An optical fiber consists of a core with refractive index n_{core} and a cladding with refractive index n_{cladding} . For a signal to be transmitted via total internal reflection, which condition must be met regarding the refractive indices and the incident angle \theta at the core-cladding interface?

n_{core} > n_{cladding} and \theta > \arcsin(n_{cladding} / n_{core})

n_{core} = n_{cladding} and \theta = 90^\circ

n_{core} and \theta > \arcsin(n_{core} / n_{cladding})

n_{core} > n_{cladding} and \theta

In the context of rainbow formation, which combination of optical phenomena accurately describes the behavior of sunlight within a single spherical raindrop to produce the primary arc?

Refraction at the entry, dispersion of colors, one internal reflection at the back surface, and refraction at the exit.

Refraction at the entry, two internal reflections at the back surface, and refraction at the exit.

Total internal reflection at the entry, dispersion of colors, and refraction at the exit.

Dispersion at the entry, diffraction around the edges of the drop, and total internal reflection at the exit.

A ray of light is propagating inside a diamond with a refractive index of n = 2.42 towards an interface with air ( n = 1.00 ). If the light ray strikes the interface at an angle of 30^\circ relative to the normal, which of the following best describes the path of the light?

The light undergoes total internal reflection because the incident angle is greater than the critical angle.

The light travels along the boundary between the diamond and the air.

The light refracts into the air and bends away from the normal.

The light refracts into the air and bends toward the normal.

An equilateral glass prism has a refractive index of n = 1.50 . A light ray is incident internally on one of the side faces at an angle of 40^{\circ} to the normal. Given that the surrounding medium is air ( n = 1.00 ), what is the behavior of the light ray at this interface?

The light undergoes total internal reflection because the critical angle is approximately 41.8^{\circ} , which is greater than the angle of incidence.

The light travels along the boundary because the critical angle is exactly 40^{\circ} .

The light refracts out of the prism because the incident angle of 40^{\circ} is less than the critical angle.

The light refracts toward the normal as it exits the denser glass medium into the air.

A ray of light is propagating through a diamond ( n = 2.42 ) and strikes an interface with water ( n = 1.33 ). What is the critical angle c for this interface, and what happens if the light hits the interface at an angle of incidence of 30^\circ ?

c \approx 33.3^\circ ; the light refracts into the water.

c \approx 56.7^\circ ; the light refracts into the water.

c \approx 24.4^\circ ; the light undergoes total internal reflection.

c \approx 33.3^\circ ; the light undergoes total internal reflection.

A ray of light is contained within a rectangular glass slab ( n = 1.52 ) surrounded by air. If the ray strikes the top surface from within the glass at an angle of 45^{\circ} to the normal, which phenomenon will occur?

Refraction at an angle greater than 45^{\circ} into the air

The ray will travel exactly along the surface boundary

Refraction at 45^{\circ} into the air

An underwater light source emits a ray that strikes the water-air interface ( n_{water} = 1.33 , n_{air} = 1.00 ). What is the maximum angle of incidence \theta for which the light can still escape into the air, and what happens at an angle of 50^{\circ} ?

\theta_c \approx 48.8^{\circ} ; at 50^{\circ} the ray undergoes total internal reflection.

\theta_c \approx 60.1^{\circ} ; at 50^{\circ} the ray refracts into the air.

\theta_c \approx 41.2^{\circ} ; at 50^{\circ} the ray refracts into the air.

\theta_c \approx 48.8^{\circ} ; at 50^{\circ} the ray refracts along the boundary.

7.2.1 Critical Angle 1. The critical angle is the angle of incidence at which the refracted ray: A. Bends toward the normal B. Bends away from the normal C. Travels along the boundary D. Is totally reflected Answer: C 7.2.2 Snell’s Law & Critical Angle 2. Which formula correctly represents the critical angle c when light travels from medium 1 to medium 2? A. n1cos⁡c=n2 B. n2sin⁡c=n1 C. n1sin⁡c=n2 D. n1sin⁡c=n2sin⁡90∘ Answer: D 7.2.3 Total Internal Reflection 3. Total internal reflection occurs only when: A. Light travels from air to glass B. Angle of incidence is less than the critical angle C. Light travels from a denser to a rarer medium D. Refractive index of the second medium is greater Answer: C 4. Which condition is not required for total internal reflection? A. Light must travel from a denser medium B. Angle of incidence must exceed the critical angle C. Refractive index of second medium must be lower D. Light must strike at 90° Answer: D 7.2.4 Ray Diagrams & Angle Calculations 5. A ray in water (n = 1.33) hits the surface at 40°. Critical angle = 48.8°. What happens? A. Refraction only B. Total internal reflection C. No refraction D. Light stops Answer: A 7.2.5 Snell’s Law in Glass Blocks & Prisms 6. A ray enters glass (n = 1.5) from air at 30°. Which statement is correct? A. It bends away from the normal B. It bends toward the normal C. It travels straight D. It undergoes total internal reflection Answer: B 7. In a prism, the deviation of light occurs mainly because: A. Light slows down in glass B. Light speeds up in glass C. Light reflects internally D. Light cannot pass through glass Answer: A 7.2.6 Mirages 8. A mirage appears on a hot road because: A. Light reflects off the sky B. Light refracts through layers of air with different densities C. Light undergoes dispersion D. Light travels in straight lines only Answer: B 7.2.7 Dispersion Through a Prism 9. Dispersion occurs because: A. All colors refract equally B. Different wavelengths refract differently C. The prism reflects light D. White light cannot be refracted Answer: B 7.2.8 Rainbow Formation 10. A rainbow is formed due to: A. Refraction only B. Total internal reflection only C. Dispersion only D. Refraction + TIR + dispersion Answer: D 7.2.9 Optical Fibers 11. Optical fibers work mainly due to: A. Refraction B. Diffraction C. Total internal reflection D. Dispersion Answer: C 12. Which is an advantage of optical fibers? A. High signal loss B. Immune to electromagnetic interference C. Very heavy D. Slow data transmission Answer: B

The cytoskeleton is a network of thin tubes and filaments that crisscrosses the cytosol. The tubes and filaments give shape to the cell from the inside in the same way that tent poles support the shape of a tent. The cytoskeleton also acts as a system of internal tracks, shown in Figure 4-18, on which items move around inside the cell. The cytoskeleton’s functions are based on several struc- tural elements. Three of these are microtubules, microfilaments, and intermediate filaments, shown and described in Table 4-2. Microtubules Microtubules are hollow tubes made of a protein called tubulin. Each tubulin molecule consists of two slightly different subunits. Microtubules radiate outward from a central point called the centrosome near the nucleus. Microtubules hold organelles in place, maintain a cell’s shape, and act as tracks that guide organelles and molecules as they move within the cell. Microfilaments Finer than microtubules, microfilaments are long threads of the beadlike protein actin and are linked end to end and wrapped around each other like two strands of a rope. Microfilaments con- tribute to cell movement, including the crawling of white blood cells and the contraction of muscle cells. Intermediate Filaments Intermediate filaments are rods that anchor the nucleus and some other organelles to their places in the cell. They maintain the inter- nal shape of the nucleus. Hair-follicle cells produce large quantities of intermediate filament proteins. These proteins make up most of the hair shaft. 84 CHAPTER 4 TABLE 4-2 The Structure of the Cytoskeleton Property Microtubules Microfilaments Intermediate filaments Structure hollow tubes made of two strands of intertwined protein fibers coiled into coiled protein protein cables Protein subunits tubulin, with two subunits: å actin one of several types of and ∫ tubulin fibrous proteins Main function maintenance of cell shape; cell maintenance and changing of maintenance of cell shape; motility (in cilia and flagella); cell shape; muscle contraction; anchor nucleus and other chromosome movement; movement of cytoplasm; cell organelles; maintenance of organelle movement motility; cell division shape of nucleus Shape Microtubules provide a path for organelles and molecules as they move throughout the cell. FIGURE 4-18 Microtubules Nucleus Endoplasmic reticulum Mitochondrion Ribosomes Copyright © by Holt, Rinehart and Winston. All rights reserved. Copyright © by Holt, Rinehart and Winston. All rights reserved. CELL STRUCTURE AND FUNCTION 85 1. Explain how the fluid mosaic model describes the plasma membrane. 2. List three cellular functions that occur in the nucleus. 3. Describe the organelles that are found in a eukaryotic cell. 4. Identify two characteristics that make mitochon- dria different from other organelles. 5. Contrast three types of cytoskeletal fibers. CRITICAL THINKING 6. Relating Concepts If a cell has a high energy requirement, would you expect the cell to have many mitochondria or few mitochondria? Why? 7. Analyzing Information How do scientists think that mitochondria originated? Why? 8. Analyzing Statements It is not completely accurate to say that organelles are floating freely in the cytosol. Why not? SECTION 3 REVIEW During cell division, centrioles organize microtubules that pull the chromosomes (orange) apart. The centrioles are at the center of rays of microtubules, which have been stained green with a fluorescent dye. FIGURE 4-20 Cilia and Flagella Cilia (SIL-ee-uh) and flagella (fluh-JEL-uh) are hairlike structures that extend from the surface of the cell, where they assist in movement. Cilia are short and are present in large numbers on certain cells, whereas flagella are longer and are far less numerous on the cells where they occur. Cilia and flagella have a membrane on their outer surface and an internal structure of nine pairs of micro- tubules around two central tubules, as Figure 4-19 shows. Cilia on cells in the inner ear vibrate and help detect sound. Cilia cover the surfaces of many protists and “row” the protists through water like thousands of oars. On other protists, cilia sweep water and food particles into a mouthlike opening. Many kinds of protists use flagella to propel themselves, as do human sperm cells. Centrioles Centrioles consist of two short cylinders of microtubules at right angles to each other and are situated in the cytoplasm near the nuclear envelope. Centrioles occur in animal cells, where they organize the microtubules of the cytoskeleton during cell division, as shown in Figure 4-20. Plant cells lack centrioles. Basal bodies have the same structure that centrioles do. Basal bodies are found at the base of cilia and flagella and appear to organize the devel- opment of cilia and flagella.

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

Slide 1 Growing Up in the 21st Century: Challenges and Opportunities Slide 2 Introduction: What Does It Mean to Grow Up? • Growing up: The process of maturing physically, mentally, and emotionally • Transition from childhood to adulthood • Unique challenges and opportunities in the 21st century • Importance of mental growth alongside physical development Slide 3 The Journey of Self-Discovery • Exploring personal identity • Understanding values and beliefs • Developing a sense of purpose • Embracing individuality while finding community Slide 4 Mental Growth: A Key Aspect of Maturity • Emotional intelligence and self-awareness • Critical thinking and problem-solving skills • Adaptability and resilience • Importance of continuous learning and personal development Slide 5 Challenges of Growing Up in the Digital Age • Information overload and digital literacy • Social media pressure and online identity • Cyberbullying and online safety • Balancing screen time with real-life experiences Slide 6 21st Century Skills for Success • Technological proficiency • Communication and collaboration • Creativity and innovation • Global awareness and cultural competence Slide 7 Navigating Relationships in a Connected World • Building and maintaining friendships • Romantic relationships in the digital era • Family dynamics and independence • Professional networking and mentorship Slide 8 Education and Career Pathways • Evolving job market and emerging industries • Importance of lifelong learning • Balancing academic success with practical skills • Exploring unconventional career paths Slide 9 Financial Literacy and Independence • Understanding personal finance • Budgeting and saving strategies • Student loans and debt management • Investing for the future Slide 10 Mental Health and Well-being • Recognizing and managing stress • Importance of self-care and work-life balance • Seeking help and support when needed • Destigmatizing mental health issues Slide 11 Physical Health in a Changing World • Importance of regular exercise • Nutrition and healthy eating habits • Sleep hygiene and its impact on well-being • Avoiding harmful substances and addictive behaviors Slide 12 Environmental Awareness and Sustainability • Understanding climate change and its impacts • Developing eco-friendly habits • Participating in community environmental initiatives • Sustainable career opportunities Slide 13 Civic Engagement and Social Responsibility • Understanding political systems and processes • Importance of voting and civic participation • Volunteering and community service • Advocating for social justice and equality Slide 14 Cultural Competence in a Global Society • Appreciating diversity and inclusion • Developing intercultural communication skills • Opportunities for travel and cultural exchange • Embracing multilingualism Slide 15 Time Management and Productivity • Setting goals and priorities • Effective study and work habits • Balancing academics, extracurriculars, and personal life • Avoiding procrastination and developing discipline Slide 16 Dealing with Failure and Setbacks • Reframing failure as a learning opportunity • Building resilience and grit • Developing a growth mindset • Seeking feedback and continuous improvement Slide 17 Technology and Ethics • Understanding digital footprint and online reputation • Responsible use of social media and technology • Privacy concerns and data protection • Ethical considerations in a tech-driven world

Slide 1: • Title slide with the presentation topic: "Understanding Context in Film Analysis" Slide 2: • Introduction to the importance of context in film analysis. • Engaging visuals to capture students' attention. • Emphasize that context provides a deeper understanding of a film's meaning. Slide 3: • Definition of Context: • Context refers to the surrounding circumstances or conditions that influence the creation, interpretation, and reception of a film. • Analyzing context helps uncover layers of meaning, societal influences, and enhances critical thinking skills. Slide 4: • Historical Context: • Definition: Historical context refers to the specific time period in which a film was created and/or set. • Importance: Understanding the historical context helps us connect the film to its time period and comprehend the influence of historical events, social norms, and cultural movements. • Example: Analyzing the historical context of "Gone with the Wind" (1939) allows us to appreciate how the film reflects the post-Civil War era in the United States and addresses themes of race, class, and gender. Slide 5: • Social Context: • Definition: Social context refers to the social structures, norms, and values prevalent during the time of a film's creation and/or setting. • Importance: Examining the social context helps us understand how societal attitudes and values shape the film's narrative, characters, and themes. • Example: Analyzing the social context of "The Breakfast Club" (1985) reveals how the film explores the social dynamics and stereotypes within a high school setting, reflecting the cultural climate of the 1980s. Slide 6: • Political Context: • Definition: Political context refers to the political climate and ideologies present during the time of a film's creation and/or setting. • Importance: Understanding the political context helps us uncover political messages, power dynamics, and social commentary within the film. • Example: Examining the political context of "V for Vendetta" (2005) allows us to appreciate how the film critiques totalitarianism and explores themes of government control and individual freedom. Slide 7: • Authorial/Directorial Context: • Definition: Authorial/Directorial context refers to the background, artistic choices, and intentions of the director or filmmaker. • Importance: Analyzing this context helps us understand the director's unique vision, influences, and storytelling techniques, which shape the film's style and thematic focus. • Example: Exploring the authorial/directorial context of "Pulp Fiction" (1994) reveals Quentin Tarantino's nonlinear storytelling, pop culture references, and exploration of morality and violence. Slide 8: • Importance of considering multiple contexts together: • Analyzing multiple contexts together provides a comprehensive understanding of a film's meaning and impact. • Exploring the interplay between historical, social, political, and authorial/directorial contexts deepens our insights and enhances critical analysis skills. Slide 9: • Case studies: • Present two different films as case studies. • Example 1: Analyzing the historical context, social context, and authorial/directorial context of "Black Panther" (2018) provides insights into its exploration of Afrofuturism, cultural identity, and representation. • Example 2: Examining the historical context, political context, and authorial/directorial context of "Citizen Kane" (1941) reveals its commentary on power, media, and the American dream. Slide 10: • Summary slide: • Recap the main points about context in film analysis. • Encourage students to apply these concepts to their own analysis.

1. Menurut Oktarini & Evri (2020), literasi berhubungan erat dengan kemampuan? a. Menggambar dan menulis b. Menghitung dan berbicara c. Membaca, menulis, berbicara, dan mengolah informasi d. Mendengarkan dan mengarang 2. Menurut Dhina (2020), literasi tidak hanya membaca dan menulis, tetapi juga mencakup? a. Keterampilan berbicara di depan umum b. Keterampilan berpikir kritis memanfaatkan sumber pengetahuan cetak, visual, maupun digital c. Keterampilan berhitung cepat d. Kemampuan menghafal teks 3. Literasi di sekolah dasar tidak hanya sebatas kemampuan membaca dan menulis, tetapi juga membekali siswa dengan keterampilan abad ke-21 4. Jika seorang guru ingin mengembangkan kemampuan Critical Thinking siswa melalui kegiatan literasi, strategi mana yang paling tepat dilakukan? a. Membiasakan siswa membaca 15 menit sebelum pelajaran dimulai. b. Mengajak siswa menyalin kembali isi bacaan ke dalam catatan harian. c. Memberi tugas siswa untuk membandingkan dua teks berbeda dan menarik kesimpulan. d. Mengadakan lomba membaca cepat antar siswa untuk meningkatkan motivasi. 4. Gerakan Literasi Sekolah (GLS) terdiri atas tahap pembiasaan, pengembangan, dan pembelajaran berbasis literasi. Seorang guru meminta siswa menanggapi isi bacaan dengan menghubungkannya ke pengalaman pribadi. Kegiatan tersebut termasuk pada tahap … a. Pembiasaan b. Prabaca c. Pengembangan d. Pembelajaran berbasis literasi (Jawaban: C) 5. Menurut Clay (2001) dan Ferguson, komponen literasi informasi terdiri dari berikut, kecuali? a. Literasi dini b. Literasi visua c. Literasi finansial d. Literasi teknologi (Jawaban: C) 6. Literasi dini merupakan kemampuan untuk? a. Menyimak, memahami bahasa lisan, dan berkomunikasi melalui gambar dan lisan b. Menggunakan katalog dan indeks perpustakaan c. Membaca dalam hati selama 15 menit d. Mengetahui etika dalam memanfaatkan teknologi 7. Literasi digital menjadi semakin penting di era banjir informasi. Jika seorang siswa menemukan informasi yang meragukan di internet, langkah literasi digital yang paling sesuai untuk melatih kemampuan berpikir kritis adalah … a. Menyebarkan informasi tersebut di media sosial untuk meminta pendapat teman. b. Membandingkan informasi tersebut dengan sumber terpercaya sebelum menyimpulkan. c. Mengabaikan informasi karena dianggap tidak relevan dengan pelajaran. d. Menyimpannya untuk dijadikan bahan presentasi tanpa verifikasi. 8. Literasi visual mengintegrasikan teknologi dan media. Jika guru meminta siswa menafsirkan pesan dari poster literasi yang ada di kelas, maka keterampilan yang diasah adalah … a. Menulis kreatifb. Memahami makna non-verbal c. Membuat sistem katalog perpustakaan d. Menganalisis data kuantitatif (Jawaban: B) 9. Peran literasi dalam keterampilan abad ke-21 meliputi 4C. Jika siswa mampu mengajukan ide baru setelah membaca beberapa referensi, maka keterampilan yang muncul adalah … a. Critical Thinking b. Collaboration c. Creativity d. Communication (Jawaban: C) 10. Gerakan Literasi Sekolah menekankan kerja sama berbagai pihak. Jika dukungan orang tua minim dalam kegiatan literasi, maka strategi paling tepat untuk mengatasinya adalah … a. Menghapus kewajiban literasi di rumah. b. Mengurangi jumlah bacaan yang diberikan pada siswa. c. Melibatkan orang tua dalam pelatihan literasi sederhana. d. Membatasi kegiatan literasi hanya di sekolah.

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