Let's address each question one by one:
1.5 Distance travelled by the car in 5 seconds
-
During the first 2 seconds, the car accelerates from rest to 0.8 m/s with a constant acceleration. The distance \( d_1 \) during this time can be calculated using the formula: \[ d_1 = \frac{1}{2} a t^2 \] Since final velocity \( v = a \cdot t \), where \( v = 0.8 , \text{m/s} \) and \( t = 2 , \text{s} \): \[ a = \frac{0.8 , \text{m/s}}{2 , \text{s}} = 0.4 , \text{m/s}^2 \] Now calculate \( d_1 \): \[ d_1 = \frac{1}{2} (0.4) (2^2) = \frac{1}{2} (0.4) (4) = 0.8 , \text{m} \]
-
In the next 3 seconds, the car travels at a constant speed of 0.8 m/s: \[ d_2 = v \cdot t = 0.8 , \text{m/s} \cdot 3 , \text{s} = 2.4 , \text{m} \]
-
Total distance: \[ \text{Total distance} = d_1 + d_2 = 0.8 , \text{m} + 2.4 , \text{m} = 3.2 , \text{m} \]
Answer: B 3.2 m
1.6 Mass of an object with weight 2620 N
- Weight (W) is related to mass (m) by the equation: \[ W = m \cdot g \] where \( g \) is the acceleration due to gravity (approximately \( 9.81 , \text{m/s}^2 \)). Rearranging: \[ m = \frac{W}{g} = \frac{2620}{9.81} \approx 267.3 , \text{kg} \]
Closest Value: B 262 kg (since the original question may require rounding or approximation).
1.7 Property of sound that pitch depends on
- Pitch is determined by the frequency of sound waves.
Answer: C frequency
1.8 Unit of charges
- The SI unit of electric charge is the coulomb.
Answer: B coulombs
1.9 Characteristics of the image in front of a converging lens
- A converging lens produces a real, inverted image if the object is placed outside the focal length.
Answer: A inverted and real
1.10 Copper wire resistance
- The longer and thinner the wire, the higher the resistance. Thus, the wire that is longest and most slender would have the highest resistance.
Note: Need more information about the specific wires to provide a precise answer.
1.11 Charge distribution on an uncharged ball near a negatively charged rod
- The negatively charged rod will repel electrons in the ball, causing a positive charge to accumulate on the side of the ball closest to the rod and a negative charge on the side furthest from the rod.
Note: Need to see diagrams to provide a precise answer.
1.12 Energy transferred in a resistor
-
The energy (E) transferred can be calculated using the formula: \[ E = V \cdot I \cdot t \] where:
- \( V = 12 , \text{V} \)
- \( I = 5.0 , \text{A} \)
- \( t = 1 , \text{minute} = 60 , \text{s} \)
Now calculate: \[ E = 12 \cdot 5.0 \cdot 60 = 3600 , \text{J} \]
Answer: D 3 600 J
1.13 Total resistance in a circuit of three identical resistors
- If the resistors are connected in series, total resistance \( R_{total} = R + R + R = 3R \).
- If they are connected in parallel, \[ \frac{1}{R_{total}} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R} \Rightarrow R_{total} = \frac{R}{3} \]
Note: Need to see the diagram to provide a precise answer regarding their arrangement.
If you have any specific diagrams for questions 1.10, 1.11, or 1.13, feel free to share them!
2 answers