Answer:
a. unit of length: [L]
b. unit of volume: [[tex]L^3[/tex]]
c. unit of pressure:[tex]P=\frac{F}{A} \equiv\frac{[MLT^{-2}]}{[L^2]}[/tex] [tex][ML^{-1}T^{-2}][/tex]
d. unit of power: [tex]N.m.s^{-1}\equiv [ML^2T^{-3}][/tex]
e. unit of force: [tex][kg.m/s^2]\equiv [MLT^{-2}][/tex]
f. unit of power: [tex]N.m.s^{-1}\equiv [ML^2T^{-3}][/tex]
Force: [tex]F=m.a=m.\frac{v}{t}=m.\frac{x}{t}\div t[/tex]
Power: [tex]P=\frac{W}{t}=\frac{F.x}{t}[/tex]
where:
F = force
A = area
W = work
t = time
a = acceleration
v = velocity
x = displacement
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
How many numbers multiple of 3 are in the range [2,2000]?
To prevent the bubble from popping, a second bubble is made with more total fluid. This makes the walls of the bubble thicker.
a. True
b. False
3
Current is measured in units called
Answer:
current is measured in Ampere (A)
Answer:
Ampere (A)
Explanation:
The ampere is defined so the elementary charge e is 1.602 176 634 × 10−19 C or A•s.
Hope this helps <3
Microsoft Project là phần mềm có sẵn trong bộ Office 365, đúng (True) hay sai (False)?
Increasing following distance to
when encountering other motorists who follow too closely
is an example of appropriate implementation of the IPDE defensive driving strategy for the
maintenance of an appropriate Safety Cushion.
Two-seconds
Enree-seconds
Four-seconds
Twenty-seconds
Answer:
Increasing following distance to Four-seconds when encountering other motorists who follow too closely is an example of appropriate implementation of the IPDE defensive driving strategy for the maintenance of an appropriate Safety Cushion.
Explanation:
Maintaining the required safety cushion by utilizing the IPDE defensive driving strategy to manage the nine to fifteen space driving zones involves continuous scanning. Therefore, motorists should be able to identify objects and hazards in the driving scene, line of sight, and path of travel. They should predict points of driving conflicts. They should determine appropriate and safe driving actions to take, when, and where. Finally, action is required to ensure that conflicts are avoided.
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] (%)
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!
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.
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
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!)
isormophous phase diagram
Answer:
Phase diagrams represent the relationship between temperature and the composition of phases present at equilibrium. An isomorphous system is one in which the solid has the same structure for all compositions. The phase diagram shown is the diagram for Cu-Ni, which is an isomorphous alloy system.
Hope it help you friend
At what depth in water is the increased pressure five times greater than atmospheric pressure (101 kPa)?
Explanation:
40.4m
Explanation:
Pressure at depth is given as
P = P, + pgh
Final pressure at depth h= 5 Po
5Po= Po + pgh
pgh = 4Po = 4 x 1.01 x 10^5
h = (4.04×10^5)/ (1000x10)
h=40.4m
động cơ không đồng bộ 3 pha là gì
Answer:
uhhhhhhh lemme think
Explanation:
On Beverly's last project, the team identified only a few lessons learned. Which approach to lessons learned
can she implement on her current project to identify most of the lessons learned?
Select an answer.
reminding team members to be positive
providing an anonymous method for submitting lessons learned
making lessons learned a regular part of meetings
holding lessons learned meetings without managers present
Option C (making lessons learned a regular part of meetings) is the correct approach.
As nothing more than a general rule, typically construction companies only plan lessons that have been learned exercises or initiatives towards the end of a particular endeavor or segment.As almost a result of the team knowing, valuable lessons are intended to increase the comprehensive implementation of quality management practices as well as deadlines.
Aside from this, none of the choices are viable methods to learning lessons or gaining knowledge from the past. As a result, the methodology outlined above is the appropriate one.
Learn more about project teamwork here:
https://brainly.com/question/14279121
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.
Question # 1
Fill in the Blank
Complete the following sentence.
Mobility refers to the ability to _____.
to move or walk freely and easly
Question in Probability and Statistics 2
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.
: 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
Theo Anh / Chị, để đáp ứng yêu cầu phát triển nền kinh tế thị trường định hướng Xã hội Chủ nghĩa ở Việt Nam trong bối cảnh thời đại hiện nay, cần chú trọng giải quyết những vấn đề gì ?
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
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%
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
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.
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
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
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 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]
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
