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4. (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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4.-(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

4. (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 ... show full transcript

Worked Solution & Example Answer:4. (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

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 traveling down a steep straight slope is C 20 m/s, as it represents a reasonable speed for competitive cycling.

Step 2

Calculate the change in gravitational potential energy of the cyclist between the top and the bottom of the slope.

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Answer

To calculate the change in gravitational potential energy (GPE), we use the formula:

GPE=mimesgimesh\text{GPE} = m imes g imes h

Where:

  • m = mass of the cyclist = 75 kg
  • g = gravitational field strength = 10 N/kg
  • h = height of the slope = 20 m

Substituting in the values:

GPE=75kg×10N/kg×20m=15000J\text{GPE} = 75 \, \text{kg} \times 10 \, \text{N/kg} \times 20 \, \text{m} = 15000 \, \text{J}

Thus, the change in gravitational potential energy of the cyclist is 15000 J (joules).

Step 3

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

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Answer

To find the distance, x, we can use the formula:

x=v2u22ax = \frac{v^{2} - u^{2}}{2a}

Where:

  • v = final speed = 80 m/s
  • u = initial speed = 0 m/s
  • a = acceleration = 4 m/s²

Substituting the values into the equation gives:

x=(80)2(0)22×4=64008=800mx = \frac{(80)^{2} - (0)^{2}}{2 \times 4} = \frac{6400}{8} = 800 \, \text{m}

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

Step 4

State the extra piece of apparatus needed to determine the average speed.

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Answer

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

Step 5

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. The other end of the string should be fastened to the trolley. When the weight is released, it will pull the trolley along the bench, causing it to accelerate.

Step 6

State one other measurement that the student must make to determine the acceleration of the trolley.

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Answer

The student must measure the time taken for the trolley to travel between the two marks to determine the acceleration.

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