The distance from the earth to the moon is 3.84 x 105 km
a) Find this distance in metres.
b) How long would it take a spaceship to travel to the moon from earth if its average speed was 400 ms l?​

The distance from the earth to the moon is 384 x 105 kma Find this distance in metres b How long would it take a spaceship to travel to the moon from earth if i class=

Respuesta :

Part (a)

Answer:  [tex]3.84 \times 10^8 \text{ meters}[/tex]

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Explanation:

Multiply by 1000 to go from km to meters.

This is because 1000 m = 1 km.

Multiplying by 1000 in scientific notation means we add 3 to the exponent 5. The 3 is because 1000 = 10^3.

So the [tex]3.84 \times 10^5 \text{ km}[/tex] becomes [tex]3.84 \times 10^8 \text{ meters}[/tex]

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Part (b)

Answer in standard form:  960,000 seconds

Answer in scientific notation: [tex]9.6\times10^{5} \text{ seconds}[/tex]

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Work Shown:

[tex]\text{distance} = \text{rate}*\text{time}\\\\\text{time} = \frac{\text{distance}}{\text{time}}\\\\\text{time} = \frac{3.84 \times 10^8 \text{ meters}}{400 \ \text{ m}/\text{s}}\\\\\text{time} = \frac{3.84 \times 10^8}{4 \times 10^2} \text{ seconds}\\\\\text{time} = \frac{3.84}{4}*\frac{10^{8}}{10^2} \text{ seconds}\\\\\text{time} = 0.96 \times 10^{8-2} \text{ seconds}\\\\\text{time} = 0.96\times 10^{6} \text{ seconds}\\\\[/tex]

[tex]\text{time} = (9.6\times10^{-1})\times10^{6} \text{ seconds}\\\\\text{time} = 9.6\times(10^{-1}*10^{6}) \text{ seconds}\\\\\text{time} = 9.6\times10^{-1+6} \text{ seconds}\\\\\text{time} = 9.6\times10^{5} \text{ seconds}\\\\\text{time} = 960,000 \text{ seconds}\\\\[/tex]

To go from the scientific notation to standard form, move the decimal point 5 spaces to the right. The 5 is from the exponent.