Question

which expression(s) would be most appropriate to describe the distance from earth to the sun? select all that apply. a) $1.496 \times 10^{8}$ km b) $1.496 \times 10^{14}$ mm c) $5.890 \times 10^{12}$ in. d) $5.890 \times 10^{14}$ cm e) $9.296 \times 10^{7}$ mi

Explanation:

Step1: Recall the average distance from Earth to Sun

The average distance from Earth to the Sun (astronomical unit, AU) is approximately \( 1.496\times 10^{8}\) km, \( 9.296\times 10^{7}\) miles, \( 1.496\times 10^{11}\) meters. Let's convert the given options to a common unit (e.g., meters) or check known values.

Step2: Analyze Option A

Option A: \( 1.496\times 10^{8}\) km. Since \( 1\) km \( = 1000\) m, \( 1.496\times 10^{8}\) km \( = 1.496\times 10^{11}\) m, which matches the known AU in meters (scaled). This is a standard way to express AU, so A is correct.

Step3: Analyze Option B

Option B: \( 1.496\times 10^{14}\) mm. Since \( 1\) m \( = 1000\) mm, \( 1.496\times 10^{14}\) mm \(=\frac{1.496\times 10^{14}}{1000}\) m \( = 1.496\times 10^{11}\) m, which is correct (since \( 1.496\times 10^{8}\) km \( = 1.496\times 10^{11}\) m and \( 1.496\times 10^{14}\) mm \( = 1.496\times 10^{11}\) m). Wait, let's check conversion: \( 1\) km \( = 10^{6}\) mm (since \( 1\) km \( = 1000\) m, \( 1\) m \( = 1000\) mm, so \( 1\) km \( = 1000\times1000 = 10^{6}\) mm). Then \( 1.496\times 10^{8}\) km \( = 1.496\times 10^{8}\times 10^{6}\) mm \( = 1.496\times 10^{14}\) mm. So B is correct.

Step4: Analyze Option C

Option C: \( 5.890\times 10^{12}\) in. Let's convert inches to meters. \( 1\) inch \( = 0.0254\) m. So \( 5.890\times 10^{12}\) in \( = 5.890\times 10^{12}\times0.0254\) m \( \approx 1.496\times 10^{11}\) m (since \( 5.890\times0.0254\approx0.1496\), so \( 0.1496\times 10^{12}\times 10^{0}\)? Wait, no: \( 5.890\times 10^{12}\times0.0254 = 5.890\times0.0254\times 10^{12}\approx0.1496\times 10^{12}=1.496\times 10^{11}\) m. Wait, but known AU in inches: Let's check with miles. \( 1\) mile \( = 63360\) inches. \( 9.296\times 10^{7}\) miles \( = 9.296\times 10^{7}\times63360\) inches \( \approx 9.296\times6.336\times 10^{11}\approx58.9\times 10^{11}=5.89\times 10^{12}\) inches. Oh! Wait, option C is \( 5.890\times 10^{12}\) in. So C is correct? Wait, no, let's recalculate: \( 9.296\times 10^{7}\) mi \( \times 63360\) in/mi \( = 9.296\times63360\times 10^{7}\) in. \( 9.296\times63360\approx9.296\times6.336\times 10^{4}\approx58.9\times 10^{4}=5.89\times 10^{5}\)? No, wait \( 63360 = 6.336\times 10^{4}\), so \( 9.296\times 10^{7}\times6.336\times 10^{4}=9.296\times6.336\times 10^{11}\approx58.9\times 10^{11}=5.89\times 10^{12}\) in. So C is correct? Wait, but let's check option D.

Step5: Analyze Option D

Option D: \( 5.890\times 10^{14}\) cm. \( 1\) m \( = 100\) cm, so \( 5.890\times 10^{14}\) cm \( = 5.890\times 10^{12}\) m. But AU is \( 1.496\times 10^{11}\) m, so \( 5.890\times 10^{12}\) m is about 10 times larger? Wait, no, let's convert km to cm. \( 1\) km \( = 10^{5}\) cm. So \( 1.496\times 10^{8}\) km \( = 1.496\times 10^{8}\times 10^{5}\) cm \( = 1.496\times 10^{13}\) cm. But option D is \( 5.890\times 10^{14}\) cm, which is \( 58.9\times 10^{13}\) cm, which is wrong. Wait, maybe miscalculation. Wait, \( 1\) km \( = 1000\) m \( = 1000\times100\) cm \( = 10^{5}\) cm. So \( 1.496\times 10^{8}\) km \( = 1.496\times 10^{8}\times 10^{5}\) cm \( = 1.496\times 10^{13}\) cm. Option D is \( 5.890\times 10^{14}\) cm \( = 58.9\times 10^{13}\) cm, which is not equal. So D is incorrect.

Step6: Analyze Option E

Option E: \( 9.296\times 10^{7}\) mi. The average distance from Earth to Sun is approximately \( 93\) million miles, which is \( 9.3\times 10^{7}\) miles. So \( 9.296\times 10^{7}\) mi matches this, so E is correct.

Wait, earlier mistake with option C: Let's recheck. \( 9.296\times 10^{7}\) mi \( \times 63360\) in/mi \( = 9.296\times63360\times 10^{7}\) in. \( 63360 = 6.336\times 10^{4}\), so \( 9.296\times6.336\times 10^{11}\approx(9\times6.336 + 0.296\times6.336)\times 10^{11}\approx(57.024 + 1.875)\times 10^{11}\approx58.899\times 10^{11}=5.8899\times 10^{12}\) in, which is \( 5.890\times 10^{12}\) in. So C is correct. Wait, but earlier when converting option B: \( 1\) km \( = 10^{6}\) mm, so \( 1.496\times 10^{8}\) km \( = 1.496\times 10^{8}\times 10^{6}\) mm \( = 1.496\times 10^{14}\) mm, which is correct. So let's re-express:

- A: \( 1.496\times 10^{8}\) km (correct, standard AU in km)
- B: \( 1.496\times 10^{14}\) mm (since \( 1\) km \( = 10^{6}\) mm, \( 1.496\times 10^{8}\) km \( = 1.496\times 10^{14}\) mm, correct)
- C: \( 5.890\times 10^{12}\) in (from \( 9.296\times 10^{7}\) mi \( \times 63360\) in/mi \( \approx 5.890\times 10^{12}\) in, correct)
- D: \( 5.890\times 10^{14}\) cm (convert km to cm: \( 1\) km \( = 10^{5}\) cm, \( 1.496\times 10^{8}\) km \( = 1.496\times 10^{13}\) cm, which is not \( 5.890\times 10^{14}\) cm, so D is incorrect)
- E: \( 9.296\times 10^{7}\) mi (correct, as 93 million miles is \( 9.3\times 10^{7}\) mi)

Wait, but initially I thought D was wrong, which it is. Let's confirm with known values:

- AU in km: \( 1.496\times 10^{8}\) km (A is correct)
- AU in miles: \( 9.296\times 10^{7}\) mi (E is correct)
- Convert km to mm: \( 1\) km \( = 10^{6}\) mm, so \( 1.496\times 10^{8}\) km \( = 1.496\times 10^{8}\times 10^{6}\) mm \( = 1.496\times 10^{14}\) mm (B is correct)
- Convert miles to inches: \( 1\) mile \( = 63360\) inches, so \( 9.296\times 10^{7}\) miles \( = 9.296\times 10^{7}\times 63360\) inches \( = 9.296\times 6.336\times 10^{11}\) inches \( \approx 58.9\times 10^{11}\) inches \( = 5.89\times 10^{12}\) inches (C is correct)
- Convert km to cm: \( 1\) km \( = 10^{5}\) cm, so \( 1.496\times 10^{8}\) km \( = 1.496\times 10^{8}\times 10^{5}\) cm \( = 1.496\times 10^{13}\) cm, which is not \( 5.890\times 10^{14}\) cm (D is incorrect)

Answer:

A. \( 1.496 \times 10^{8}\) km, B. \( 1.496 \times 10^{14}\) mm, C. \( 5.890 \times 10^{12}\) in, E. \( 9.296 \times 10^{7}\) mi