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3 years ago

An atomic mass unit is equal to

Physics

2 answers:

Tresset [83]3 years ago

8 0

Answer: c. one-twelfth the mass of a carbon-12 atom.

Explanation: An atomic mass unit ( amu) is defined as $\frac{1}{12}$ of the mass of an atom of carbon with mass number 12 , which is the most abundant isotope of carbon. Thus the carbon-12 atom has six protons and six neutrons in its nucleus.

1 amu is approximately equal to $1.67377\times 10^{-27}kg$ or $1.67377\times 10^{-24} g$ . The mass of an atom in amu is roughly equal to the sum of the number of protons and neutrons in the nucleus.

amu is used to differentiate between isotopes as they would have different mass.

elixir [45]3 years ago

6 0

One-twelfth the mass of carbon-12 atom

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Thermodynamic Processes: An ideal gas is compressed isothermally to one-third of its initial volume. The resulting pressure will

djyliett [7]

Answer:

The resulting pressure is 3 times the initial pressure.

Explanation:

The equation of state for ideal gases is described below:

$P\cdot V = n \cdot R_{u}\cdot T$ (1)

Where:

$P$ - Pressure.

$V$ - Volume.

$n$ - Molar quantity, in moles.

$R_{u}$ - Ideal gas constant.

$T$ - Temperature.

Given that ideal gas is compressed isothermally, this is, temperature remains constant, pressure is increased and volume is decreased, then we can simplify (1) into the following relationship:

$P_{1}\cdot V_{1} = P_{2}\cdot V_{2}$ (2)

If we know that $\frac{V_{2}}{V_{1}} = \frac{1}{3}$ , then the resulting pressure of the system is:

$P_{2} = P_{1}\cdot \left(\frac{V_{1}}{V_{2}} \right)$

$P_{2} = 3\cdot P_{1}$

The resulting pressure is 3 times the initial pressure.

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2 years ago

A 300N box on a 43 degree angle.

Ksenya-84 [330]

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2 years ago

Two steel balls, of masses m1=1.00 kg and m2=2.00 kg, respectively, are hung from the ceiling with light strings next to each ot

Zolol [24]

Answer:

(a) The maximum height achieved by the first ball, m₁ is 0.11 m

(a) The maximum height achieved by the second ball, m₂ ball is 0.44 m

Explanation:

Given;

mass of the first ball, m₁ = 1 kg

mass of the second ball, m₂ = 2 kg

The velocity of the first when released from a height of 1 m before collision;

u₁² = u₀² + 2gh

u₀ = 0, since it was released from rest

u₁² = 2gh

u₁² = 2 x 9.8 x 1

u₁² = 19.6

u₁ = √19.6

u₁ = 4.427 m/s

The velocity of the second ball before collision, u₂ = 0

Apply the principle of conservation of linear momentum, to determine the velocity of the balls after an elastic collision.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

where;

v₁ is the final velocity of the first ball after an elastic collision

v₂ is the final velocity of the second ball after an elastic collision

m₁u₁ + m₂(0) = m₁v₁ + m₂v₂

m₁u₁ = m₁v₁ + m₂v₂

1 x 4.427 = v₁ + 2v₂

v₁ + 2v₂ = 4.427

v₁ = 4.427 - 2v₂ ----- equation (1)

one directional velocity;

u₁ + v₁ = u₂ + v₂

u₂ = 0

u₁ + v₁ = v₂

v₁ = v₂ - u₁

v₁ = v₂ - 4.427 ------ equation (2)

Substitute v₁ into equation (1)

v₂ - 4.427 = 4.427 - 2v₂

3v₂ = 4.427 + 4.427

3v₂ = 8.854

v₂ = 8.854 / 3

v₂ = 2.95 m/s (→ forward direction)

v₁ = v₂ - 4.427

v₁ = 2.95 - 4.427

v₁ = - 1.477 m/s

v₁ = 1.477 m/s ( ← backward direction)

Apply the law of conservation of mechanical energy

$mgh_{max} = \frac{1}{2}mv_{max}^2$

(a) The maximum height achieved by the first ball (v₁ = 1.477 m/s)

$mgh_{max} = \frac{1}{2}mv_{max}^2 \\\\gh_{max} = \frac{1}{2}v_{max}^2\\\\ h_{max} = \frac{1}{2g}v_{max}^2\\\\ h_{max} = \frac{1}{2*9.8}(1.477^2)\\\\ h_{max} = 0.11 \ m$

(b) The maximum height achieved by the second ball (v₂ = 2.95 m/s)

$mgh_{max} = \frac{1}{2}mv_{max}^2 \\\\gh_{max} = \frac{1}{2}v_{max}^2\\\\ h_{max} = \frac{1}{2g}v_{max}^2\\\\ h_{max} = \frac{1}{2*9.8}(2.95^2)\\\\ h_{max} = 0.44 \ m$