we have:

Fe ---> Fe²⁺ + 2e⁻

Number density of conduction electrons = 2 conduction electrons/1 atom of iron

Number density of conduction electrons = 2 × 8.49 × 10²⁸

Number density of conduction electrons = 1.70 × 10²⁹ conduction electrons/m³.

For the drift speed of conduction electrons, we have:

V = I/(n × A × Q)

V = 30/(1.70 × 10²⁹ × 1.602 × 10⁻¹⁹ × 5 × 10⁻⁶)

Drift speed, V = 2.21 × 10⁻⁴ m/s.

Read more on drift speed here: brainly.com/question/15219891

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Complete Question:

An iron wire has a cross-sectional area of 5.00 x 10-6 m2. Carry out steps (a) through (e) to compute the drift speed of the conduction electrons in the wire.

(a) How many kilograms are there in 1 mole of iron?

(b) Starting with the density of iron and the result of part (a), compute the molar density of iron (the number of moles of iron per cubic meter).

(d) Obtain the number density of conduction electrons given that there are two conduction electrons per iron atom.

(e) If the wire carries a current of 30.0 A, calculate the drift speed of conduction electrons.

4 0

2 years ago

At a certain elevation, the ________ , the air becomes saturated and water-vapor molecules ________.

At a certain elevation, the , the air becomes saturated and water-vapor molecules .