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2 years ago

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You are asked to measure nondestructively the yield strength and tensile strength of an annealed 65–45–12 cast iron structural m

ember. Fortunately, a small hardness indentation in this structural design will not impair its future usefulness, which is a working definition of nondestructive. A 10-mm-diameter tungsten carbide sphere creates a 4.26-mm-diameter impression under a 3000-kg load. What are the yield and tensile strengths?

1 answer:

anzhelika [568]2 years ago

5 0

Answer:

The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.

Explanation:

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Gina is about to use a fire extinguisher on a small fire. What factor determines the type of extinguisher she should use

amm1812

There’s 5 different types of fire extinguishers that you can differentiate by their color codes.

Red - Water based

Creme - Foam based

Blue - Powder based

Black - CO2 or carbon dioxide based

Yellow - Wet chemical based

What would determine the type of fire extinguisher used would be the class of fire it is.

Class A - Combustible materials ( i.e. paper, wood) Extinguishers to use - Red, Creme, Blue, and Yellow. (Do not use Black)

Class B - Flammable liquids ( i.e. paint, petrol, alcohol) Extinguishers to use - Creme, Blue, and Black. (Do not use Red or Yellow)

Class C - Flammable gases ( i.e. butane, methane) Extinguishers to use - Blue (Do not use Red, Creme, Black or Yellow)

Class D - Flammable metals ( i.e. lithium, potassium) Extinguishers to use - Blue (Do not use Red, Creme, Black or Yellow)

Class F - Deep fat fryers ( i.e. chip pans) Extinguishers to use - Yellow (Do not use Red, Creme, Blue or Black)

Electrical - any sort of electrical equipment

( i.e. computers, generators) Extinguishers to use - Blue and Black (Do not use Red, Creme or Yellow)

8 0

2 years ago

Read 2 more answers

A thin, flat plate that is 0.2 m × 0.2 m on a side is oriented parallel to an atmospheric airstream having a velocity of 40 m/s.

Leto [7]

Answer:

The rate of heat transfer from both sides of the plate to the air is 240 W

Explanation:

Given;

area of the flat plate = 0.2 m × 0.2 m = 0.04 m²

velocity of atmospheric air stream, v = 40 m/s

drag force, F = 0.075 N

The rate of heat transfer from both sides of the plate to the air:

q = 2 [h'(A)(Ts -T∞)]

where;

h' is heat transfer coefficient obtained from Chilton-Colburn analogy

$h' = \frac{C_f}{2} \rho u C_p P_r^{-2/3}\\\\\frac{C_f}{2} = \frac{\tau'_s}{2*\rho u^2/2}$

Properties of air at 70°C and 1 atm:

ρ = 1.018 kg/m³, cp = 1009 J/kg.K, Pr = 0.7, v = 20.22 x 10⁻⁶ m²/s

$\frac{C_f}{2} = \frac{(0.075/2)/(0.2)^2}{2*(1.018)(40)^2/2} = 5.756*10^{-4}\\\\Thus,\\h' = 5.756*10^{-4} (1.018*40*1009)*(0.7)^{-2/3}\\\\h' = 30 \ W/m^2.K$

Finally;

q = 2 [ 30(0.04)(120 - 20) ]

q = 240 W

Therefore, the rate of heat transfer from both sides of the plate to the air is 240 W

4 0

2 years ago

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) A given system has four sensors that can produce an output of 0 or 1. The system operates properly when exactly one of the sen

Rashid [163]

A given system has four sensors that can produce an output of 0 or 1. The system operates proper . An alarm must be raised when two or more sensors have the output of 1. Design the simplest circuit that can be used to raise the alarm ly when exactly one of the sensors has its output equal to Repeat problem #4 for a system that has 7 sensors. Hint: Before you slog through a truth table with 128 rows in it, think about whether SOP or POS might be a better approach.

7 0

2 years ago

A spring-loaded toy gun is used to shoot a ball of mass m = 1.50 kg straight up in the air. The spring has spring constant k = 6

adell [148]

Answer:

1) a) Mechanical energy is conserved because no dissipative forces perform work on the ball.

2) The muzzle velocity of the ball is approximately 5.272 meters per second.

3) The maximum height of the ball is 1.417 meters.

Explanation:

1) Which of the following statements are true?

a) Mechanical energy is conserved because no dissipative forces perform work on the ball.

True, statement indicates that there is no air resistence and no friction between ball and the inside of the gun because the first never touches the latter one.

b) The forces of gravity and the spring have potential energies associated with them.

False, force of gravity do work on the ball and spring receives a potential energy at being deformated by the ball.

c) No conservative forces act in this problem after the ball is released from the spring gun.

False, the absence of no conservative forces is guaranteed for the entire system according to the statement of the problem.

2) According to the statement, we understand that spring is deformed and once released and just after reaching its equilibrium position, the muzzle velocity is reached. As spring deformation is too small in comparison with height, we can neglect changes in gravitational potential energy. By Principle of Energy Conservation, we describe the motion of the ball by the following expression:

$U_{k, 1}+K_{1}=U_{k,2}+K_{2}$ (Eq. 1)

Where:

$U_{k,1}$ , $U_{k,2}$ - Initial and final elastic potential energies of spring, measured in joules.

$K_{1}$ , $K_{2}$ - Initial and final translational kinetic energies of the ball, measured in joules.

After using definitions of elastic potential and translational kinetic energies, we expand the equation above as:

$\frac{1}{2}\cdot m\cdot (v_{2}^{2}-v_{1}^{2}) = \frac{1}{2}\cdot k\cdot (x_{1}^{2}-x_{2}^{2})$

And the final velocity is cleared:

$m\cdot (v_{2}^{2}-v_{1}^{2}) = k\cdot (x_{1}^{2}-x_{2}^{2})$

$v_{2}^{2}-v_{1}^{2} =\frac{k}{m}\cdot (x_{1}^{2}-x_{2}^{2})$

$v_{2}^{2} =v_{1}^{2}+\frac{k}{m}\cdot (x_{1}^{2}-x_{2}^{2})$

$v_{2} = \sqrt{v_{1}^{2}+\frac{k}{m}\cdot (x_{1}^{2}-x_{2}^{2}) }$ (Eq. 2)

Where:

$v_{1}$ , $v_{2}$ - Initial and final velocities of the ball, measured in meters per second.

$k$ - Spring constant, measured in newtons per meter.

$m$ - Mass of the ball, measured in kilograms.

$x_{1}$ , $x_{2}$ - Initial and final position of spring, measured in meters.

If we know that $v_{1} = 0\,\frac{m}{s}$ , $k = 667\,\frac{N}{m}$ , $m = 1.50\,kg$ , $x_{1} = -0.25\,m$ and $x_{2} = 0\,cm$ , the muzzle velocity of the ball is:

$v_{2} =\sqrt{\left(0\,\frac{m}{s} \right)^{2}+\left(\frac{667\,\frac{N}{m} }{1.50\,kg} \right)\cdot [(-0.25\,m)^{2}-(0\,m)^{2}]}$

$v_{2}\approx 5.272\,\frac{m}{s}$

The muzzle velocity of the ball is approximately 5.272 meters per second.

3) After leaving the toy gun, the ball is solely decelerated by gravity. We construct this model by Principle of Energy Conservation:

$U_{g,2}+K_{2} = U_{g,3}+K_{3}$ (Eq. 3)

Where:

$U_{g,2}$ , $U_{g,3}$ - Initial and gravitational potential energies of the ball, measured in joules.

$K_{2}$ , $K_{3}$ - Initial and final translational kinetic energies of the ball, measured in joules.

After applying definitions of gravitational potential and translational kinetic energies, we expand the equation above and solve the resulting for the final height:

$m\cdot g \cdot (h_{3}-h_{2}) = \frac{1}{2}\cdot m \cdot (v_{2}^{2}-v_{3}^{2})$

$h_{3}-h_{2}=\frac{v_{2}^{2}-v_{3}^{2}}{2\cdot g}$

$h_{3} = h_{2} +\frac{v_{2}^{2}-v_{3}^{2}}{2\cdot g}$ (Eq. 4)

$h_{2}$ , $h_{3}$ - Initial and final heights of the ball, measured in meters.

$v_{2}$ , $v_{3}$ - Initial and final velocities of the ball, measured in meters per second.

$g$ - Gravitational acceleration, measured in meters per square second.

If we get that $v_{2} = 5.272\,\frac{m}{s}$ , $v_{3} = 0\,\frac{m}{s}$ , $h_{2} = 0\,m$ and $g = 9.807\,\frac{m}{s^{2}}$ , the maximum height of the ball is:

$h_{3} = 0\,m+\frac{\left(5.272\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$h_{3} = 1.417\,m$

The maximum height of the ball is 1.417 meters.

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Evan notices a small fire in his workplace. Since the fire is small and the atmosphere is not smoky he decides to fight the fire

Norma-Jean [14]

Answer:

not calling the firemean

Explanation:

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