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4 years ago

The open-ended check is to check for ________.

Engineering

1 answer:

Step2247 [10]4 years ago

7 0

Answer:

any abnormalities

Explanation:

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Match each situation with the type of material (conductor or inductor) you would want to use in it. You need to connect a recent

Marina CMI [18]

Answer: for the following process, I will explain each process and where the material is best to be used.

1. You need to connect a recently landed plane to the Earth in order to ground it and remove the built-up precipitation static. You would want to use this kind of material:

Answer: Conductor

Explanation: for you to ground the plane, you need a conductor that can be able to direct the current down to the earth. Because electron can only flow freely in a conductor.

2. You need to move a live power line out of a puddle of water. There is a lot of charge moving through this line, and if any makes it to your hands it's going to hurt. You would want to use this kind of material:

Answer: Inductor

Explanation: an Inductor resist the flow of electric current through it. You have to use an inductor, to avoid been electrocuted by the live wire. If a conductor is used current will flow through it, which may lead to electrocutions, as the water is also a conductor of electricity.

3. You need a smooth sphere to put a sensitive piece of equipment inside that will minimize any sparks between the sphere and pieces of equipment outside the sphere. You would want to use this kind of material:

Answer: Inductor

Explanation: to avoid spark, an inductor should be used, because when they is a friction between a conductors and an electric current, they will be a spark. So an inductor should be used to avoid spark. Inductors does not give a spark when in friction with an electric current

AN INDUCTOR IS ANY MATERIAL THAT RESIST THE FLOW OF ELECTRIC CURRENT THROUGH IT.

A CONDUCTOR IS ANY MATERIAL THAT ALLOWS THE FLOW ELECTRIC CURRENT THROUGH IT.

5 0

4 years ago

Use Newton's method to determine the angle θ, between 0 and π/2 accurate to six decimal places. for which sin(θ) = 0.1. Show you

kirza4 [7]

Answer:

x3=0.100167

Explanation:

Let's find the answer.

Because we are going to find the solution for sin(Ф)=0.1 then:

f(x)=sin(Ф)-0.1 and:

f'(x)=cos(Ф)

Because 0<Ф<π/2 let's start with an initial guess of 0.001 (x0), so:

x1=x0-f(x0)/f'(0)

x1=0.001-(sin(0.001)-0.1)/cos(0.001)

x1= 0.100000

x2=0.100000-(sin(0.100000)-0.1)/cos(0.100000)

x2=0.100167

x3=0.100167

3 0

4 years ago

The end of a cylindrical liquid cryogenic propellant tank in free space is to be protected from external (solar) radiation by pl

prohojiy [21]

If the temperature of the shield is 338 kelvin. Then the heat flux through the tank will be 25.3 Watt per square meter.

The increase in heat energy movement through a particular surface is known as heat flux, and the heat flux density is the absolute temperature per unit area.

Assume the view factor between the tank and the shield is unity; all surfaces are diffuse and gray, and the surroundings are at 0 K.

It is given that T= 100 K, ε₁ = ε₂ = 0. 10, $\varepsilon_t$ = 0.20, and GS= 1250 W/m².

Then we have

The temperature of the shield will be

$\rm \alpha _sG_s - \varepsilon _1 E_b (T_s) - \dot{q}_{ST} = 0$ ...1

and

$\rm q''_{12}=\dfrac{ \sigma (T_{1}^{4} - T_{2}^{2})}{\frac{1}{\varepsilon _1 }+ \frac{1}{\varepsilon _2} -1}}$ ...2

Then from equations 1 and 2, we have

$\rm \alpha _sG_s - \varepsilon _1 E_b (T_s) - \dfrac{ \sigma (T_{1}^{4} - T_{2}^{2})}{\frac{1}{\varepsilon _1 }+ \frac{1}{\varepsilon _2} -1}} = 0$

Then the value of $\rm T_s$ will be

$\rm T_s =\left [ \dfrac{\alpha _sGs+\left ( \dfrac{\sigma T_1^4}{\frac{1}{\varepsilon _1}+\frac{1}{\varepsilon _2} - 1} \right )}{\sigma \left ( \varepsilon _1 + \dfrac{1}{\frac{1}{\varepsilon _1} + \frac{1}{\varepsilon _2}-1} \right )} \right ] ^{\dfrac{1}{4}}$

Put all the values, then we have

$\rm T_s = \left [ \dfrac{0.05 \times + \left ( \dfrac{\sigma (100)^4}{\frac{1}{0.1}+\frac{1}{0.05}-1} \right )}{\sigma \left ( 0.05 + \dfrac{1}{\frac{1}{0.1}+\frac{1}{0.05} - 1} \right )} \right ]^{\dfrac{1}{4}} \\\\\\T_s = 338 \ K$

Then the heat flux will be

$\rm q"_{ST}=\dfrac{\sigma (T_S^4 - T_t^4)}{\frac{1}{\varepsilon _1} + \frac{1}{\varepsilon _2} - 1} \\\\\\q"_{ST}=\dfrac{5.67 \times 10^{-8}(388^4-100^4)}{\frac{1}{0.1}+\frac{1}{0.05}-1}\\\\\\q"_{ST} = 25.3 \ W/m^2$

More about the heat flux link is given below.

brainly.com/question/12913016

#SPJ1

8 0

3 years ago

A thermodynamicist claims to have developed a heat pump with a COP of 1.7 when operating with thermal energy reservoirs at 273 K

Vinil7 [7]

Answer:

cfcghvjvct

Explanation:

3 0

3 years ago

Compute the volume percent of graphite, VGr, in a 3.1 wt% C cast iron, assuming that all the carbon exists as the graphite phase

Arte-miy333 [17]

Answer:

The volume percent of graphite is 9.9%

Explanation:

Given;

weight percent of graphite, C = 3.1wt%

density of ferrite, $\rho _f$ = 7.9 g/cm³

density of graphite, $\rho _g$ = 2.3 g/cm³

Determine the mass fraction;

$W_f = \frac{C_g - C_0}{C_g -C_f} \\\\W_f =\frac{100 - 3.1}{100-0}\\\\W_f = 0.969\\\\\\W_g = \frac{C_0 - C_a}{C_g -C_f} \\\\W_g =\frac{3.1-0}{100-0}\\\\W_g = 0.031$

Determine the volume fraction;

$V = \frac{W_g/ \rho_g}{W_f / \rho_f \ + \ W_g / \rho_g} *100 \%\\\\V = \frac{0.031/ 2.3}{0.969 / 7.9 \ + 0.031 / 2.3}*100\%\\\\V = 9.9\%$

Therefore, the volume percent of graphite is 9.9%

8 0

3 years ago