Answer:
[tex]D=4.02[/tex]
Explanation:
From the question we are told that:
Part Dimension [tex]x=4.00in[/tex]
Generally the equation for the corresponding dimension to which the mold cavity should be machined is mathematically given by
[tex]D=x-(xS)+(4S^2)[/tex]
Where
S= Value for Skinage
From Table pf AB's
[tex]S=0.006[/tex]
Therefore
[tex]D=4-(4*0.006)+(4(0.006)^2)[/tex]
[tex]D=4.02[/tex]
Biết op-amp có Vsat = 12V, R1=R2=R3=R4=R, dạng sóng điện áp Vi(t) được cho như
hình 2.16b
a. Tính V0 theo Vi, độ lợi áp AV
b. Vẽ lại dạng sóng điện áp V0(t) khi Vm= 3
8 V
c. Vẽ lại dạng sóng điện áp V0(t) khi Vm = 5V
Answer:
Hello bro
Explanation:
I think i can help you something but can you translate it on english plzz
The input sin(20) is sampled at 20 ms intervals by using impulse train sampling: i. Construct the input and sampled signal spectra.
Solution :
Let [tex]$x(t) = \frac{\sin (20 \pi t)}{\pi t}$[/tex]
[tex]$T_s = 20$[/tex] ms, so [tex]$f_s=\frac{1}{T_s}[/tex]
[tex]$=\frac{1}{20}$[/tex]
= 0.05 kHz
[tex]$f_s=50 $[/tex] Hz , ws = [tex]$2 \pi f_s = 100 \pi$[/tex] rad/s
We know that,
FT → [tex]$\frac{\sin (20 \pi \omega)}{\pi \omega}$[/tex]
The sampled signal is :
[tex]$XS(\omega) = \frac{1}{T_s} \sum_{k=- \infty}^{\infty}X (\omega-k\omega S)[/tex]
So, [tex]$XS(\omega) = \frac{1}{20 \times 10^{-3}} \sum_{k=- \infty}^{\infty}X (\omega-100 k \pi)[/tex]
[tex]$XS(\omega) = 50 \sum_{k=- \infty}^{\infty}X (\omega-100 k \pi)[/tex]
Consider a laminar forced flow inside a pipe with constant wall temperature, the heat flux will have a higher value near the ____________ of the pipe.
Answer:
Inlet
Explanation:
Consider a laminar forced flow inside a pipe with constant wall temperature, the heat flux will have a higher value near the INLET of the pipe.
This is because the friction factor is experienced at the highest level when a laminar forced flow is at the tube inlet where the thickness of the boundary layer is zero. Also, the friction factor decreases step by step at a lower rate to the fully augmented value.
What are three types of land reform
Answer:
Abolition of intermediaries (rent collectors under the pre-Independence land revenue system); Tenancy regulation (to improve the contractual terms including the security of tenure); A ceiling on landholdings (to redistributing surplus land to the landless);
Types of Land Reform
Abolition of Intermediaries
The first step taken by the Indian government under land reforms post-independence was passing the Zamindari Abolition Act. The primary reason of a backward agrarian economy was the presence of intermediate entities like, jagirdars and zamindar who primarily focussed on collecting sky-rocketing rents catering to their personal benefits, without paying attention to the disposition of farms and farmers. Abolition of such intermediaries not only improved conditions of farmers by establishing their direct connection with the government but also improved agricultural production.
Regulation of Rents
This was in direct response to the unimaginably high rents which were charged by intermediaries during British rule, which resulted in a never-ending cycle of poverty and misery for tenants. Indian government implemented these regulations to protect farmers and labourers from exploitation by placing a maximum limit on the rent that could be charged for land.
Tenure Security
Legislations were passed in all states of the country to grant tenants with permanent ownership of lands and protection from unlawful evictions on expiry of the lease. This law protects tenants from having to vacate a property immediately after their tenure is over unless ordered by law. Even in that case, ownership can be regained by tenants with the excuse of personal cultivation.
At steady state, a thermodynamic cycle operating between hot and cold reservoirs at 1000 K and 500 K, respectively, receives energy by heat transfer from the hot reservoir at a rate of 1500 kW, discharges energy by heat transfer to the cold reservoir, and develops power at a rate of (a) 1000 kW, (b) 750 kW, (c) 0 kW. For each case, apply Eq. 5.13 on a time-rate basis to determine whether the cycle operates reversibly, operates irreversibly, or is impossible.
Answer:
a. impossible
b. possible and reversible
c. possible and irreversible
Explanation:
a. 1000kw
Qh - Wnet
we have
QH = 1500
wnet = 1000
1500 - 1000
= 500kw
σcycle = [tex]-[\frac{QH}{TH} -\frac{QC}{TC} ][/tex]
Qh = 1500
Th = 1000
Tc = 500
Qc = 500
[tex]-[\frac{1500}{1000} -\frac{500}{500} ][/tex]
solving this using LCM
= -0.5
the cycle is impossible since -0.5<0
b. 750Kw
Qc = 1500 - 750
=750Kw
Qh = 1500
Th = 1000
Tc = 500
Qc = 750
σ-cycle
[tex]-[\frac{1500}{1000} -\frac{750}{500} ]\\= 1.5 -1.5\\= 0[/tex]
This cycle is possible and it is also reversible
c. 0 kw
Qc = 1500-0
= 1500
Qh = 1500
Th = 1000
Tc = 500
Qc = 1500
σ- cycle
[tex]-[\frac{1500}{1000} -\frac{1500}{500} ]\\-(1.5-3)\\-(-1.5)\\= 1.5[/tex]
1.5>0
so this cycle is possible and irreversible
The purification of hydrogen gas is possible by diffusion through a thin palladium sheet. Calculate the number of kilograms of hydrogen that pass per hour (in kg/h) through a 4.0-mm thick sheet of palladium having an area of 0.25 m^2 at 500°C. Assume a diffusion coefficient of 6.0 x 10^-8 m^2/s, that the concentrations at the high- and low-pressure sides of the plate are 3.5 and 0.25 kg/m^3 (kilogram of hydrogen per cubic meter of palladium), and that steady-state conditions have been attained. (clearly show your solution step by step, pay attention to units otherwise you will lose points!)
what is the term RF exiciter?
Different metabolic control systems have different characteristic time scales for a control response to be achieved. Match the time scale with the control system.
a. Covalent modification
b. Allosteric control
c. Gene expression
1. Seconds to minutes
2. Milliseconds
3. Hours
Answer:
a. Covalent modification = Seconds to minutes
b. Allosteric control = Milliseconds
c. Gene expression = Hours
Explanation:
Covalent modifications refer to the addition and/or removal of chemical groups by the action of particular enzymes such as methylases, acetylases, phosphorylases, phosphatases, etc. For example, histones are chromatin-associated proteins covalently modified by enzymes that add methyl groups (histone methylation), acetyl groups (histone acetylation), phosphate groups (histone phosphorylation), etc. Moreover, allosteric control, also known as allosteric regulation, is a type of regulation of the enzyme activity by binding an effector molecule (allosteric modulator) at a different site than the enzyme's active site, thereby triggering a conformational change on the enzyme upon binding of an effector. Finally, gene expression encompasses the cellular processes by which genetic information flows from genes to proteins (i.e., transcription >> translation). In metabolic pathways, enzymes that are able to catalyze irreversible reactions represent sites of control (for example, during glycolysis, pyruvate kinase is an enzyme that catalyzes an irreversible reaction, thereby serving as a control site). In turn, enzymatic activity is modulated by covalent modifications or reversible binding of allosteric effectors. Finally, metabolic pathways are also modulated by gene regulatory mechanisms that control the transcription of specific enzymes required for such pathways. During these processes, the times required for allosteric regulation, covalent modification (e.g., phosphorylation) and transcriptional control can be counted in milliseconds, seconds, and hours, respectively.
Race cars at the Indianapolis Speedway average speeds of 185 mi/h. After determining the altitude of Indianapolis, find the Mach number of these cars and estimate whether compressibility might affect their aerodynamics.
Answer:
- the Mach number is 0.24.
- Compressibility becomes effective when Mach number is greater than 0.3, the Mach number of the race cars is less than 0.3, hence, compressibility will not affect their aerodynamics.
Explanation:
Given the data in the question;
Average speed V = 185 miles per hour = ( 185 /2.237 ) m/s = 82.7 m/s
From Almanac, we can find that Indianapolis is at 220 m altitude.
So from table, at that altitude, the standard speed of sound will be 339.4 m/s .
Mach number of the race car will be;
Mach Number = Velocity / sound speed
we substitute
Mach Number = ( 82.7 m/s ) / ( 339.4 m/s )
Mach Number = 0.24
Therefore the Mach number is 0.24.
We know that, compressibility becomes effective when the Mach number is greater than 0.3.
Since the Mach number of the race cars is less than 0.3, compressibility will not affect their aerodynamics.
An apple, potato, and onion all taste the same if you eat them with your nose plugged
Answer:
I didn't understand your question or is it a fun fact
định khoản nghiệp vụ sau : tạm ứng cho nhân viên A đi công tác bằng tiền mặt 50.000
Answer:
report on a fight you have witnessed
what is the term Self-induced EMF (Induction)
Self induction and Mutual induction
If the current through the coil changes, the flux through the coil changes and an emf is induced. Since the emf is induced in the coil due to its own field, it is called self–induced emf and the phenomenon is called self–induction.
Answer:
Self induced EMF is the electro-motive force that is generated due to change in flux in the coil caused by the change in current in the same coil.
Question in Probability and Statistics 2
According to the reading who was named the second most disruptive company in 2018 by CNBC
Answer:
Crowdstrike.com.
Explanation:
This question can't be answered correctly without the reading. However, according to crowdstrike.com, "CrowdStrike Named to CNBC's Disruptor 50 List for Second Consecutive Year. Sunnyvale, CA – May 22, 2018 – CrowdStrike® Inc., the leader in cloud-delivered endpoint protection, today announced that it was named to the 2018 CNBC Disruptor 50 list for the second consecutive year."
Hope This Helps!
A binary system of species 1 and 2 consists of vapor and liquid phases in equilibrium
at temperature T. The overall mole fraction of species 1 in the system is z1 = 0.65. At
temperature T, lnγ1 = 0.67 x2
2; lnγ2 = 0.67 x1
2; P1
sat = 32.27 kPa; and P2
sat = 73.14 kPa.
Assuming the validity of Eq. (13.19),
Final PDF to printer
13.10. Problems 511
smi96529_ch13_450-523.indd 511 01/06/17 03:27 PM
(a) Over what range of pressures can this system exist as two phases at the given T and z1?
(b) For a liquid-phase mole fraction x1 = 0.75, what is the pressure P and what molar
fraction of the system is vapor?
(c) Show whether or not the system exhibits an azeotrope
Explain the LWD process why is it important in drilling operations?
Answer:
Logging while drilling (LWD) is a technique of conveying well logging tools into the well borehole downhole as part of the bottom hole assembly (BHA). ... In these situations, the LWD measurement ensures that some measurement of the subsurface is captured in the event that wireline operations are not possible.
Explanation:
pls mark brainliest
What is protection scheme?
Answer:
The objective of a protection scheme is to keep the power system stable by isolating only the components that are under fault, whilst leaving as much of the network as possible still in operation.
Explanation:
The devices that are used to protect the power systems from faults are called protection devices.
Use a truth table to verify the first De Morgan law ¬(p ∧ q) ≡ ¬p ∨ ¬q.
Answer:
p q output ¬(p ∧ q)
0 0 1
0 1 1
1 0 1
0 0 0
p q output ¬p ∨ ¬q
0 0 1
0 1 1
1 0 1
0 0 0
Explanation:
We'll create two separate truth tables for both sides of the equation, and see if they match.
The expressions in the question use AND, OR and NOT operators.
The AND operation needs both inputs to be 1 to return a 1.The OR operation needs at least 1 of the inputs to be 1 to return a 1. The NOT operation takes a 1 and turns it into a 0, or takes a 0 and turns it into a 1.Let's start with ¬(p ∧ q)
NOT (0 AND 0) = NOT (0) = 1NOT (0 AND 1) = NOT (0) = 1NOT (1 AND 0) = NOT (0) = 1NOT (1 AND 1) = NOT (1) = 0Now let's move on to the second expression ¬p ∨ ¬q
NOT(0) OR NOT(0) = 1 OR 1 = 1NOT(0) OR NOT(1) = 1 OR 0 = 1NOT(1) OR NOT(0) = 0 OR 1 = 1NOT(0) OR NOT(0) = 0 OR 0 = 0Therefore we can say the two expressions are equivalent.
Attached the truth table to verify the first De Morgan's law ¬(p ∧ q) ≡ ¬p ∨ ¬q:
What is the explanation of the truth table?As you can see from the attached truth table, the truth values for ¬(p ∧ q) and ¬p ∨ ¬q are the same for all combinations of p and q, confirming the validity of the first De Morgan's law.
De Morgan's law is a fundamental principle in propositional logic.
It states that the negation of a conjunction (AND) is equivalent to the disjunction (OR) of the negations of the individual propositions.
Learn more about truth table at:
https://brainly.com/question/28605215
#SPJ3
: A cyclical load of 1500 lb is to be exerted at the end of a 10 in. long aluminium beam (see Figure below). The bar must survive for at least 10° cycles. What is the minimum diameter of the bar?
Answer:
the minimum diameter of the bar is 1.634 in
How many numbers multiple of 3 are in the range [2,2000]?
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
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.
Future solution for air pollution in new zealand
Answer:
New Zealand may use some of these solutions to prevent air pollution
Explanation:
Using public transports.
Recycle and Reuse
No to plastic bags
Reduction of forest fires and smoking
Use of fans instead of Air Conditioner
Use filters for chimneys
Avoid usage of crackers
What is working fluid and mention five example of working fluid.
Answer:
For fluid power, a working fluid is a gas or liquid that primarily transfers force, motion, or mechanical energy. ... Examples without phase change include air or hydrogen in hot air engines such as the Stirling engine, air or gases in gas-cycle heat pumps, etc.
Explanation:
Answer:
A working fluid is a gas or liquid that primarily transfers force, motion or mechanical energy
Examples:Air, pentane, chlorofluorocarbons, butane, propane and ammonia
Technician A says that a continuously variable transmission is an automatic transmission that does not shift gears. Technician B says that a continuously variable transmission has a maximum of six distinct gear ranges. Which technician is correct?
Answer:
Only Technician A
Explanation:
Continuously variable transmission (CVT) does not shift gears because it simply doesn't have gears. It instead uses 2 pulleys which change constantly and continuously in size, and they are linked by a belt. Continuously variable transmission (CVT) is automatic transmission as mentioned in the question above, largely due to the power the belt continuously transmits, making the vehicle's engine work in the optimum power range.
While Technician B saying, that a continuously variable transmission has a maximum of six distinct gear ranges, which is not correct because continuously variable transmission (CVT) works with an unlimited number of gear ratios within a fixed range.
The host at the end of the video claims that ___________ is crucial to his success as a driver. A. Reaction time B. A safe space C. His seat belt
Answer:
answer is C. his seat belt
Hot air is to be cooled as it is forced to flow through the tubes exposed to atmospheric air. Fins are to be added in order to enhance heat transfer. Would you recommend attaching the fins inside or outside the tubes? Why? When would you recommend auaching fins both inside and outside the tubes?
Answer:
Fins should be attached outside the tube Fins can be attached on both sides when convection coefficient of air inside the tube is equal to the convection coefficient of atmospheric air outside the tubeExplanation:
The main function of the fins that are to be added is to ensure the speedy transfer of heat from the Hot air.
The fins should be attached outside the tube because the convection coefficient of air is higher inside the tube than the convection coefficient of the outside air ( atmospheric air ), BUT
When convection coefficient of air inside the tube is equal to the atmospheric air outside the tube, it is recommended that the fins can be added on both sides of the tube ( i.e. in and outside the tube )
Carbon dioxide at a temperature of 0oC and a pressure of 600 kPa (abs) flows through a horizontal 40-mm- diameter pipe with an average velocity of 2 m/s. Determine the friction factor if the pressure drop is 235 N/m2 per 10-m length of pipe.
Answer:
f = 0.04042
Explanation:
temperature = 0°C = 273k
p = 600 Kpa
d = 40 millemeter
e = 10 m
change in P = 235 N/m²
μ = 2m/s
R = 188.9 Nm/kgk
we solve this using this formula;
P = ρcos*R*T
we put in the values into this equation
600x10³ = ρcos * 188.9 * 273
600000 = ρcos51569.7
ρcos = 600000/51569.7
=11.63
from here we find the head loss due to friction
Δp/pg = feμ²/2D
235/11.63 = f*10*4/2*40x10⁻³
20.21 = 40f/0.08
20.21*0.08 = 40f
1.6168 = 40f
divide through by 40
f = 0.04042
Question # 1
Fill in the Blank
Complete the following sentence.
Mobility refers to the ability to _____.
to move or walk freely and easly
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] (%)
