Skema JawapanSoalan 1 : SMK PSP tak beri skema jawapan Soalan 2 : 2 (a) (i) (ii ) (b ) (c) (i) (ii ) (d ) 1 1 Petua genggaman tangan kanan / The right - hand grip rule Tinggi / Higher Jauh / Further 1 1 P 1 Total 6 Slope of the gradient is higher. (b) Q= Pt = (1000W) x (6x60 s) = 360000 J @ 360 kJ (c) Kecerunan semakin meningkat.(a) Pressure is the force acting on per unit area normal to the force. so the shorter the time to boil the water.0 × 104 Pa (b) (i) different of pressure / perbezaan tekanan Soalan 4 : 4 (a) Haba pendam tentu pengewapan suatu bahan ditakrifkan sebagai jumlah haba yang diperlukan untuk mengubah 1kg bahan itu daripada fasa cecair kepada fasa gas tanpa perubahan suhu. Maka masa lebih singkat untuk air mendidih. Soalan 3 : Tekanan adalah daya yang bertindak per unit luas normal (ii)Water pressure at the kepada daya itu. OR Tekanan air pada dasar atau takungan air. P=hρg = 3 × 1 000 × 10 = 3. Specific latent heat ofvaporization of a substance is defined as the amount of heat required to change 1kg of the substance from liquid phase to gas phase without change in temperature. Because pressure on the mountain top is lower cause the boiling point of water decrease. (1) . Kerana tekanan di puncak gunung rendah. base of reservoir.(1) Specific heat capacity of pot: small (1) because heat that needed is little to cook the egg. (d) Muatan haba tentu perikuk : rendah (1) kerana haba yang diperlukan sedikit sahaja untuk memasak telur . Menyebabkan takat didih air menjadi kurang. 8 m f = 5 Hz Frekuensi gelombang di kawasan air dalam = Frekuensi gelombang air cetak.Perambatan cahaya mestilah daripada medium ketumpatan tinggi kepada mediun ketumpann rendah. b. Gentian optic -Gentian optik terdiri daripada dua bahagian iaitu bahagian teras dan bahagian luarnya.2 c i. Dua syarat Pantulan Dalam Penuh. Sudut tuju 6. -Bahan bagi bahagian teras adalah lebih tumpat ( indeks biasan / n lebih besar) berbanding bahagian luar teras ( penyalut) -Sudut tuju cahaya yang masuk lebih besar daripada sudut gentingnya. . cii.1 < sudut tuju 6. 1.Sudut tuju > sudut genting. -Maka pantulan dalam penuh akan berlaku. Soalan 6 : 6a Sudut genting adalah sudut tuju apabila sudut pembiasannya adalah 90o . 2.Soalan 5 : Soalan 5 (a) (b) (c) (d)(i) (ii) Skema Pemarkahan Pembiasan λ = v/f = 4/5 = 0. 1 -Sebab sebahagian tenaga elektrik digunakan untuk mengatasi rintangan dalam sel bateri. 1 (b) 2 . ketegangan kurang method in diagram 2.45 cm T 60 0 C 60 0 C T (d) 2T = W 2T = 600 T = 300N (f) kaedah 2. 1 -Nilai bacaan voltmeter yang dicatatkan dalam litar terbuka menjadi kurang daripada nilai dge . tension is lower Soalan 8 : 8 (a) (i) (ii) beza keupayaan yang merentasi sel apabila litar berada dalam keadaan terbuka atau tiada arus yang mengalir dalam litar.Soalan 7 : (a) daya paduan / daya bersih kosong resultant force / net force is zero (b) W = mg = 60 x 10 = 600 N (c) T = 345 N L = 3.45 cm T = 345 N L = 3. pembiasan cahaya lebih Dan panjang fokus kanta menjadi lebih pendek. . Therefore the thicker the lens. the greater the refraction of light and with that the shorter the focal length of a lens will be (b) (i) 1 Sebagai kanta pembesar. Distance from the optical centre to a focal point. 1 5 Lens K is thicker than lens J Light ray is refracted more in lens K than lens J. Kanta K lebih tebal dari kanta J Sinar cahaya dibiaskan lebih dalam kanta K dari kanta J Panjang fokus kanta K lebih pendek dari kanta J Lebih tebal kanta. Focal length of lens K is shorter than lens J.Boleh beri markah jika tiada reostat (c)(i) (ii) berkadar terus 1 1 R = 32/8 = 4 Ω 1 (d) (i) Q 1 (ii) P 1 - 1 Rintangan rendah Mengurangkan kehilangan tenaga 1 JUMLAH 12 Soalan 9 : 9(a) (i) (ii) Jarak dari pusat optik ke pusat fokus. pembesaran ↑ High power lens has a 4 . Guna kanta cembung berkuasa tinggi sebagai kanta mata. fo . (ii) Pembesaran Songsang Maya 3 Enlarge/magnified Inverted/upside down Virtual (c)(i) Kanta S sebagai kanta objektif Kanta Q sebagai kanta maata Kanta S diletakkan depan kanta Q Dua kanta disusun supaya adalah dalam penyelarasan normal dimana jarak antara dua kanta = fo + fe (ii) Choose lens S as objective lens Choose lens Q as eyepiece Lens S is placed in front of lens Q The two lenses are adjusted so that they are in normal adjustment where distance between the two lenses is equal to (fo + fe) Pengubahsuaian/Modificati on 1. Gunakan kanta cembung berkuasa rendah sebagai kanta objektif. fo . kanta berkuasa rendah mempunuai panjang fokus yang lebih. 2. Use high power convex lens as the eye lens Penjelasan/Explanation . magnification . Low power lens has a longer focal length. fe .Pembesaran teleskop/Magnification of fo fe telescope = .As a magnifying glass. Kanta yang berkuasa tinggi mempunyai panjang fokus yang lebih pendek. Use low power convex lens as the objective lens. pembesaran . [1 m] (c) – Soft wood or a thick curtain is placed in front of a hard wall.e. diffraction and interference. is reflected and received by a receiver. The wave hits a target. [1 m] ii. which can travel through a medium( water and solid) [1 m] – The wave is generated from an instrument. reflection. – A wall with holes or lined with objects with holes. More light permitted to enter the telescope and a clearer image is seen 6 20 Soalan 10 : 10 (a) i. – The wave used is the longitudinal wave. [1 m] – Reflected waves [1 m] – Reflected waves occur when incident waves hit a reflector and are reflected. [1 m] (b) – An example of a longitudinal wave is the sound wave whereas a transverse wave does not need a medium to travel. [1 m] – Both types of waves show the properties of waves. [1 m] – The soft surface can absorb sound and reduce the reflection of sound. for example egg trays. can reduce . Gunakan kanta yang berdiameter lebih besar sebagai kanta objektif.3. [1 m] – The angle of incidence is equal to the angle of reflection. fe . refraction. [1 m] – The reflection of sound produces interference of sound. [1 m] – Both types of waves transfer energy from one end to another end. measure the depth of the sea. i. magnification ↑ Lebih banyak cahaya dibenarkan masuk ke teleskop dan imej yang lebih jelas akan dapat dilihat. [1 m] – A longitudinal wave needs a medium to travel whereas a transverse wave does not need a medium to travel. Sonar is an echo system that uses ultrasonic waves to detect objects in the sea and to . Use bigger diameter of objective lens shorter focal length. Prinsip Bernoulli 1 Characteristic A shape of cross section which is upper side is longer than the bottom/aerofoil Explanation To produce the speed of airflow above the wings to be higher than the speed of airflow below Large the area of the wing The larger the lift force Low density of the wing materials Less weight // produce more upward resultant force The higher the difference in speed of air The higher the difference in pressure 1 1 1 1 1 1 1 1 1 1 1 1 . [1 m] – Sound with high frequency will break the glass in the auditorium. [1 m] – The glass used must not be fragile. The computer will generate sound waves to produce destructive interference with the sound. [1 m] – Normally. – The sitting area is arranged in the antinode region only so that louder sound can be heard. The amplifier acts as the coherent source. the audience gets to enjoy the performances without any disturbance. [1 m] – Sound waves produced by the audience can be distracting. [1 m] Soalan 11 : No. When the speed of moving air is higher the pressure is lower. This causes certain parts to receive soft sounds. Constructive interference and destructive interference occur. Hence air pressure below the wings is higher compare to above the wings. 11 (a) (b) (i) (ii) (c) Aerofoil 1 The shape of cross section of the wing causes the speed of airflow above the wings to be higher than the speed of airflow below.the effect. [1 m] – The microphone can detect the sound and send the information to a computer. Therefore. two amplifiers are fixed in a concert auditorium. large the area of the wing. -Putaran turbin akan menghasilkan arus elektrik. ( 2 M) Perubahan tenaga yang terlibat ialah. Tenaga keupayaan tenaga kinetic tenaga elektrik 12b. the low density of the wing materials and the high the difference in speed of air. Kuasa adalah kadar melakukan kerja. i. Np = 500 Ns = 1000 ( 6 Markah ) . Transformer P : -Transformer injak naik Formula : 12 = 500 24 1000 Transformer Q -Transformer injak turun ( 2 M) Transformer R -transformer injak turun Vp = Np Vs Ns 24 = 1200 8 400 8 = 160 6 120 Np = 1200 Ns = 400 Np = 160 Ns = 120 Maka. ii -Pengaliran air (daripada air terjun /empangan) akan memutarkan turbin. i. Because it has a shape of cross section which is upper side is longer than the bottom.The most suitable wing is Y. (d)(i) 1 1 1 P=F/A F = 400 x 50 F = 20000 N (ii) 1 1 Resultant Force = 20 000 – 900(10) = 11000 N 1 Direction of force : upwards Total 20 Soalan 12 : 12a. 12b.4)2 ( 30 ) = 4. teras besi lembut mudah dicas dan dinyahcaskan / berlamina untuk mengurangkan ‘arus eddy’ ( 1 Markah ) 12b. ii.8 W ( 1 Markah ) ( 1 Markah ) ( 2 Markah ) ( 3 Markah dengan unit yang betul) . ii. I = V/R = 12/30 = 0. i. Jenis wayar : Kuprum Kerana. Jenis teras : Teras besi lembut berlamina ( 1 Markah ) Kerana . 12c. iii. Mempunyai rintangan yang rendah 12c.4 A P = I2 R = (0.
Report "skema jawapan Fizik PPT 2016 kertas 2.docx"