Video: Pack 1 β€’ Paper 2 β€’ Question 2

Pack 1 β€’ Paper 2 β€’ Question 2

05:28

Video Transcript

The time, in seconds, it takes for light to travel a given distance, 𝑑, in metres can be calculated using the formula 𝑑 is equal to 𝑑 over three times 10 to the eight. The distance between the Sun and the Earth is 1.49 times 10 to the 11 metres. Work out the ratio of the time it takes light to travel across an anti-UV layer of 3.85 times 10 to the negative nine metres to the time it takes light to travel between the Sun and the Earth. Write your answer in the form one to 𝑛.

Let’s start by finding the time it takes light to travel across an anti-UV layer. And we can call this 𝑑 one. In order to find this time, we’ll be using the formula given in the question. So the distance 𝑑 which the light has travelled will be 3.85 times 10 to the negative nine metres. So we get that 𝑑 one is equal to 3.85 times 10 to the negative nine over three times 10 to the eight. And we can separate this fraction up to get 3.85 over three timesed by 10 to the negative nine over 10 to the eight.

Now we have that 10 to the negative nine over 10 to the eight is also equal to 10 to the power of negative nine minus eight, which is the same as 10 to the negative 17. We obtain that 𝑑 one is equal to 1.283 recurring timesed by 10 to the negative 17 seconds. At this point, it’s important to keep 𝑑 one as accurate as possible. By not rounding the recurring decimal, we can ensure that we will not lose any accuracy in our final answer.

Next, let’s calculate the time it takes for light to travel between the Sun and the Earth, which we’ll call 𝑑 two, which we can find by again using the formula given in the question, along with the distance 1.4 times 10 to the 11 metres.

We have that 𝑑 two is equal to 1.4 times 10 to the 11 over three times 10 to the eight. This is the same as 1.4 over three timesed by 10 to the 11 over 10 to the eight. Then 10 to the 11 over 10 to the eight is equal to 10 to the power of 11 minus eight, which also equals 10 to the three. Using this, we obtain that 𝑑 two is equal to 0.496 recurring times 10 to the three seconds.

And we can convert this to standard form by taking one of the tens from the 10 to the three and multiplying it by the 0.496 recurring. We get that 𝑑 two is equal to 4.96 recurring times 10 to the two seconds. Again, it’s important to keep 𝑑 two as accurate as possible so we can maintain this accuracy in our answer.

Now we note that the question asked us to find the ratio of the time it takes light to travel across an anti UV-layer, so that’s 𝑑 one, to the time it takes light to travel between the Sun and Earth, so that’s 𝑑 two. Since we have calculated the values of 𝑑 one and 𝑑 two, we can simply substitute them into the ratio, to obtain the ratio that 𝑑 one to 𝑑 two is equal to 1.283 recurring times 10 to the negative 17 to 4.96 recurring times 10 to the two.

Now we have found the ratio of 𝑑 one to 𝑑 two. We simply need to get this in the form of one to 𝑛. When looking at the left-hand side of both of these ratios, we can see that the number changes from 1.283 recurring times 10 to the negative 17 to one. In order to make this jump, the operation we are performing is dividing by 1.283 recurring times 10 to the negative 17.

And so in order to go from 4.96 recurring times 10 to the two to 𝑛, we need to perform the same operation. So we divide by 1.283 recurring times 10 to the negative 17. This tells us that 𝑛 is equal to 4.96 recurring times 10 to the two over 1.283 recurring times 10 to the negative 17. This can also be written as 4.96 recurring over 1.283 recurring timesed by 10 to the two over 10 to the negative 17. 10 to the two over 10 to the negative 17 is equivalent to 10 to the power of two minus negative 17, which is equal to 10 to the power of 19. From this, we can obtain that 𝑛 is equal to 3.870129 all recurring times 10 to the 19.

Now since we don’t need to perform any more calculations, it’s okay to round this value to three significant figures. Since the fourth figure in this number is zero, which is less than five, that means that we round down to three significant figures. And so we end up with 𝑛 is equal to 3.87 timesed by 10 to the 19. Finally, we can substitute this value of 𝑛 back into our ratio, to give us a solution of one to 3.87 times 10 to the 19.

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