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3 years ago

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A cooking pan whose inner diameter is 20 cm is filled with water and covered with a 4-kg lid. If the local atmospheric pressure

is 101 kPa, determine the temperature at which the water starts boiling when it is heated

2 answers:

sashaice [31]3 years ago

8 0

Answer:

$T=100.2^0C$

Explanation:

The boiling point of the water depends only on the absolute pressure p. the atmospheric temperature is given by P=101000 pa and the pressure from the lid we calculate by dividing its weight with the surface area of lid. the weight is product of mass and the gravitational acceleration. surface area is calculated by given diameter.

$D=20cm=20cm.\frac{1m}{100cm} =0.2m\\A=D^2\pi /4\\p_{l} =W/A=\frac{m.g}{D^2\pi/4 }$

The absolute pressure p is sum of atmospheric pressure and the pressure from the weight of lid.

$p=p_{a} +p_{l} \\\\p=p_{a} +\frac{m.g}{D^2\pi /4} \\\\p=101000Pa+\frac{4kg.9.81m/s^2}{(0.2m^2)^2.\pi /4} \\p=102249Pa$

At this pressure the boiling temperature of water is 100.2.

Arisa [49]3 years ago

7 0

Explanation:

Below is an attachment containing the solution.

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A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are 30.00 a

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Answer:

The original length of the specimen $l_{o} = 104.7 mm$

Explanation:

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Final length $l_{1}$ = 105.20 mm

Elastic modulus E = 65.5 G pa = 65.5 × $10^{3}$ M pa

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We know that the relation between the shear modulus & elastic modulus is given by

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$25.5 = \frac{65.5}{2 (1 + \mu)}$

$\mu = 0.28$

This is the value of possion's ratio.

We know that the possion's ratio is given by

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${\frac{change \ in \ length}{l_{o} } = \frac{\frac{0.04}{30} }{0.28}$

${\frac{change \ in \ length}{l_{o} } =$ 0.00476

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Final length $l_{o}$ = 105.2 m

Original length

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This is the original length of the specimen.

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Answer:

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As per standard ANSI B4.1 :

Desired Tolerance: FN 2

Tolerance TZone: H7S6

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Design is acceptable If compressive stress induced due to selected dimensions and load is less than compressive strength of the material.

Explanation:

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The thickness of aluminium needed to stop the beam electrons, protons and alpha particles at the given dfferent kinetic energies is 1.5 x 10⁻¹⁴ m.

<h3>Thickness of the aluminum</h3>

The thickness of the aluminum can be determined using from distance of closest approach of the particle.

$K.E = \frac{2KZe^2}{r}$

where;

Z is the atomic number of aluminium = 13
e is charge
r is distance of closest approach = thickness of aluminium
k is Coulomb's constant = 9 x 10⁹ Nm²/C²

<h3>For 2.5 MeV electrons</h3>

$r = \frac{2KZe^2}{K.E} \\\\r = \frac{2 \times 9\times 10^9 \times 13\times (1.6\times 10^{-19})^2}{2.5 \times 10^6 \times 1.6 \times 10^{-19}} \\\\r = 1.5 \times 10^{-14} \ m$

<h3>For 2.5 MeV protons</h3>

Since the magnitude of charge of electron and proton is the same, at equal kinetic energy, the thickness will be same. r = 1.5 x 10⁻¹⁴ m.

<h3>For 10 MeV alpha-particles</h3>

Charge of alpah particle = 2e

$r = \frac{2KZe^2}{K.E} \\\\r = \frac{2 \times 9\times 10^9 \times 13\times (2 \times 1.6\times 10^{-19})^2}{10 \times 10^6 \times 1.6 \times 10^{-19}} \\\\r = 1.5 \times 10^{-14} \ m$

Thus, the thickness of aluminium needed to stop the beam electrons, protons and alpha particles at the given dfferent kinetic energies is 1.5 x 10⁻¹⁴ m.

Learn more about closest distance of approach here: brainly.com/question/6426420

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