Answer:
4.25
Explanation:
The mean deviation is how far, on average, all values are from the middle. To do that, we subtract each element from the mean of the elements, take their absolute values, and then divide by the number of elements in the set. Here's the formula: [tex]\sum\frac{|x-\mu|}{n}[/tex]
Here the mean is 9.5, the count n = 8, so the mean deviation is 4.25
The base of an aluminum block, which is fixed in place, measures 90 cm by 90 cm, and the height of the block is 60 cm. A force, applied to the upper face and parallel to it, produces a shear strain of 0.0060. The shear modulus of aluminum is . What is the displacement of the upper face in the direction of the applied force
The question is incomplete. The complete question is :
The base of an aluminum block, which is fixed in place, measures 90 cm by 90 cm, and the height of the block is 60 cm. A force, applied to the upper face and parallel to it, produces a shear strain of 0.0060. The shear modulus of aluminum is [tex]3.0 \times 10^{10} \ Pa[/tex] . What is the displacement of the upper face in the direction of the applied force?
Solution :
The relation between shear modulus, shear stress and strain,
[tex]$\text{Shear modulus, S =} \frac{\text{Shear stress}}{\text{shear strain}}$[/tex]
Shear stress = shear modulus (S) x shear strain
[tex]$=3 \times 10^{10} \times 0.0060$[/tex]
[tex]$=1.80 \times 10^8$[/tex] Pa
[tex]$=180 \times 10^6$[/tex] Pa
[tex]$=180 \ MPa$[/tex]
The length represents the distance between the fixed in place portion and where the force is being applied.
Therefore,
[tex]$\text{Displacement} = \text{shear strain} \times \text{length}$[/tex]
= 0.006 x 60 cm
= 0.360 cm
= 3.6 mm
Thus, the displacement of the upper face is 3.6 mm in the direction of the applied force.
A security engineer deploys a certificate from a commercial CA to the RADIUS server for use with the EAP-TLS wireless network. Authentication is failing, so the engineer examines the certificate's properties:
Issuer: (A commercial CA)
Valid from: (yesterday’s date)
Valid to: (one year from yesterday’s date)
Subject: CN=smithco.com
Public key: RSA (2040 bits)
Enhanced key usage: Client authentication (1.3.6.1.5.7.3.2)
Key usage: Digital signature, key encipherment (a0)
Which of the following is the MOST likely cause of the failure?
A. The certificate is missing the proper OID.
B. The certificate is missing wireless authentication in key usage.
C. The certificate is self-signed.
D. The certificate has expired.
Answer:
The MOST likely cause of the failure is:
A. The certificate is missing the proper OID.
Explanation:
OID (Object Identifier) is a globally unique group of characters, alphanumeric or numeric identifier, registered under the ISO registration standard to reference a specific object or object class (an entity). In computing, OID allows a server or end-user to retrieve an object without identifying the specific physical data location. Since OID is system-generated, it is immutable and can only be assigned to one object. It cannot be shared. Its presence in the certificate would have enabled the security engineer to authenticate the certificate.
5-2 discussion: the role of communication in success
Answer:
Answer to the following question is as follows;
Explanation:
It is critical to communicate well during negotiations and conversation in order to attain your objectives. Communication also becomes essential in business and corporation . Effective communication may help you as well as your employees develop a strong professional relationship, which can boost morale and efficiency.
what is the term RF exiciter?
2 Air enters the compressor of a cold air-standard Brayton cycle at 100 kPa, 300 K, with a mass flow rate of 6 kg/s. The compressor pressure ratio is 10, and the turbine inlet temperature is 1400 K. The turbine and compressor each have isentropic efficiencies of 80%. For k 5 1.4, calculate (a) the thermal efficiency of the cycle. (b) the back work ratio. (c) the net power developed, in kW. (d) the rates of exergy destruction in the compressor and turbine, respectively, each in kW, for T0 5 300 K.
answer this qustion plz
Answer:
click here and I'm so happy I have been doing this weekend I can see the same way you could come back in a bit and then we are all of you to know you do for a bit of time with me or
Consider the need to insulate a flat wall using three different insulating materials, each having the same thickness. The materials have constant thermal conductivities, k, of 1, 2, and 3 Wm-1 K-1 . You are asked to advise workers on the proper installation order of the three materials so that the final composite will be the most effective at reducing heat loss from the wall. Which statement is true
Answer: The options related to your question is missing attached below is the missing option
a.) The insulation must be arranged in order of increasing k starting with the lowest k material at the wall surface.
b.) The insulation must be arranged in order of decreasing k starting with the highest k material at the wall surface.
c.) The insulation can be arranged in any order.
d.) It is impossible to decide because the temperatures are not specified and the order will depend on the ∆T.
e) The insulation with the lowest k should be nearest the wall, and the order of the other two is irrelevant.
answer:
The insulation must be arranged in order of increasing k starting with the lowest k material at the wall surface. ( A )
Explanation:
The correct statement is ; The insulation must be arranged in order of increasing k starting with the lowest k material at the wall surface.
Because the insulation has to be arranged following the order of it's increasing thermal conductivity ( K-value )
A recessed luminaire bears no marking indicating that it is ""Identified for Through- Wiring."" Is it permitted to run branch-circuit conductors other than the conductors that supply the luminaire through the integral junction box on the luminaire?
Answer:
No it is not permitted
Explanation:
It is not permitted because as per NEC 410.21 policy no other conductor is allowed to be passed through integral junction box luminaries unless such conductor supply recessed luminaries.
The marking will show that the Luminaries is of the right construction or right installation to ensure that the the conductors ( in the outer boxes ) will not be exposed to temperatures greater than the conductor rating, hence the lack of marking makes it not to be permitted.
what is an OTG USB? how is it useful
Answer:
An OTG or On The Go adapter (sometimes called an OTG cable, or OTG connector) allows you to connect a full sized USB flash drive or USB A cable to your phone or tablet through the Micro USB or USB-C charging port
Explanation:
pls mark brainliest
thì nghiệm nén xác định cường độ của bê tông trên ba mẫu thí nghiệm hình trụ HxD=300x150(mm). kết quả thu được lực phá hoại P1=45200daN, P2=46800daN, P3=46000daN. hãy xác định cường độ chịu nén của bê tông theo TCNV 3118:1993
spanish
Explanation:
the above question is written in spanish
True or false: You can create a network with two computers.
Answer
True
Write the code using the do-while loop to force the user to enter a number in the range [20,50]
Answer:
Mark as brainlist pls hello
______ is not a type of digital signaling technique
Answer:
Data Rate Signaling
an atom that gained an electron is called
Answer:
Hey mate......
Explanation:
This is ur answer.....
When an atom gains/loses an electron, the atom becomes charged, and is called an ion.Hope it helps!
Brainliest pls!
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Troy must keep track of the amount of refrigerant he uses from a 50-pound cylinder to ensure that accurate
records are kept. He used 13 pounds on a systein for Ms. Jones and 9 pounds on a system for a commercial
client. How many pounds should he have left in the cylinder?
tof
Troy should have
pounds of refrigerant left in the cylinder.
baon naid Thamar basic
محمود احمد مجد
12
اهداء ما در
Answer:
Amount of gas still in cylinder = 28 pound
Explanation:
Given:
Amount of gas in cylinder = 50 pound
Amount of gas used in Ms. Jones system = 13 pound
Amount of gas used in client system = 9 pound
Find:
Amount of gas still in cylinder
Computation:
Amount of gas still in cylinder = Amount of gas in cylinder - Amount of gas used in Ms. Jones system - Amount of gas used in client system
Amount of gas still in cylinder = 50 - 13 - 9
Amount of gas still in cylinder = 28 pound
A steel component with ultimate tensile strength of 800 MPa and plane strain fracture toughness of 20 MPam is known to contain a tunnel (internal) crack of length 1.4 mm. This alloy is being considered for use in a cyclic loading application for which the design stresses vary from 0 to 410 MPa. Would you recommend this alloy for this application
Complete question:
A steel component with a tensile strength of 800 MPa and fracture toughness Kic=20 MPa Nm is known to contain internal cracks (also called tunnel cracks) with the maximum length of 1.4 mm. This steel is being considered for use in a cyclic loading application for which the designed stresses vary from 0 to 420 MPa. Would you recommend using this steel in this application?
a. Not sure. Because cyclic loading is applied. Fatigue test is needed in order to make the recommendation.
b. Yes, this because the tensile strength of steel is much higher than the applied highest stress of 420 MPa.
c. Yes, this because the calculated critical stress to fracture for the cracks is higher than the highest applied stress of 420 MPa and the steel can withstand the stress of 420 MPa.
d. No. Although the calculated critical stress to fracture for the cracks is slightly higher than the highest applied stress of 420 MPa and the steel may withstand the static stress of 420 MPa, the cyclic loading may cause rapid fatigue fracture.
Answer:
A. Not sure. Because cyclic loading is applied. Fatigue test is needed in order to make the recommendation.
Explanation:
we are not sure if to recommend this alloy for this application given that this material has already been left to experience fatigue degradation. the cyclic load application brings about a growth in the crack. We know that cyclic loading is continuous loading that is useful for the testing of fatigue. Therefore the answer to this question is option a. We cannot make recommendations except fatigue testing has been carried out.
thank you!
P9.28 A large vacuum tank, held at 60 kPa absolute, sucks sea- level standard air through a converging nozzle whose throat diameter is 3 cm. Estimate (a) the mass flow rate through the nozzle and (b) the Mach number at the throat.
Answer:
a) [tex]m=0.17kg/s[/tex]
b) [tex]Ma=0.89[/tex]
Explanation:
From the question we are told that:
Pressure [tex]P=60kPa[/tex]
Diameter [tex]d=3cm[/tex]
Generally at sea level
[tex]T_0=288k\\\\\rho_0=1.225kg/m^3\\\\P_0=101350Pa\\\\r=1.4[/tex]
Generally the Power series equation for Mach number is mathematically given by
[tex]\frac{p_0}{p}=(1+\frac{r-1}{2}Ma^2)^{\frac{r}{r-1}}[/tex]
[tex]\frac{101350}{60*10^3}=(1+\frac{1.4-1}{2}Ma^2)^{\frac{1.4}{1.4-1}}[/tex]
[tex]Ma=0.89[/tex]
Therefore
Mass flow rate
[tex]\frac{\rho_0}{\rho}=(1+\frac{1.4-1}{2}(0.89)^2)^{\frac{1.4}{1.4-1}}[/tex]
[tex]\frac{1.225}{\rho}=(1+\frac{1.4-1}{2}(0.89)^2)^{\frac{1.4}{1.4-1}}[/tex]
[tex]\rho=0.848kg/m^3[/tex]
Generally the equation for Velocity at throat is mathematically given by
[tex]V=Ma(r*T_0\sqrt{T_e}[/tex])
Where
[tex]T_e=\frac{P_e}{R\rho}\\\\T_e=\frac{60*10^6}{288*0.842\rho}[/tex]
[tex]T_e=248[/tex]
Therefore
[tex]V=0.89(1.4*288\sqrt{248})\\\\V=284[/tex]
Generally the equation for Mass flow rate is mathematically given by
[tex]m=\rho*A*V[/tex]
[tex]m=0.84*\frac{\pi}{4}*3*10^{-2}*284[/tex]
[tex]m=0.17kg/s[/tex]
Question 1: Determine the maximum load P the steel bracket can withstand if the steel bracket has a circular cross section with a diameter of 1.2 in, and has an allowable normal stress of allow
Complete Question
Complete Question is attached below
Answer:
[tex]P=1124.2ibf[/tex]
Explanation:
From the question we are told that:
Diameter [tex]d=1.2in[/tex]
Allowable Normal stress [tex]\sigma=27.5ksi=27.5 * 10^3 psi[/tex]
Generally the equation for Bending Stress is mathematically given by
[tex]\phi= \frac{32M}{ \pi d^3}[/tex]
[tex]\phi= {32 * 4 P}{\pi * 1.2^3}[/tex]
[tex]\phi=23.58 psi[/tex]
Generally the equation for Direct Normal Stress is mathematically given by
[tex]\sigma'=\frac{4P}{ \phi * 1.2^2}[/tex]
[tex]\sigma'= 0.88P psi[/tex]
Therefore
Total Normal stress
[tex]\sigma_T=23.58 + 0.88[/tex]
[tex]\sigma_T=24.46P[/tex]
Generally the equation for Allowable Stress is mathematically given by
[tex]\sigma=\sigma_T P[/tex]
[tex]P=\frac{\sigma}{\sigma_T}[/tex]
[tex]P=\frac{27.5 * 10^3}{24.46P}[/tex]
[tex]P=1124.2ibf[/tex]
Consider CO at 500 K and 1000 kPa at an initial state that expands to a final pressure of 200 kPa in an isentropic manner. Report the final temperature in units of K and using three significant digits.
Answer:
[tex]T_2=315.69k[/tex]
Explanation:
Initial Temperature [tex]T_1=500K[/tex]
Initial Pressure [tex]P_1=1000kPa[/tex]
Final Pressure [tex]P_2=200kPa[/tex]
Generally the gas equation is mathematically given by
[tex]\frac{T_2}{T_1}=\frac{P_2}{P_1}^{\frac{n-1}{n}}[/tex]
Where
n for [tex]CO=1.4[/tex]
Therefore
[tex]\frac{T_2}{500}=\frac{200}{1000}^{\frac{1.4-1}{1.4}}[/tex]
[tex]T_2=315.69k[/tex]
A 40-mm-diameter solid steel shaft, used as a torque transmitter, is replaced with a hollow shaft having a 40-mm outer diameter and a 36-mm inner diameter. If both materials have the same strength, what is the percentage reduction in torque transmission
Answer:
65.61%
Explanation:
we have the following information to answer this question
diameter of the solid steel shaft = 40 mm
outer diametr of the hollow shaft = 40mm
inner diametr pf the hollow shaft = 36mm
[tex]percentage reduction in torque transmission = \frac{Tsolid-Thollow}{Tsolid} *100[/tex]
= (40³ - (40⁴-36⁴)/40)/40³ * 100
= (40³ - 22009.6)/40³ * 100
= 41990.4/64000 * 100
= 0.6561 x 100
= 65.61%
percentage reduction in torque transmission = 65.61%
the heat treatment process depends on 3 steps according to the time-temperature cycle. Name and Briefly describe each one
Explanation:
there are three stages of heat treatment
1.hit the metal slowly to ensure that the metal maintains a uniform temperature
2.soak or hold the metal at a specific temperature for a alloted period of time
3.cool the metal to room temperature
8.50 nC of charge is uniformly distributed along a thin rod of length L = 9.20 cm, which is then bent into a semicircle. What is the magnitude of the electric field at the center of the circle
Maggie discovered that a pipe in her basement has sprung a leak. She calls a plumber but in the meantime she grabs a roll of duct tape and wraps it around the pipe to stop the water from leaking. The duct tape in this situation is similar to _____ in that it/they _____.
Answer:
The answer is "Option a".
Explanation:
Myelination was the myelinization mechanism of a neuron axon. The endothelium is enveloped all around the axon and isolates the axon that inhibits the neuronal message from leaking with the other neuronal axons. Inside this example, therefore, its tubes tape worked similarly to those of myelin sheath, which stops brain transmission.
The Myelination was the myelinization mechanism of the neuron axon. Thus the option A is correct.
What is a Myelination?It is the process by which the brain's oligodendrocytes make layers of myelin and is wound around the neural axons and is seen as a layer of insulation for the transmission electrical potential down to the neuronal axon.
Find out more information about the sentence.
brainly.com/question/5636726
Describe the main component in an electronic control unit (ECU) used in an automobile. Indicate the function of each ingredient.
A steam turbine receives steam at 1.5MPa and 220oC, and exhausts at 50kPa, 0.75 dry. Neglecting heat losses and changes in kinetic and potential energy, estimate the work output per kg steam.
If when allowance is made for friction, radiation and leakages losses, the actual work of that estimated in (a), calculate the power output of the turbine when consuming 600kg of steam per minute.
Answer:
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Calculate the pressure of dry O2 if the total pressure of O2 generated over water is measured to be 698 Torr and the temperature is 30.1 oC. P(H2O) = 19.8 torr.
If the volume of the O2 sample in the question above was 56.3 ml, what volume would the dry O2 occupy at 755 torr (assume the temp was unchanged).
Answer:
[tex]V_2=46mL[/tex]
Explanation:
From the question we are told that:
Pressure over Water [tex]P=698 Torr[/tex]
Temperature [tex]T= 30.1 \textdegree C[/tex]
Pressure of Water [tex]P(H2O) = 19.8 torr.[/tex]
Volume of O2 [tex]O_2=56.3[/tex]
Pressure of Dry O2 [tex]P_(0)=755torr[/tex]
Generally the equation for Total Pressure is mathematically given by
[tex]P_t = P_O + P_H[/tex]
Therefore
[tex]P_O=P_t-P_H[/tex]
[tex]P_O=638-19.8[/tex]
[tex]P_O=618.2torr[/tex]
Generally the equation for Ideal gas is mathematically given by
[tex]P_1*V_1 = P_2*V_2[/tex]
[tex]V_2=\frac{P_1*V_1}{P_2}[/tex]
Therefore
[tex]V_2=\frac{ 618.2*56.3}{755}[/tex]
[tex]V_2=46mL[/tex]
Hence,The volume would the dry O2 occupy at 755 torr
[tex]V_2=46mL[/tex]
An Otto cycle with air as the working fluid has a compression ratio of 8.2. The ambient temperature is 298K. The maximum temperature of the Otto cycle is 2000K. Under cold air standard conditions, the thermal efficiency of this cycle is ____
Answer:
The correct answer is "57%".
Explanation:
Given:
Temperature:
[tex]T_a = 298 \ K[/tex]
[tex]T_3 = 2000 \ K[/tex]
Compression ratio (r),
= [tex]8.2[/tex]
For Otto cycle, the thermal efficiency will be:
⇒ [tex]\eta =1-\frac{1}{(r)^{\nu-1}}[/tex]
By substituting the values, we get
[tex]=1-\frac{1}{(8.2)^{1.4-1}}[/tex]
[tex]=1-\frac{1}{(8.2)^{0.4}}[/tex]
[tex]=57[/tex] (%)
If a corporation is socially responsible, it will develop and implement a sustainability plan and communicate it to stakeholders.
True
False
Answer:
True
Explanation:
All big companies are pretty much required in today's day and ages to complete these reports whether they truly believe it.
determine the number of flipflops required to build a binary counter that count from 0 to 2043
Answer:
10 flip -flops are required to build a binary counter circuit to count to from 0 to 1023 .
Explanation:
A 40kg steel casting (Cp=0.5kJkg-1K-1) at a temperature of 4500C is quenched in 150kg of oil (Cp=2.5kJkg-1K-1) at 250C. If there are no heat losses, what is the change in entropy of?
(i) The casing.
(ii) The oil.
(iii) Both considered.
