how long does it take in minutes for light to reach Saturn from the sun,a distance of 1.466*10^9 km?

Question

anonymous · Answer

the speed of light is c = 2.99x10^8 so using dimensional analysis we can get $$(1.466x10^{12}m)/(299x10^8 m/s) = $$ The meters cancel out and you get seconds and that's how long it takes.

anonymous · Answer

i did that and the answer came out wrong.is there another way of solving,what are the units of speed of light?

unklerhaukus · Answer

$$s=\frac dt\quad\implies t=\frac ds$$
$$d=a_{\hbar}=1.466\times10^9[\text {km}]$$$$s=c_0=2.998\times10^8[\text {m/s}]$$

$$ t=\frac ds=\frac {a_\hbar}{c_0}=\frac{1.466\times10^{9}[\text {km}]}{2.998\times10^8[\text {m/s}]}$$
$$\qquad\qquad=\frac{1.466\times10^{9}[\text {km}]}{2.998\times10^8[\text {m/s}]}\times\frac{1000[\text m]}{[\text {km}]}\times\frac{[\text{min}]}{60[\text s]}$$
$$\qquad\quad=\frac{1.466\times10^9\times 10^{3}}{2.998\times10^8\times60}[\text{min}]$$

whpalmer4 · Answer

@saralee4 did your answer have the wrong number of zeros, but the right "number"? If you just typed in the expression that @azolotor gave before the equals sign, he neglected to put in the decimal point between 2 and 9, which means your answer would be off by a factor of 100.  Also, you would have to convert the answer from seconds (which is what you would get with his expression) to minutes, by multiplying your answer by $(1 [\text {min}]/60 [\text s])$.  If you do that, and put in the missing decimal point (plus a few more digits for grins):
$$\frac{1.466*10^{12} [\text m]}{2.99792458*10^8[\text m/\text s]}*\frac{1 [\text {min}]}{60[\text s]} $$ will give you the correct answer, just like @UnkleRhaukus gets.

unklerhaukus · Answer

it's a  bit longer than an hour

whpalmer4 · Answer

As it turns out, I had opportunity to compare the response speed of the customer service organization at a vendor with the time it would take to bounce my query off a satellite orbiting Pluto a few days ago.  Gotta love Wolfram|Alpha — I can walk down the street and still easily get answers to questions like "distance to Pluto in light seconds" :-)