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The acceleration remains constant for the first 10 seconds before increasing  decreasing constant rate from 10 seconds to 20 seconds

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The bicycle had a constant acceleration for the first 1 seconds.
From 10 to 20 seconds, the bicycle had constant decreasing acceleration.

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1a. There is a constant acceleration for the first 10s. Between 10-20s, the acceleration of the man decreases.

bi. 3.87m/s^2 👍
ii. 40.7m/s 👍

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3.87m/s^2 👍
1bii)40.7m/s 👍

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1a. From 0s to 10s, the man accelerates at a constant rate. (constant acceleration)

From 10s to 20s, the man accelerates at a decreasing rate (decreasing acceleration)

1bi. 3.87ms^-2 👍
1bii. 21.3ms^-1 X (40.7ms^-1) 👍

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3.87m/s2 👍
40.7m/s 👍

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2a. Point where diagonal AC and BD intersect above B 👍
2b. Line of action of weight directly above B 👍
2c. The block can be tilted to a larger angle about B. At a lower height, the centre of gravity is lowered. This increases the stability of the block and hence a larger moment about B is required to create a toppling moment. 👍

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At the point where the line of action of weight is directly above point B. 👍

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3a. 103000Pa X (122000Pa)
3b. As the flask the starts to cool, air molecules have lower kinectic energy ✔ and move slower. Hence, the frequency of collisions between the inner surface of the flask decreases 👌 and the molecules exert a lower force 👍on the inner surface of the flask. Since pressure is defined as a force per unit area, a decrease in the magnitude of the force leads to a decrease in pressure in the flask ✔

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2a. Place the 'X' at the point where the diagonals of AC and BD intersect 👍

b. The line of action of weight, w is directly above B 👌

c. When the height of the block decreases, the centre of gravity is lowered, causing the block to be more stable. This increases the moment used to topple the block  The block can be tilted more (or at greater angle) before it topples over.

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Since the height is lowered, the center of gravity is lower thus the block is more stable. Therefore, the angle at which the box can be tilted before toppling over is greater. 👌

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Where the line of action of the weight is directly above point B 👍

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122000Pa

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The lower height causes the centre of gravity to be lowered thus the block becomes more stable and a force can tilt the rectangle at a larger angle is needed to push it topple it over

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3a. 130000 Pa X (122000Pa)

b. As the air molecules in the flask starts to cool, they have a lower kinetic energy 👌 so the move slower. Therefore the frequency of collisions between the air molecules and the wall of the flask decreases.👍 The air molecules also exert a smaller force on the walls of the flask.✔ As pressure is force per unit area, when the force decreases, the pressure decreases. This cause a drop in mercury level.👍

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As air in the flask cools, the air molecules have a lower kinetic energy 👍 thus they move slower. Since pressure is the force acting per unit area, the air molecules will hit the wall of the container with a lower force 👌and have less frequent collisions with the walls of the flask✔ thus resulting in a decrease in pressure. the decrease in pressure in the flask causes the difference in mercury levels to decrease. 👌

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4a Work is when a force produces a motion perpendicular to the force

Product of force and distance travelled by object
in the direction of the force.

4b. 16.3ms^-1 👌
4ci. 80.0N ❌ (100J)
4cii. 255J ❌  (165J)
4d. Height AB ❌ (above AB)

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work is when a force acting on an object produces a motion X

Product of force and distance travelled by object
in the direction of the force.

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As the air in the flask cools, the molecule lose kinetic energy 👍 and move slower, causing the molecule to collide with the wall of the flask less often👌, thus exerting a lower net force on the flask👍. Since pressure is defined as a force per unit area, this reduces the pressure on the walls of the flask which will cause the difference in mercury level to decrease

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6a. Highest: 1×10^9 Hz
Lowest: 4.17×10^8 Hz ❌ (check your powers of 10)     (4.17x10^8 Hz)

6bi. Infrared  👌
6bii. Remote controls 👌

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20.5m/s ❌(16.3m/s)

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4a.  Work is when an applied force moves an object X

Product of force and distance travelled by object
in the direction of the force.

b. 15.5m/s ❌ (16.3m/s)

ci.  100J ✔
ii.  165J ✔

d. The same height as the line AB ❌ (above AB)

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Work is the energy transferred when a force acts on an object and moves it? X

Product of force and distance travelled by object
in the direction of the force.

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100J ✔

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8.31m/s ❌

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100 J ✔
165 J ✔

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5ai. 3V is the potential differnece across a beaker of liquid of unknown temperatue and the surrounding temperature. Since the temperature of the liquid id unknown, it is not possible to determine the temperature of surrounding air 👌

5aii. 24°C ✔
5b. The thermometer will show a higher  LOWER temperature. This is because the capillary has a smaller WIDER BORE space which means some mercury may be pushed upwards. Hence, a higher temperature may be displayed

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7ai. 22.1° 👌
7aii. The angle of incidence at R2 is greater than the critical angle. Light was also travelling from an optically denser to an optically less dense medium. Hence, the light ray underwent total internal reflection, explaining why the light ray at R2 does not exit the bottle. 👌
7b. In fig 7.2, light undergoes total internal reflection as they hit the inner surface of the plastic bottle causing ALL of the light entering the bottle to exit via the base of the bottle and brightens the entire room.  which allows light to spread out in all directions. In fig 7.3, light just shines into the house without undergoing total internal reflection so light is not spread out in all directions. So light only brighten certain portion of the house
7c. Radiation. There is no medium and only radiation does not require a medium. So thermal energy could travel through space to reach the boy from the sun 👌

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165J ✔

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8bi. Iron 👌
8bii. Iron is a soft magnetic material so it is able to be magnitised and demagnetise easily. Hence when current passes through thr coil, the iron gets magnetised quickly, allowing sound to be projected out of the megaphone quickly

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9a. 0.0825m 👌
9b. Minimum distance= 75000m
Sound wave emmision device is better becuase the beeping only starts closer to the wall so the driver knows when he is approaching the wall 👌

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5ai.  As the cold junction is placed in a liquid with and unknown temperature and the hot junction is placed in the surrounding air, the temperature of the unknown liquid cannot be found as the temperature of the surrounding air and liquid are both unknown , hence the p.d of 3V is useless ❌ The P.D. makes use of the difference in temperature between the hot and cold junction and telling temperature is impossible since temperature of liquid is unknown.

ii. 24 degrees celsius ✔

b. The thermometer would reflect an inaccurate reading as the space between the capillary changes from small to large so the temperature would be lower 👌

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at AB ❌(C?) (above AB)

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12 m ❌ (above AB)

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The temperature of the liquid in the beaker is unknown therefore the 3V which is the p.d across the beaker of water and surrounding air is not useful.❌ The P.D. makes use of the difference in temperature between the hot and cold junction and telling temperature is impossible since temperature of liquid is unknown.

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24 degree celsius ✔

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The non-uniform capillary walls may cause the temperature reading not to increase proportionally with the temperature 👌 The temperature reading after the fault will be lower than the actual reading

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The capillary walls are not evenly distanced and the smaller WIDER space between the walls causes the reading to be higher LOWER. Therefore this leads to an inaccurate reading.

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As the liquid is of an unknown temperature, we cannot use the thermocouple to measure the temperature of the surrounding air through the potential difference ❌The P.D. makes use of the difference in temperature between the hot and cold junction and telling temperature is impossible since temperature of liquid is unknown.

24C ✔

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4.17(10^5)Hz ❌ (check your powers of 10)

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4.17x108 ❌ (check your powers of 10)

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infrared 👌

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Used in remote controls. 👌

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6a.  4.17x10^8 Hz ❌ (check your powers of 10) (4.17 x 10^14 Hz)

bi. Infrared 👌

ii. thermometer 👌

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22.1degrees 👌

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Light is travelling from an optically denser medium to an optically less dense medium and the angle of incidence of R2 is larger than the critical angle. R2 undergoes total internal reflection thus it does not exit the bottle after the first refraction. 👌

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In fig 7.2, the light that enters the bottle undergoes total internal reflection causing it to causing ALL of the light entering the bottle to exit via the base of the bottle and brightens the entire room.  spread in different directions around the room whereas in Fig7.3, light does not undergo total internal reflection and so brightens only part of the room depending on the position of the sun. does not spread around the room.

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Infrared 👌
Remote control  👌

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22.1 degree 👌
The angle of incidence is greater than the critical angle and the light is travelling from the optically denser to the optically less dense medium, thus the light ray undergoes total internal reflection and does not exit the bottle 👌

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In figure 7. 2 the light undergoes total internal reflection causing ALL of the light entering the bottle to exit via the base of the bottle and brightens the entire room. and spreads out ❌ but in figure 7.3 the light does not undergo total internal reflection and thus doesn't spread out  so brightens only part of the room depending on the position of the sun.

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Radiation as it does not require a medium to travel in. Thus the thermal energy is able to travel from the sun through space, where there is no medium, and into the house on earth to reach the boy. 👌

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the empty space between the field lines.

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Radiation can travel without a medium to reach the boy because the infra-red rays must pass through vacuum in outerspace between the sun and the boy.

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10a. 35.1Ω 👌
10b. $4.29 👌

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Iron 👌

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it is a soft magnetic material and is able to be magnetized quickly.  👌

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0.0825m 👌

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7ai. 22.1 degrees 👌

ii. The angle of incidence, i is geater than the critical angle, c so the light ray undergoes total internal reflection. Moreover, the light ray is also travelling from an optically denser medium to an optically less dense medium 👌

b. For fig 7.2 as the light ray is travelling from an optically less dense medium to an optically denser medium, and the angle of incidence is greater than the critical angle, total internal reflection occurs and causing ALL of the light entering the bottle to exit via the base of the bottle and brightens the entire room  it spreads out over the entire house area ❌. However, for fig 7.3,  the light ray passes through the plastic sheet and shines straight into the house without spreading out 👌

7c. Radiation as it does not require a medium to reach the boy so thermal energy could travel through a vacuum in outerspace from the sun. 👌

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8aii. The tangent line in the middle

bi. Soft iron 👌

ii.  Soft iron is a material that is easy to magnetize and demangetise so when current flows through the coil, A gets magnetised easily and this enable the sound to be transmitted quickly. 👌

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a. 4.17x10^14 Hz 👌
bi. Infrared 👌
ii. remote controls 👌

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11ai. 2381kg 👌
11aii. 3.10x10^8J 👌
11aiii. 25800s 👌
b) It takes 6h to heat up the water in FIji but the average daily sunshine is 4h. Hence there is insufficient solar energy to heat up the water in FIji 👌
c. A metal. Becuase metals are good conductors of heat 👌

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The sound wave emission device is better as the minimum distance at which the sensor can detect the sound delay for the equipment that  transmits and receives radio waves is 75000m which is much larger as compared to 0.0825m.  👌

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ai. 22.1 degrees 👌

ii. The angle of incidence is greater than the critical angle and the light ray is travelling from an optically denser medium to an optically less dense medium, thus total internal reflection occurs. 👌

b. In Fig. 7.2, total internal reflection occurs and causing ALL of the light entering the bottle to exit via the base of the bottle and brightens the entire room the light is able to spread out over the house. However, in Fig 7.3, total internal reflection does not occur and thus the light is not able to be spread out  so brightens only part of the room depending on the position of the sun

c. Radiation as it can travel through vacuum ❌ Too brief

Radiation as it does not require a medium to reach the boy so thermal energy could travel through a vacuum in outerspace from the sun

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in the middle between the two magnets

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iron 👌
soft magnetic material that can be easily magnetised and demagnetised 👌

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0.0825m 👌

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9a. 0.0825m 👌

b. The sound wave emission device is better because the sensor is able to produce a beeping sound when the driver is very close to the wall than compared to the radio waves emission device which produces a beep is as far as 75000m 👌

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35.1ohm 👌

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$4.29 👌

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12a. Ohm's Law states that the current passing through a conductor is directly proportional to the potential difference between its ends provided that the physical conditions and temperature remain constant 👌
12b. The I/V graph shows a curve instead of a linear line 👌
c) The resistance increases. As V increases, the gradient which is 1/R  decreases thus resistance will increase 👌
di) rheostat 👌
dii) The rheostat can be set to a higher resistance to increase the effective resistance. Since V=RI and the voltage remains the same, the current will decrease 👍
ei) 0.05A ✔
eii)0.0125A ✔

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the sound wave emission device is better because it can transmit at a smaller distance

Minimum distance
= 0.5(300000000m/s × 0.0005s)
= 75 000 m
75000 m  minimum distance is NOT practical for parking purpose.

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bi. Iron 👌
ii. It is easily magnetized and demagnetized 👌

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a. 0.0825m 👌
b. The sound wave emission device is better as it only transmits a signal when the car is 0.0825m away from the wall as compared to the radio wave emission device that transmits a signal when the car is 75000m away from the wall. 👌

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13a) DCBA. Using the Fleming Left Hand Rule, the thumb which represents the direction of induced force points upwards (along DC) and downwards (along BA) and the second index finger which represents the direction of magnetic field is to the right. Hence, the middle finger which represents direction of current points into the paper for the left side of the coil (D-->C) and (B-->A) on the right side of the coil. Hence, current flows in DCBA

c) Component X is the split ring commutator that reverses the direction of current every 1/2 revolution (when the coil is in a vertical position)  such that a continuous upwards force on the left side of the coil and a continuous downward force on the right side allowing the coil to complete full rotation.

d) The battery has been flipped over such that the positions of their terminals are switched and a stronger battery was used 👍

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35.1 ohm 👌

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2381kg 👌

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10a. 35.1 ohm 👌

b. $4.29 👌

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13ai) Positively charged 👌
13aiii) Gravitational pull form Earth ✔✔
bi)  Electrons from earth are attracted up to the ball 👌
bii) The sphere moves towards the positively charged plate. The ball is has a net negative charge and since unlike charges attract, the ball is attracted to the plate 👍
iii) negative charges spread out evenly 👌

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$4.29 👌

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a. 35.1 ohm 👌
b. $4.29 👌

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11ai. 2381kg  👌

ii. 3.10x10^8 J 👌

iii. 25800s

b. In Fiji, it uses 6 hours to heat up the water. However, the average daily sunlight is only 4 hours so there is insufficient heat energy absorbed to heat up the water. 👌

c. Iron as iron is a good conductor of heat to it can absorb a large amount to heat to heat the water 👌(Note: Iron tend to rust in contact with water so better to use copper)

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2381 kg 👌
3.10x10⁸ J 👌
25800 sec 👌
There is not enough hours of sunlight to provide sufficient energy to heat up the water as the daily sunshine is four hours but six hours are needed to heat up water

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3.10 x 10^8 J 👌

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25800s 👌

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It takes 6hours to heat up water. However the average daily sunshine is only 4 hours. thus there is insufficient solar energy to heat up the water. 👌

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a. The current passing through a conductor is directly proportional to the potential difference across the two points.

b. The graph shows a curve instead of a straight line

c. When the potential difference increases, the resistance also increases.

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stainless steel. it is a good conductor of heat. (note: steel tends to rust and corrode in contact with water and oxygen)

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Ohm's law states that the current passing through a conductor is directly proportional to the potential difference  between its ends provided all other physical conditions (e.g. temperature) stays constant.

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the graph is not a linear line and is a curve. 👍

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ai. 2381kg 👌
ii. 3.10x10^8J 👌
iii. 25800s 👌

b. In Fiji, 6h is required to heat up the water , but Fiji only receives an average of 4h of sunshine a day. Therefore, not enough solar energy is able to be absorbed and the heating system is unable to maintain the indoor temperature of 25 degrees celsius. 👌

c. Copper as it is a good conductor of heat. 👌

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when the potential difference increases, the resistance of the lamp also increases.

Using Fig.12.2, state and explain the change to the resistance of the lamp

Note: Ratish, you need to state the changes in the gradient from the graph and link it to the resistance. Hence, as the potential difference increases, the gradient (1/R) of the graph decreases (become gentler). Hence, the value of R increases.

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12a. Ohm's law states that the current passing through a conductor is directly proportional to the potential difference between the two points provided the other physical conditions remain constant (e.g. temperature) Note: last part about physical condition is impt

b. The graph is not a straight line passing through the origin but a curve.  Note: The graph passes thru origin ....

c.  When the potential difference increases, the resistance of the lamp increases. You need to make use of the graph!!

From the question "Using Fig.12.2, state and explain the change to the resistance of the lamp"

Note: Yun Xin, you need to state the changes in the gradient from the graph and link it to the resistance. Hence, as the potential difference increases, the gradient (1/R) of the graph decreases (become gentler). Hence, the value of R increases.

di. Rheostat 👌

ii. As V=IR, when the effective resistance increases as the rheostat's resistance increases, the potential difference would remain constant, so the current would decrease 👍

ei. 0.05A ✔

ii. 0.0125A ✔

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metal as it is a good conductor of heat 👌

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Rheostat 👍

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a. Ohm's law states that the current passing through a conductor between two points is directly proportional to the potential difference across the two points provided all other physical conditions (e.g. temperature) stays the same.

b. The graph is a curve and not a straight line👍

c. The resistance increases. This is because when the potential difference increases, the gradient, 1/R, decreases. Therefore, the resistance R increases 👍

di. A rheostat 👌

dii. Since V=RI, when component X is set to a higher resistance, the effective resistance will increase. Therefore, the current will decrease while the potential difference stays constant ✔

ei. 0.05A ✔

eii. 0.0125A ✔

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The voltage across a conductor is directly proportional to the current that flows through it provided all other physical conditions stay the same (e.g. temperature) 👌

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X can be set to a higher resistance. this will increase the total effective resistance. Since V=RI, with the current and voltage remaining the same, the current will decrease. 👌

(note: how can voltage stay the same and decrease at the same time)

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The graph is not a straight line✔ and does not pass through the origin

Note - The graph does pass thru origin. Just the gradient is not constant i.e. graph is a curve. Hence, the filament lamp does NOT obey Ohm's Law

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0.05A 👌
12eii)
0.0125 A ✔