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(a) (i) Which of these would be a typical speed for a racing cyclist travelling down a steep straight slope? A 0.2 m/s B 2 m/s C 20 m/s D 200 m/s (iii) A cyclist travels down a slope - Edexcel - GCSE Physics Combined Science - Question 4 - 2019 - Paper 1

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(a)-(i)-Which-of-these-would-be-a-typical-speed-for-a-racing-cyclist-travelling-down-a-steep-straight-slope?--A-0.2-m/s--B-2-m/s--C-20-m/s--D-200-m/s--(iii)-A-cyclist-travels-down-a-slope-Edexcel-GCSE Physics Combined Science-Question 4-2019-Paper 1.png

(a) (i) Which of these would be a typical speed for a racing cyclist travelling down a steep straight slope? A 0.2 m/s B 2 m/s C 20 m/s D 200 m/s (iii) A cyclis... show full transcript

Worked Solution & Example Answer:(a) (i) Which of these would be a typical speed for a racing cyclist travelling down a steep straight slope? A 0.2 m/s B 2 m/s C 20 m/s D 200 m/s (iii) A cyclist travels down a slope - Edexcel - GCSE Physics Combined Science - Question 4 - 2019 - Paper 1

Step 1

(a) (i) Which of these would be a typical speed for a racing cyclist travelling down a steep straight slope?

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Answer

The typical speed for a racing cyclist would likely be 20 m/s (Option C). This speed is realistic for competitive cyclists, especially when traveling downhill.

Step 2

(a) (iii) Calculate the change in gravitational potential energy of the cyclist between the top and the bottom of the slope.

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Answer

Calculation of Gravitational Potential Energy

  1. Formula: The change in gravitational potential energy (GPE) can be calculated using the formula: GPE=mgh\text{GPE} = mgh Where:

    • mm = mass (75 kg)
    • gg = gravitational field strength (10 N/kg)
    • hh = height difference (20 m)
  2. Substituting the values: GPE=75kg×10N/kg×20m\text{GPE} = 75\,\text{kg} \times 10\,\text{N/kg} \times 20\,\text{m} GPE=15000J\text{GPE} = 15000\,\text{J}

Thus, the change in gravitational potential energy of the cyclist is 15,000 J.

Step 3

(b) Calculate the distance, x, travelled by the aircraft while it is accelerating.

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Answer

Using the Provided Formula

  1. Formula: The distance travelled during acceleration is given by: x=v2u22ax = \frac{v^{2} - u^{2}}{2a} Where:

    • vv = final velocity (80 m/s)
    • uu = initial velocity (0 m/s)
    • aa = acceleration (4 m/s²)
  2. Substituting the values: x=(80)2(0)22×4x = \frac{(80)^2 - (0)^2}{2 \times 4} x=64008x = \frac{6400}{8} x=800mx = 800\,\text{m}

The distance travelled by the aircraft while it is accelerating is 800 m.

Step 4

(c) (i) One piece of apparatus is missing from the diagram.

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Answer

The extra piece of apparatus needed to determine the average speed is a stopwatch.

Step 5

(c) (ii) Describe how the student can make the trolley accelerate along the bench.

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Answer

The student can make the trolley accelerate by attaching a weight to one end of the string that is connected to the trolley. By allowing the weight to drop, the force of gravity will pull on the string, causing the trolley to move along the bench.

Step 6

(c) (iii) State one other measurement that the student must make to determine the acceleration of the trolley.

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Answer

The student must also measure the time taken for the trolley to travel between the two marks. This measurement can be used alongside the distance to calculate acceleration.

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