Epithelial tissue ______________.

is the most abundant tissue in the body

Connective tissue ____________.

is the most abundant tissue in the body

is found very little in the body

is not found in the circulatory system

is only found in the joints of the body

Cardiac muscle tissue is found in the ______________.

lining of blood vessels, heart & capillaries

walls of the heart and lining of blood vessels

Smooth muscle is found in ___________.

blood vessels and hollow organs

03 FINAL EXAM REVIEW: Tissues

LEVEL 03 FINAL EXAM

In this video we take a look at the 0:02 fetch to code 0:03 execute cycle including its effect on 0:06 the various registers we've previously 0:12 [Music] 0:14 discussed a computer is defined Definition 0:17 as an electronic device that takes an 0:20 input 0:22 processes data 0:25 and delivers output 0:29 in this simple example you can see we're 0:31 taking the input 5 0:35 we're multiplying it by 2 that's our 0:37 process 0:39 and we're outputting 10. 0:44 but this could be way more complex for 0:46 example of a game console 0:48 the input could be the buttons you press 0:50 on a controller 0:53 the processes would then be carried out 0:55 by the console itself 0:59 and the output would be some form of 1:01 update to a monitor 1:02 and sound out for a speaker possibly 1:04 vibration feedback through the 1:06 controller 1:10 to process data a computer follows a set 1:13 of instructions 1:14 known as a computer program 1:18 if we take the lid off a typical desktop 1:20 computer we can identify 1:22 two critical components the memory 1:26 that stores the program and the central 1:29 processing unit or processor 1:31 which is under this large fan and 1:33 carries out the instructions 1:37 a computer carries out its function by 1:40 fetching 1:41 instructions decoding them and then 1:43 executing them 1:44 in a continuous repetitive cycle 1:46 billions of times a second 1:48 let's look at each of these stages in a 1:50 little more detail Fetch 1:53 so let's start with the fetch stage the 1:55 very first thing that happens 1:57 is the program counter is checked as it 2:00 holds the address 2:01 of the next instruction to be executed 2:07 the address stored is then copied into 2:09 the memory address register 2:14 the address is then sent along the 2:16 address bus to main memory 2:18 where it waits to receive a signal from 2:21 the control 2:22 bus so it knows what to do 2:27 as we want to read the data that's 2:29 stored in memory address 2:30 0 0 0 0 the control unit sends 2:34 a read signal along the control bus to 2:36 main memory 2:41 now main memory knows the data needs to 2:44 be read 2:45 the content stored in memory address 000 2:49 can be sent along the data bus to the 2:51 memory data register 2:56 now as we're currently in the process of 2:58 fetching an instruction 3:00 the data received by the memory data 3:03 register gets copied 3:04 into the current instruction register 3:11 the instruction effectively has now been 3:14 fetched from memory 3:16 just before we proceed to the decode 3:18 phase we now 3:19 increment the program counter so that 3:22 the address it contains 3:24 points to the address of the next 3:26 instruction which will need to be 3:30 executed 3:32 the instruction now being held in the 3:33 current instruction register 3:35 is ready to be decoded 3:39 now as we mentioned in the previous 3:41 video the instruction is made up of two 3:43 parts 3:44 we have the op code that's what it is we 3:47 need to do 3:50 and we have the operand what are we 3:53 going to do it to 3:55 now the operand could contain the actual 3:57 data 3:58 or indeed it could contain an address of 4:01 where the data is to be found 4:06 by decoding this instruction we can see 4:08 the operation we need 4:10 is a load operation so we need to load 4:14 the contents of memory location0101 4:18 into the cpus accumulator 4:25 in the exam a simple model will be used 4:27 to describe the 4:29 structure of any given instruction 4:32 you're not going to be expected to 4:34 define how an opcode is made up 4:36 but simply to interpret opcodes in the 4:39 given context of an exam 4:40 question in the example here 4:44 you can see there's a total of 16 4:46 different opcodes available 4:48 and this is because we're using four 4:50 bits for our representation 4:56 so now we've fetched the instruction and 4:59 we've decoded it so we know what we need 5:00 to do 5:01 we're finally ready to execute it 5:05 so we now send address 0101 5:08 to the memory dress register 5:13 now we're in the memory address register 5:15 we can finally send the address 5:18 down the address bus to main memory 5:24 this time we want to read the data 5:26 that's stored in memory 5:28 and so the control unit again sends a 5:30 read signal along the control bus 5:36 so main memories now receive an address 5:38 and a read signal 5:40 so the content stored at memory location 5:43 0101 5:44 can now be sent along the data bus back 5:46 to the cpu 5:47 and into the memory data register 5:54 finally the contents of the memory data 5:56 register are copied to the accumulator 5:59 and this is one of a number of general 6:00 purpose registers found in the cpu 6:04 this first instruction is now complete Branching 6:11 so what does this program actually do 6:14 you should be able to work it through 6:16 carefully and figure it out 6:19 we're now pointing instructions zero 6:21 zero zero one in the program counter 6:23 and we're ready to fetch the second 6:25 instruction 6:27 at the end of this video we're gonna 6:29 provide you with the answer 6:34 so let's talk a second about programs 6:37 that branch 6:40 on the left here we have a very simple 6:42 piece of pseudo code 6:44 line zero says first execute this line 6:46 of code 6:47 line 1 now execute this line and then 6:50 line 2 says 6:52 if the age is greater than 18 then 6:56 we're going to execute lines 3 and 4 6:58 otherwise 6:59 we're going to execute lines six and 7:02 seven 7:03 so this program doesn't necessarily 7:05 follow strictly in sequence from line 7:07 zero through to seven there's a chance 7:10 here the program may branch and jump 7:14 around 7:16 so we're going to pretend that this 7:17 program has been loaded into memory 7:20 each line of code on the left here has 7:23 ended up 7:24 as a location in memory now this is not 7:27 strictly how this would happen in this 7:28 one-to-one way 7:29 but for the purpose of example it's 7:31 absolutely fine 7:35 so the program counter starts by 7:37 pointing to memory address zero 7:39 and we fetch the first instruction 7:41 decode it and execute it 7:44 it then updates and tells us the next 7:47 instruction 7:48 is zero zero zero one because remember 7:50 the program counter is being incremented 7:52 so we fetch it decode it and we execute 7:55 line one of our program 7:59 we then fetch line two which in binary 8:01 is one 8:02 zero 8:06 now at this point depending on what 8:10 happens during the execution 8:11 of line two the program may be required 8:15 to fetch line three from memory or 8:18 line five from memory 8:25 so let's look at how this actually works 8:27 because we've said the program counter 8:28 simply gets incremented 8:31 well in the current instruction register 8:33 we have an instruction with the op code 8:36 0 1 1 0. 8:41 now when we look this up in the decode 8:43 unit we discover that this 8:45 code means branch always 8:51 this replaces the value held in the 8:54 program counter 8:56 with the contents of the operand that's 8:58 the second part of the instruction 9:01 from the current instruction register so 9:03 this case 9:04 one zero zero one 9:09 now when the next fetch cycle begins the 9:12 program counter is obviously checked 9:14 and as its contents have been previously 9:16 updated to a new memory location 9:19 and not simply incremented the program 9:22 effectively is able to jump 9:24 around memory 9:28 so having watched this video you should 9:30 be able to answer the following key 9:32 question 9:33 how does a cpu work 9:39 okay so let's um answer the question we 9:41 posed 9:42 earlier what did that program actually 9:48 do 9:50 so this is the first fetch to code 9:53 execute cycle 9:55 and this is the one that we ran through 9:57 in detail earlier 9:58 it effectively loaded the contents of 10:01 the memory 10:02 stored at location location0101 10:05 into the accumulator in other words 10:08 the dna number 3 is moved 10:11 from memory into the cpu 10:18 we then proceed onto the second fetch 10:20 decode execute cycle 10:23 now this one adds the contents of memory 10:27 located at 0 1 1 0 10:30 to the current contents of the 10:32 accumulator 10:34 so in other words the dna number one 10:38 because that's what's stored at address 10:40 zero one one zero 10:43 is added to the number three that was in 10:45 the accumulator 10:46 the results are stored back over the 10:48 accumulator 10:49 so effectively we've done three plus one 10:53 equals four 10:58 the third fetch to code execute cycle 11:00 stores the contents which are in the 11:02 accumulator 11:03 into memory location zero one one one 11:07 and that's because the op code the first 11:09 part of this current instruction 11:10 zero zero one one is the command to 11:13 store when we look it up in the decoder 11:15 unit 11:16 so in other words the result of the 11:17 previous calculation three plus one 11:19 equals four 11:20 is now written back into main memory 11:28 the fourth fetch decode execute cycle 11:30 outputs the contents of the accumulator 11:33 remember they were copied into main 11:34 memory but they're still held in the 11:35 accumulator 11:37 so in this simple abstraction the number 11:40 four is now 11:41 output to the user so they can see the 11:43 result of the calculation 11:49 the fifth and final fetch code execute 11:51 cycle 11:52 brings a halt to the current program 11:58 so this very simple program which has 12:01 five 12:02 fetch decode execute cycles has 12:04 performed the calculation 12:06 three plus one is then stored the result 12:09 in main memory 12:10 and displayed the result four to the 12:12 user 12:13 and in a high-level language this may 12:15 look something very similar to the 12:17 following two lines of code 12:20 sum variable equals num1 plus num2 12:24 print sum to the user 12:27 so you can start to get an appreciation 12:29 here of how the high level code you 12:32 write actually ends up being fetched 12:34 decoded 12:35 and executed inside a processor 12:38 of course your processor is doing 12:40 billions and billions of these 12:42 operations a second 12:43 which when you think about it is really 12:45 very impressive 12:52 [Music] 13:03 you. make 10 questions for a standerd of a level

Чиняева Аделина 18.03 - Le test final sur la "nourriture"

EVALUACION FINAL CURSO: MANEJO SEGURO DE CARGAS MANUALES. NOM-006-STPS-2014 Manejo y almacenamiento de materiales- Condiciones de seguridad y salud en el trabajo. NOM-036-1-STPS-2018 Factores de riesgo ergonómico en el Trabajo-Identificación, análisis, prevención y control. Parte 1: Manejo manual de cargas. 17/03/2023. SIGE

Riferimenti normativi per il settore residenziale: Art 81: Accesso alla rete viaria—>> il cancello deve essere arretrato di almeno 4,5m dal filo esterno del marciapiede Art 82: Passo carrabile—>> larghezza non inferiore a 4,5m e non superiore a 6,5m Art 83:Pendenza Rampa —>> max 16% Art 97: Superficie minima degli ambienti - cucina —>> min 5mq - studio—>> min 7mq - soggiorno—>> min 14 mq - soggiorno spazi di cottura—>> 17 -camera (1posto letto)—>> 8mq - camera (2posti letto)—>> 12mq Superficie alloggio totale —>> non inferiore a 28 mq Art 95: Altezze minime- cucina, soggiorno, camera e studio—>> min 2,70m -locali accessori—>> min 2,40 m (bagno, lavanderie) - locali di servizio—>> min 2,10m (disimpegni, riposti.) - soppalchi—>> min 2,10m - parapetti—>> non inf. a 1,1m (10 cm cordolo) (92) Art 86: Distanze - negli edifici di nuova costruzione la distanza degli edifici dal confine con proprietà di terzi — nei NAF—>> non inf. a 3 m — altri ambiti —>> non inf. a 5m Art 89: Scale (R.E. ) —alzate—>> max 12 consecutive — a chiocciola—>> consentite solo all’interno di un’unità abit. — illuminazione—>> se collegano più di due piani devono . essere areati con lucernario. Dim: 0,3 mq . per ogni piano servito (R.I.: 0,4 mq x piano) — areazione —>> non ci può essere areazione verso i vani scala . ( Lo dice anche il regolamento d’igiene ) (R.I) —>> superficie non inferiore a 1mq per piano servi. — larghezza (R.I.) —>> deve garantire la possibilità di soccorso e . trasporto di persone Art 88: Locali sotterranei —>> non possono MAI essere adibiti ad abitazione Locali seminterrati—>> possono ma devono rispettare determinati . requisiti - l’altezza media deve essere > di 2,7m Art 91: Copertura—>> istallazione di apparati tecnici non deve essere visibile . dalle pubbliche vie Art 98: Bagni —>> ambiente contenente il vaso deve essere disimpegnato . dalla cucina (R.I: disimpegnato dai locali abitabili, esclusione Secondo bagno se è a servizio esclusivo di una camera) (R.I)—>> deve essere dotato di vaso, lavabo, bidet doccia o vasca —>> il lavabo può essere ubicato nell’antibagno Art 100: Areazione —>> riscontro d’aria deve essere garantito su aperture . perpendicolari o contrapposte.(non inf. a 1\10) —>> appartamenti inf. a 60mq possono essere . monoaffaccio ( ma non esposti a nord) Art 125: Raccolta rifiuti (R.E)—>> non meno di 0,18mq per ogni abitante . virtuale —>> non meno di 5 mq —>> altezza minima 2,4m (R.I: 2 m) —>> deve avere un punto di allacciamento d’acqua Regolamento d’igiene —>> dimensione tale da poter contenere 4,5l Di rifiuti per abitante.( in ogni caso > di 2mq) —>> scarichi sifonati dall’acqua di lavaggio —>> accorgimenti che assicurino un’adeguata Difesa antimurina e antinsetti Norma UNI 10750: superficie commerciale (ciò che compriamo) , cioè la somma delle superfici coperte—>> 100% delle superfici calpestatili —>>100% delle superfici su cui poggiano . Le pareti divisorie interne non portanti —>> 50% delle pareti portanti interne perimetrali —>> 25% delle aree non abitabili Superfici scoperte—>> 25% dei balconi e delle terrazze scoperte —>> 35% dei balconi e dei terrazzi coperti (3 lati) —>> 35% dei pati e dei porticati —>> 60% delle verande —>> 15% dei giardini di appartamento —>> 10% dei giardini di ville e villini (Se un muro è al confine con un altro appartamento lo considero dalla mezzeria, se confina con uno spazio comune idem, lo considero tutto invece se da sull’esterno ) Regolamento d’igiene (su esso prevale il Re) : norme che discipilinano degli aspetti della vita quotidiana al fine di tutelare la salute dei fruitori Si occupa di : Rumori , odori , fumi, vapori. Scarichi nel sottosuolo Pulizia e decoro Malattie infettive Igenicità degli ambienti Pareti trasparenti (tenendo conto di telai e infissi)—>> deve avere un area pari a 1\8 (nazionale), 1\10 (Milano ) della superficie di pavimento Profondità di pavimento—>> non deve superare i 2,5m dalla finestra Bagno cieco solo se la superficie lorda di pavimento è inferiore ai 70mq e se è presente una sola camera da letto , oppure se è un secondo bagno (altrimenti finestra > 0,5 apribile). Superficie illuminante—>> superficie totale dell’apertura meno - superficie finale non utile (C): 60 cm - superficie superiore non utile (A) A= va considerato per intero se non ci sono aggetti o se questi sono inferiori a 150 cm. Al contrario ne considero solo un terzo. Es: con aggetto, b+ 1\3a . Se il rapporto illuminante è rispettato la profondità del locale non può essere più di 2,5 volte l’altezza del voltino . Se non è rispettato (inferiore a 1\8) allora deve essere 3,5 volte Alloggi devono essere dotati— per 1\2 perone—>> 1 spazio cottura,1 servizio igienico , 1 ripostiglio — per 3\4 persone—>> 1 cucina indipendente, 1 servizio igienico , 1 ripostiglio — per 4\5 persone —>> 1 cucina indipendente, 2 servizi igienici, 1 ripostiglio (per il secondo servizio è richiesta una superficie minima di 2mq e un lato minimo di 1,2m) Dotazione dei servizi: Cucina—>> pavimenti e pareti con superficie Di materiale impermeabile, liscio, lavabile,resist. —>> soffitto materiale traspirante —>> cappa collegata a ogni punto di cottura (Vedi bagno su) Prevenzione incendi : definisce —dimensionamenti—accessi all’area (locali di intrattenimento e di pubblico Spettacolo —>> larghezza 3,5 —>> h libera 4m —>> raggio di svolta 13m —>> pendenza non sup al 10% —>> resist al carico almeno 20t —profondità locali —>> i locali al chiuso non possono Essere ubicati oltre il secondo piano Interrato (non oltre i 10m) . Questi se Sono tra i 7,5 e i 10 m devono essere Protetti da un’impianto sprinkler e Essere dotati di uscite sicure. — carichi d’incendio — comunicazione (locali di intrattenimento e di pubb. Spettacolo —>> locali possono comunicare con altre Attività purché dotate di filtri a prova di Fumo e di porte REI (ameno 30) (queste Non vanno cont nel comp. delle vie d’uscit) — compartimentazioni — autorimesse — comportamento al fuoco

03-Ψηφιακός Μετασχηματισμός | Digital Technologies & Business Analytics

03/23/20 - Actuaries

03 Biological Macromolecules

Related quizzes

Related quizzes