What does the term 'Sira' refer to in Islamic studies?

The biography of the Prophet Muhammad (s).

The history of Islamic nations.

What event does the Hadith about the Bedouin man asking about 'the Hour' illustrate?

The importance of preparedness in faith.

The significance of prayer only.

The unimportance of the Day of Judgment.

The need for following cultural traditions.

What is meant by 'following the Sunnah' of the Prophet Muhammad (s)?

Emulating the practices and teachings of the Prophet Muhammad (s).

Respecting historical figures in Islam.

Ignoring his teachings in daily life.

Practicing rituals without understanding their meanings.

What was one of the challenges faced by early Muslims during the time of the Prophet (s)?

Difficulty in finding followers.

Indifference from the community.

Why is it said that the Prophet Muhammad (s) is a mercy to mankind?

His teachings and character serve as a model for humane living.

S:01 | EP:01 | Seerah | Introduction to the Seerah - Shaykh Mustafa Abu Rayyan Quiz 1

1.1945-1949: The immediate years after the Second World War ● At the end of 1945, Mao Zedong had come to see the USA as the greatest threat to his aspirations. a. He understood that East Asians were looking to the USA as the true liberator from Japanese imperialism. b. The USA’s support for the Kuomintang(KMT) and the restoration of U.S. authority in formerly Japanese Manchuria clashed with the CCP’s plans to use the region for its own needs in the impending civil war between the CCP and the GMD. ■ To compound matters, while the KMT was recognised internationally as the official government in China, Mao and the CCP saw the party as a puppet of U.S. imperialism. ● While Mao saw the USA as the greater threat to the CCP’s plans, Soviet actions also frustrated him. a. The USSR provided minimal and incoherent support for the Chinese Communists in Yan’an and Manchuria. b. Stalin also attempted to extract territorial and economic concessions from the Guomindang government in the Friendship and Alliance Treaty China signed in August 1945 under American and Soviet pressure in exchange for Soviet entry into the Second World War against Japan. ● The emerging superpower conflict over Europe and over American intervention in the impending civil war in China led to Mao’s ideological perception of the 8838/01 H1 History Paper 1 Theme II: The Cold War and East Asia (1945-1991) \ Page | 8 USA as an aggressive imperialist power that was hostile towards other countries, especially the USSR and China. ● In 1946, Mao promoted the theory of the intermediate zone, which envisioned a global united front against American imperialism. a. Mao saw the emerging superpower conflict as an American-Soviet contest for the intermediate zones, the capitalist, colonial and semi- colonial countries of West Europe, Africa, and Asia. b. Mao believed that the USSR was the defender of world peace. c. The intermediate zone, which included China, would not be part of the socialist camp. d. Despite the tremendous potential that U.S. aid held for China’s reconstruction, Mao’s ideological worldview and the impending civil war against the Guomindang prevented him from seeking normalised relations with the USA. In 1949, Mao decided to lean towards the side of the USSR despite two decades of unreliable support from them. e. Mao saw the anti-bourgeois campaigns in East Europe as evidence that China should isolate capitalist-bourgeois forces within it.2 f. Stalin had expelled Yugoslavia from the socialist camp as its leader, Tito was seen to have directly challenged Stalin’s authority. ■ Mao thus saw it as imperative to stress close unity to the USSR lest he was seen as a second Josip Broz Tito. At the same time, Mao sought a loose partnership with the USSR because Mao believed that China should preserve a high measure of self- reliance and zili gengsheng (自力更生) (regeneration through one’s own efforts). ● When the People’s Republic of China was formed on 1 October, 1949, relations between China’s and the USSR’s communists had improved substantially. a. However, the Chinese Communist Party (CCP) was also aware that the USSR never treated Chinese interests as a priority. What the CCP failed to fully understand was that Stalin ruled East Europe much like it was his empire and how this would have implications for China. b. In Mao’s first visit to the USSR in December 1949, Stalin was non- committal regarding the interests raised by the Chinese, and treated Mao as an underling as he feared that closer relations with the PRC would cause the USSR to lose privileges gained from the KMT. _________________________ 2 What Mao did not realise at that point was that the anti-bourgeois campaigns in East European countries were part of Stalin’s intentional design to consolidate the power of communists in them. 8838/01 H1 History Paper 1 Theme II: The Cold War and East Asia (1945-1991) \ Page | 9 A note on Sino-American relations 2. Early 1950: The USA’s hands-off policy towards Taiwan begins to change ● By early 1950, the Truman administration had written off Taiwan and believed it was only a matter of time before the island fell to the PLA. ● Two events in early 1950 changed the USA’s position on East Asia. ○ The formation of the USSR-PRC alliance in February 1950 ○ The North Korean invasion of South Korea in June 1950 3. 1950: The Sino-Soviet Friendship, Alliance and Mutual Assistance Treaty ● Signed on 14 February, 1950. 3.1Implications for Sino-Soviet relations ● Stalin saw it as a means to get concessions that he had failed to get from the Kuomintang (KMT) government in 1945. ● For Mao and the newly founded People’s Republic of China (PRC), the alliance would provide security against U.S. imperialism and allow the PRC to get economic aid for reconstruction from the USSR. ● The Chinese realised soon after the 1950 treaty had been signed that the Soviet Union was intent on exploiting the agreement in its own favour. 8838/01 H1 History Paper 1 Theme II: The Cold War and East Asia (1945-1991) \ Page | 10 ● The Sino-Soviet alliance was officially directed against Japanese militarism and its allies, especially the USA. ● The Sino-Soviet alliance comprised three elements: party, military and economic relations. ○ Party: The Chinese Communist Party (CCP) was included in the customs of communist party internationalism, such as regular exchange of party delegations to congresses of the fraternal parties in Stalin’s socialist camp. ■ This move was meant to bring the PRC’s ideological beliefs about communism into greater alignment with the USSR’s. ○ Military: The alliance was supposed to provide the newly formed and weak PRC with a strategic deterrent and military aid against the USA on three fronts: Guomindang-held Taiwan, divided Korea, and Vietnam where France attempted to reestablish its colonial control. ■ Convinced that the USA would aggressively seek ways to undermine the CCP-led PRC through Taiwan, Korea and Vietnam, Mao sought an active defence. ● While in Moscow, Mao unsuccessfully asked Stalin to provide military assistance for the liberation of Taiwan. ● At the beginning of 1950, the PRC delivered large-scale military aid to Hanoi. The PRC was the first country to grant the communist-led Democratic Republic of Vietnam diplomatic recognition on 18 January 1950; Mao persuaded Stalin to do so on 30 January 1950. ● The PRC committed itself to North Korea, where Mao saw the commitment to North Korea both as a defence against U.S. imperialism and as support for a fellow communist country. ○ Economic: During Mao’s first stay in Moscow, Stalin had personally promised the delivery of fifty projects for primary industrialisation. ■ The agreement also led to a series of supplementary ones, such as a US$ 300 million loan that the PRC would repay with a mixture of strategic materials, rubber, agricultural products, goods for daily use and hard currency. ■ Significantly, Stalin used Soviet military and economic aid to extract concessions similar to those he failed to get from the Guomindang government in 1945. ■ The USSR and PRC would disagree on the pace and extent of the PRC’s planned development. ● In the last five weeks of Stalin’s life in early 1953, he attempted to pressure the PRC to reduce the planned 8838/01 H1 History Paper 1 Theme II: The Cold War and East Asia (1945-1991) \ Page | 11 development speed to a mere annual growth of 13-14 percent, and to plan individual projects in detail beforehand. These moves would potentially result in the PRC’s economy growing at a slower rate than initially projected. ● However, after Stalin’s death on 5 March 1953, the PRC’s Zhou Enlai decided to use his visit of condolence to the USSR to press forward negotiations. ○ When talks resumed in 1 April 1953, Beijing pressed for 150 Soviet industrial projects, but Moscow reduced them to 91 on the basis of insufficient data provided by the Chinese. ■ The economic disarray after China’s civil war and the economic pressures that came with the Korean War influenced recovery and reconstruction in the early years of the PRC. ● Despite the PRC being unable to tap into Soviet economic assistance immediately, mutual trade between China and the USSR nevertheless increased 6.5 times from 1950 to 1956. ● Together with the 50 projects promised by Stalin in 1950, the final version of the First FYP for the PRC included 141 Soviet and 68 East European projects in a total of 649 planned. Three thousand Soviet advisers sent to China in subsequent years were directly linked to the First FYP. ● By 1955, over 60 percent of China’s goods exchange was with the USSR. ● Soviet economic assistance to China added up to the largest foreign development venture in the socialist camp ever. ○ The total number of planned projects amounted to between 300 and 360 projects. ○ However, the number of total finished projects ranged between 134 and 150. ● Transfers of knowledge and expertise were important to China’s economic development. ○ A study on Soviet experts counts 1,445 political advisers and 9,313 technical specialists sent to China until their sudden withdrawal in mid-1960. ■ For political reasons, the gradual withdrawal of advisers began after late 1956.

MINERAL PROCESSING-I MCQ'S TEST-01

문학 The Hitchhiker's TEST D-01

21.01.29 Ben S. Retrieval Practice

Ben S. 15/01 Retrieval Test

Mr Mike's Dot n' Dash Lesson 01

01 ESS [8.ESS.1]: The composition and properties of Earth’s interior are identified by the behavior of seismic waves

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

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