Jeemion JealiTeacher at SM Sains Kuching um SM Sains Kuching

Forces and Pressure

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• 1. FIZIKMOZAC 2010 CHAPTER 3: FORCES & PRESSURE ANSWER 3.1 PRESSURE Question 1 (a) Pascal = Nm-2 (b) A = 800 x 2 x 10-3 = 1.6 m2 P = 500 = 312.5 Pa 1.6 (c) A sharp knife has a small surface area. So it produce larger pressure on the bread. Question 2 (a) Pressure = force area (b) (i) balloon B (ii) pressure on balloon B is higher (iii) the surface of the needle in contact with the balloon is smaller than the finger. (iv) the smaller the surface area, the larger the pressure exerted on the balloon. (v) Pressure increases Question 3 (a) (i) The contact area between the wheels in Diagram 6.2 is larger (ii) weight are equal (b) Vehicle uses the wheels in Diagram 6.2. Because it has smaller pressure exerted on the soft ground and will not sink. (c) Pressure (d) 10 000 : 500 4A 2A 2,500 : 250 10 : 1 (e) When the air pressure inside the wheel lower, the contact area is larger so the pressure on the ground is smaller 3.2 LIQUID PRESSURE Question 4 (a)(i) The wall of a dam in Figure 4.2 is much thicker at the bottom than at the top and withstand the higher pressure at the bottom of the lake. (a)(ii) Pressure at B is higher than at A (b)(i) Dam in Diagram 5.2 (b)(ii) 1- When depth increases, pressure increases. 2- Thicker at the base can withstand high pressure. (c)(i) Siphon system (c)(ii) Diffrence in water level will cause different in pressure (c)(iii) Question 5 (a) Depth / density / acceleration due to gravity (b)(i) PQ > P p (b)(ii) PQ = h ρ g = 5 x 1000 x 10 = 50000 Pa (c)(i) Different in pressure (c)(ii) Water level at P is same as the water level in the house water tank // pressure is the same between at P and inside the tank.
• 2. FIZIKMOZAC 2010 No difference in pressure (d)(i) Place the concrete tank at higher place // on top of hill Higher difference of pressure. or Use water pump Increase the difference of pressure. (d)(ii) Question 6 (a)(i) Magnitude : same magnitude of atmospheric pressure Directions : atmospheric pressure and mercury are in the same direction // gas pressure direction against the direction of mercury and atmospheric pressure // atmospheric pressure acts downwards (a)(ii) Phg + Patm ,// P gas (a)(iii) Same / equal (b) Gas pressure = atmospheric Pressure + mercury pressure (c)(i) Mercury level drops and at same level in both columns (c)(ii) Same pressure // atmospheric Pressure 3.4 PASCAL’S PRINCIPLE Question 7 (a) Pascal’s principle (b) Show the correct direction (c) the liquid pressure in the main brake cylinder and the small brake cylinder are the same/equal 1. 4 2 4 106105 15     F 2. F2 = 18 N Question 8 (a) Pascal’s principle (b) P = 5/2 = 2.5 Ncm-2 = 2.5 x 104 Nm-2 (c) Same pressure (d) F2 = 2.5 x 5 = 12.5 N (e) Liquid cannot be compressed easily 3.5 ARCHIMEDES’ PRINCIPLE Question 9 (a)(i) Archimedes principle (a)(ii) upward: Buoyant force Downward: weight of Hydrometer (b)(i) the length of hydrometer submerged in oil is longer than in water. (b)(ii) Density of oil is less than water (c)(i) Buoyant force = Wair - Wwater = 0.25 – 0.22 = 0.03 N (c)(ii) volume of object = volume of water displaced 0.03 = 1000 x V x 10 V = 3 x 10-6 m3 Question 10 (a) Pacal (b) Depth (c) (i) Weight of water displaced = buoyant force = ρVg = 1010 x 2.5 x 10 = 25,250 N (ii) Tension + buoyant force = weight of object T = 125,000 – 25,250 = 99,750 N
• 3. FIZIKMOZAC 2010 Question 11 (a) Mass per volume (b) (i) Density sphere A less than B (ii) Weight A less than B (iii) The weight of water displaced by A less than B (iv) Larger weight of sphere, displaced bigger weight of water (v) Weight of water displaced = up thrust // When the weight of water displaced increase, up thrust increase (c) Archimedes’ principle (d) Submarine Question 12 (a)(i) Same Volume Net force zero (a)(ii) Y < X < Z (b)(i) Box Y floats and immersed partially / box X immersed fully and floats box Z sink (b)(ii) Greater weight means greater mass and greater density. The higher density object needs more volume to increase the buoyant force to support the weight . (c) Archimedes principle// equilibrium of forces Question 13 (a) Density is the mass per volume (b)(i) Level of the boat is higher in the sea than in the river. (the part of boat submerged in the sea is less than in the river) (ii) Water displaced in the sea is less than in the river. (b)(iii) Density of sea water is higher than river water. (c)(i) The lower the density of water, the greater /higher the volume of water displaced. (c)(ii) Weight of the boat = Weight of the water displaced (d) Archimedes’ principle (e) Ballast tank filled by sea water Weight of submarine > buoyant force Question 14 (a)(i) Function – for safety purpose/To ensure the maximum weight limit (a)(ii) F = mg = 7500 x 10 = 7.5 x 104 N (a)(iii) The mark should be higher than the sea water level (a)(iv) 1. Density of sea water is denser than the density of river water. 2. The volume of water displaced increased when density of liquid decrease (b)(i) Up thrust = Weight (b)(ii) Accelerates upwards or moves Up wards (b)(iii) 1. The weight of the air balloon is decreased 2. Buoyant force /up thrust higher than weight 3. The balloon experiences the unbalanced force. Question 15 (a)(i) Bernoulli’s principle (a)(ii) Y (b) The air moves with a high speed (c)1. The atmospheric pressure which is higher pushes the liquid up through the narrow tube.
• 4. FIZIKMOZAC 2010 2. The jet air will force the liquid to be sprayed as fine spray liquid Question 16 (a)(i) Student mark at the same level in tube - X,Y and Z- (a)(ii) Atmospheric pressure (b)(i) Water level in vertical tube P is higher than in R and higher than in P./hp > hR > hQ (b)(ii) Bernoulli’s principle. (b)(iii) P = hρg = 0.15 x 1000 x 10 = 1500 Pa Question 17 (a) Distance per time (b) (i) Before: water levels are the same and the roof stay intact. After : water levels are not the same and the roof rise up. (b)(ii) Pressure above the roof is higher compare to pressure below (b)(iii) Speed increases pressure decreases or vice versa (c) Bernoulli (d)(i) Q is slower and R is faster (d)(ii) Q is higher and R is lower Question 18: Kedah 07  The depth of the water in Diagram 9.1 is higher than in Diagram 9.2  The water spurts out in Diagram 9.1 is at a higher rate than in Diagram 9.2  The water spurts out further in Diagram 9.1 than in Diagram 9.2  The deeper the water, the further the distance of water spurt  The deeper the water, the higher the pressure of the water Question 19: Kedah 07 (a)  The pressure of water increases with the depth of the water  The bubble expands upon reaching the surface of the water//The volume of air bubble increases as the depth of water decreases (b)  Buoyant force increases as the volume of the bubble increases  The air bubble moving with increasing acceleration  (volume of air bubble = volume of water displaced)  (Buoyant force is larger than the weight of the air bubbles) Question 20: Trengganu 07  A force is applied when you squeezed at the bottom end of the toothpaste tube  Pressure is applied to the toothpaste (tube)  According to Pascal’s principle  The pressure is transmitted equally to the whole tube Question 21: Perak 07  High altitude low density of air  Less collision of molecules with surface  Low altitude high density of air  More collision of molecules with surface Question 22:SBP 07  B is denser than A.  The weight of water displaced is the same of the weight of the rod.  Weight of B is greater than weight of A  B will displace more volume of water
• 5. FIZIKMOZAC 2010 Question 23: Trengganu 07  The shape of the wing is aerofoil.  The shape of cross section of the wing causes the speed of airflow  The air move faster than above the wings than below the wing.  When the speed of moving air is higher ,the pressure is lower  Hence air pressure below the wings is higher compare to above the wings  there is difference in pressure which produce an upward resultant force.  Bernoulli’s principle Question 24: Teknik 07 (a)  Buoyant Force : Force experience when an object totally or partly immersed into the liquid (b)  Density of the gas inside the balloon less dense then air  Air is displaced by the balloon and produced buoyant force  The buoyant force is larger than the weight of the balloon and load and it rises up.  When the buoyant force is equal to weight of balloon and load, it will float still. (c) Quantitative problem: (i) Resultant force = 250 – 5 = 245 N (ii) Use F=ma 245 = 5 a a = 49 ms-2 (iii) air resistance is zero Characteristics Reason Used helium gas Its light/less dense then air Mass of load is 20 kg Total weight of balloon and the load equal to buoyant force Tension allowed of the rope is 300 N To ensure the rope not break (ii) Set C Because its used helium gas, mass of load is 20 kg and tension allowed is greater than 250 N (iii) A is not suitable because mass of the load causes weight of the load and the balloon less then buoyant force. The balloon will rise up ( Accept any other set and the reason) Question 25: Perak 07 Characteristic Reason Large tyre better stability Liquid in hydraulic system liquid cannot be compressed Large mass big inertia Large base area better stability Low centre of gravity better stability Choose – M Large tyre, liquid in hydraulic system, large mass, large base area or low centre of gravity l
• 6. FIZIKMOZAC 2010 Question 26: SBP 07 Characteristic Reason Material made from glass Glass does not corrode with acid Small diameter of capillary tube To increase the sensitivity of the hydrometer High density of shots Makes the hydrometer stays upright Big diameter of bottom bulb To obtaine a bigger upthrust Choose N N is made from glass, has small diameter of capillary tube, high density of shots and a big diameter of bottom bulb. Question 27:Trengganu 07 Characteristics Reason A shape of cross section which is upper side is longer than the bottom To produced the speed of airflow above the wings to be higher than the speed of airflow below Large surface area of the wing Produce larger lift force Less density of the wing materials Less weight // produce more upward resultant force Higher difference in speed of air The higher the difference in pressure The most suitable choice is P Because it has A shape of cross section which is upper side is longer than the bottom Large surface area of the wing Less density of the wing materials High difference in speed of air Question 28: Kedah 07 (i)  Diagram 9.3  The weight of the dam is supported by the force exerted by the water (ii)  Water in the dam can be filtered and chlorinated to be uses as public water supply  To drive turbines for the generation of hydroelectricity// For irrigation//Recreation centre Suggestion Explanation Thicker wall at the base To withstand greater pressure at the bottom as the pressure increases with depth The wall is constructed using stronger materials / Using reinforce concrete To avoid the wall from breaking / To increase the strength of the wall / To avoid leaking Equipped with the water overflow system To avoid flooding / To channel away the overflow water Question 29:Trengganu 07 modification explanation piston of bigger cross-sectional area Can support greater force (weight) Low density material Lightweight // easy to carry Non- compressible liquid Piston can be lifted up
• 7. FIZIKMOZAC 2010 Longer handle Less effort needed to press the small piston Apply released valve between small and main reservoir Liquid can flows into small reservoir Question 30: Perak 07 (i) hρg = 0.76 x 13 600 x 10 =103360 Pa (ii) hρg = 0.1 x 13 600 x 10 = 13600 Pa (iii) 0 Pa Question 31: SBP 08 (a) (i) Mass devide by volume (a) (ii)  Density of air in Diagram 9.1 is higher than in Diagram 9.2.// vice versa  The number of load in Diagram 9.1 is greater than in Diagram 9.2//vice versa  The height of the balloons in both Diagram 9.1 and Diagram 9.2 are equal  When the density of the air increase, the buoyant force increase  As the density of the air increase, the weight of the load carried increased// . As the density of the air decrease, the weight of the load carried also decreased (b)  Density of the iron nail is higher than the density of water// Average density of the cargo ship is lower than the density of water  Volume/ weight of water displaced by the iron nail is smaller  For the cargo ship, the buoyant force is equal to its weight .  For iron nail , its buoyant force is smaller than the weight modification explanation Streamline shape Decrease/ reduce the water drag/resistance thick and strong material To withstand high pressure / / pressure increase with depth Additional component - ballast tank - periscope To float or sink the submarine To observe object outside the water surface Safety feature Oxygen tank / generator For respiration Question 32:Trengganu 08 (a) (i) Gravitational force (a) (ii)  Weight lost in Diagram 9.1(b) > Diagram 9.1(c) // vise versa  Apparent weight in Diagram 9.1(c) > Diagram 9.1(b) // vise versa  Density of water > density of oil  The greater the density of liquid, the greater the weight lost / less apparent weight (iii) Up thrust /buoyant force (b)  Name two correct force (buoyant force and weight)  Buoyant force small because small volume // vise versa  Block sink because weight > buoyant force  Sheet float because weight = buoyant force
• 8. FIZIKMOZAC 2010 modification explaination Strong material Can withstand great force Low density material Lightweight Two stage plimsoll line Save in fresh and salt water Big size Can place more goods Aerodynamic shape Reduce water friction Question 33: Kelantan 08 (a) (i) Weight is the gravitational force acts on an object (a) (ii) Buoyant force = weight of the boat (a) (iii)  Sea water is denser  Boat displaced less sea water and gain the same buoyant force. Therefore boat sinks less in sea water (b) (i) Buoyant force = weight of sea water Displaced = mg = ρVg = 250 x 1080 x 10 = 2.7 x 106 N (b) (ii) 2.7 x 106 = V x 1000 x 10 V = 270 m3 (c) (i) Specification Reasons Small stem and long Increase the sensitivity where the scale divisions are far apart so that small changes in density can be detected. Glass wall Do not erode Large diameter of High upthrusts, displaced more bulb liquid to be able to float easily Lead shoots Hydrometer can stay upright. P is chosen It has small and long stem, glass wall, large diameter of bulb and lead shoots used. Question 34: N9 08 (a) Pressure is defined as the force acting normally per unit area/ Pressure = Force Area (b) 1. When the small piston is pulled up, the hydraulic oil is drawn from the reservoir into the small piston 2. When the small piston is pushed down , the hydraulic oil is exerted with force and experienced a pressure 3. The pressure is transmitted uniformly from the small piston to the big piston. 4. The forced produced raised the big piston / The system can convert a small input force into a bigger output force. Characteristics Reason Has higher boiling point So that liquid not easily boiling/ Has higher specific heat capacity So that it can’t be easily become hot
• 9. FIZIKMOZAC 2010 Has lower density So the hydraulic jack is not heavy Has lower rate of vaporisation Volume of liquid will not easily vaporise Liquid L is chosen Reasons: L has higher boiling point, higher specific heat capacity, lower density and lower rate of vaporisation Question 35: Kedah 08 (a) (i) Archimedes’ principle states that the buoyant force on an object immersed in a fluid is equal to the weight of fluid displaced by the object. (a) (ii)  The balloon acted by two forces: buoyant force and the weight of the balloon.  The density of helium gas is less than the density of surrounding air.  Buoyant force equals to the weight of the air displaced by the balloon.  Buoyant force is higher than the weight of the balloon. (c)  Large balloon  To produce bigger buoyant force// increase the volume of air displaced  Use 2 burners  To produce bigger flame // heat up the gas in the balloon faster  Synthetic nylon  Light-weight, strong and air- proof material.  High temperature of the air in the balloon  Reduce density / weight of the air in the balloon.  Hot air balloon Q is chosen  Because it is large balloon, uses 2 burners / many burners, uses synthetic nylon and has high temperature of the air in the balloon. (d) (i) Weight = buoyant force = weight of water displaced m x 10 = (10 x 2 x 10-6 ) x 1000 x 10 m = 0.02 kg (d) (ii) mg = ρVg (0.02) (10) = (0.12 x 2 x 10-4 ) ρ x 10 ρ = 833.33 kg m-3
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