The component has an exponentially distributed reliability with a mean of 2000 hours what is the probability that it will fail after 3000 hours?

Answers

Answer 1

Answer:

ABCDEFGHIJKLMNOPQRSTUVWXYZ


Related Questions

Miriam is doing a measurement with her multimeter and the LCD is showing Hz. What's she measuring?
A. Resistance
B. Amplitude
C. Voltage
D. Frequency

Answers

Answer:

voltage

Explanation:

because it is used with a moving pointer to display readings.

Why is not adjust the depth of cut in the center of work piece in facing turning?

Answers

Answer:

I am not sure it's confusing

will anyone give me some food I am hungry

Answers

Answer:

ok

Explanation:

Answer:

yeah you will have it sir

Suppose a population of rabbits is introduced to an environment that has hot summers and extremely cold winters. Every winter, many rabbits die because of the cold and the lack of food. The rabbits have a range of ear sizes. Small Ears Medium Ears Large Ears Rabbits lose a lot of their body heat through their ears. Every winter, more of the large-eared rabbits die than the others. Every summer, more of the short-eared rabbits die than the others. At first, there are an equal number of rabbits with each size of ears. After several years, what will the frequency of ear sizes in the population be like?

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Answer:

medium sized ears

Explanation:

Does anyone know this

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I tired but I can’t sorry

For the siphon shown in Figure, determine the flowrate out of the siphon and the absolute
pressure at the crest of the siphon assuming that there is no losses through the pipe.
2.00 m
Del
50-mm-diameter
Oil (s.g. =0.82)
5.00 m

Answers

The awsner would be 3.00 because you would subtract by 2

ln a circuit a voltage of 15 volts is used to dereve a current of 5amps what is the resistance of the conductor​

Answers

Answer:

Resistance = 3 Ω

Explanation:

As we know that, By Ohm's Law

V = IR

where

V is the voltage in volt

I is the current in amps

R is the resistance in ohm

Now,

Given that,

V = 15 volts

I = 5 amps

So,

R = V/I

  = 15/5

 = 3 Ω

⇒R = 3 Ω

A dryer is shaped like a long semi-cylindrical duct of diameter 1.5 m. The base of the dryer is occupied with water-soaked materials to be dried. The base is maintained at a temperature of 370K, while the dome of the dryer is maintained at 1000 K. If both surfaces behave as blackbody, determine the drying rate per unit length experienced by the wet materials.

Answers

Answer:

0.0371 kg/s.m

Explanation:

From the given information, let's have an imaginative view of the semi-cylinder; (The image is shown below)

Assuming the base surface of both ends of the cylinder is denoted by:

[tex]A_1 \ and \ A_2[/tex]

Thus, using the summation rule, the view factor [tex]F_{11[/tex] and [tex]F_{12[/tex] is as follows:

[tex]F_{11}+F_{12}=1[/tex]

Let assume the surface (1) is flat, the [tex]F_{11} = 0[/tex]

Now:

[tex]0+F_{12}=1[/tex]

[tex]F_{12}=1[/tex]

However, using the reciprocity rule to determine the view factor from the dome-shaped cylinder [tex]A_2[/tex] to the flat base surface [tex]A_1[/tex]; we have:

[tex]A_2F_{21} = A_{1}F_{12} \\ \\ F_{21} = \dfrac{A_1}{A_2}F_{12}[/tex]

Suppose, we replace DL for [tex]A_1[/tex] and

[tex]A_2[/tex] =  [tex]\dfrac{\pi D}{2}[/tex]

Then:

[tex]F_{21} = \dfrac{DL}{(\dfrac{\pi D}{2}) L} \times 1 \\ \\ =\dfrac{2}{\pi} \\ \\ =0.64[/tex]

Now, we need to employ the use of energy balance formula to the dryer.

i.e.

[tex]Q_{21} = Q_{evaporation}[/tex]

But, before that;  let's find the radian heat exchange occurring among the dome and the flat base surface:

[tex]Q_{21}= F_{21} A_2 \sigma (T_2^4-T_1^4) \\ \\ Q_{21} = F_{21} \times \dfrac{\pi D}{2} \sigma (T_2^4 -T_1^4)[/tex]

where;

[tex]\sigma = Stefan \ Boltzmann's \ constant[/tex]

[tex]T_1 = base \ temperature[/tex]

[tex]T_2 = temperature \ of \ the \ dome[/tex]

[tex]Q_{21} = 0.64 \times (\dfrac{\pi}{2}\times 1.5) \times 5.67 \times 10^4 \times (1000^4 -370^4)\\ \\ Q_{21} = 83899.15 \ W/m[/tex]

Recall the energy balance formula;

[tex]Q_{21} = Q_{evaporation}[/tex]

where;

[tex]Q_{evaporation} = mh_{fg}[/tex]

here;

[tex]h_{fg}[/tex] = enthalpy of vaporization

m = the water mass flow rate

[tex]83899.15 = m \times 2257 \times 10^3 \\ \\ m = \dfrac{83899.15}{ 2257 \times 10^3 }\\ \\ \mathbf{m = 0.0371 \ kg/s.m}[/tex]

The drying rate per unit length is 0.037 kg/S.m

Given data;

Base temperature (T1) = 370KTemperature of the dome (T2) = 1000KF[tex]_1_2[/tex] = 1.5mD = 1.5mBoltzmann's constant (δ) =  [tex]5.67 * 10 ^-^8 W/m^2.K^4[/tex]  

From the attached diagram, the surface 1 is flat, it is a view factor, f[tex]_1_1[/tex] = 0.

Applying summation rule and solving the view factor from the base surface A[tex]_1[/tex] to the cylindrical dome A[tex]_2[/tex].

[tex]f_1_1+f_1_2=1[/tex]

Put F[tex]_1_2=0[/tex]

[tex]0+f_1_2=1[/tex]

This makes [tex]f_1_2=1[/tex]

Applying reciprocal rule and solving the view factor from the cylindrical dome A[tex]_2[/tex] to the base surface A[tex]_1[/tex].

[tex]A_2F_2_1=A_1F_1_2\\F_2_1=(\frac{A_1}{A_2})F_1_2[/tex]

Where A is the area of the surface.

Substitute DL for A[tex]_1[/tex] and [tex]\frac{\pi D}{2}[/tex] for A[tex]_2[/tex]

[tex]F_2_1 = \frac{DL}{(\frac{\pi D}{2}})L *1 = \frac{2 }{\pi } =2/3.14 = 0.64[/tex]

Using the energy balance equation to the dryer,

[tex]Q_2_1=Q_e_v_a_p[/tex]

Let's calculate the radiation heat exchange between the dome and the base surface per unit length by using the equation below

[tex]Q_2_1=F_2_1A_2[/tex]δ[tex](T_2^4-T_1^4)[/tex]

[tex]Q_2_1= F_2_1 * \frac{\pi D}{2}[/tex]δ[tex](T_2^4-T_1^4)[/tex]

substitute the respective values into the equation

[tex]Q_2_1=0.64*(\frac{\pi }{2}*1.5)*5.67*10^-^8*(1000^4-370^4)\\Q_2_1=8.3899.15W/m[/tex]

Mass flow rate

Let's calculate the mass flow rate of water using the amount of heat required for drying up.

[tex]Q_2_1=Q_e_v_a_p\\Q_e_v_a_p=mh_f_g\\[/tex]

where [tex]h_f_g= 2257*10^3J/kg[/tex]

and this is the enthalpy of vaporization and mass flow rate of water.

[tex]83899.15=m*2257*10^3\\m=0.037kg/S.m[/tex]

The drying rate per unit length is 0.037kg/S.m

Learn more about mass flow rate here

https://brainly.com/question/24356835

https://brainly.com/question/24315432

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