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$lambda = 1.1 xx 10^(-38) "m"$
The de Broglie wavelength
$lambda = h/(mv)$ where:
$h = 6.626 xx 10^(-34) "J" cdot"s"$ is .$m$ is the mass of the mass-ive object in$"kg"$ .$v$ is its velocity in$"m/s"$ .
The units here are wacky, so we'll need to convert the
$"2.2 lb"/"kg"" "" "" ""5280 ft"/"1 mi"" "" ""12 in"/"1 ft"$
$"2.54 cm"/"1 in"" "" ""100 cm"/"1 m"" "" ""60 min"/"1 hr"$
$"60 s"/"1 min"$
First, convert the mass:
$5400 cancel"lbs" xx "1 kg"/(2.2 cancel"lbs") = "2454.5 kg"$
$" "" "" "" "" "" "" "" "$ (we'll round later)
Now convert the velocity:
$(57 cancel"mi")/cancel"hr" xx (5280 cancel"ft")/(cancel"1 mi") xx (12 cancel"in")/(cancel"1 ft") xx (2.54 cancel"cm")/(cancel"1 in") xx "1 m"/(100 cancel"cm") xx (cancel"1 hr")/(60 cancel"min") xx (cancel"1 min")/("60 s")$
$=$ $"25.481 m/s"$
(we'll round later)
Lastly, just solve the de Broglie relation for the wavelength.
$color(blue)(lambda) = (6.626 xx 10^(-34) cancel"kg" cdot"m"^cancel(2)"/"cancel"s")/(2454.5 cancel"kg" cdot 25.481 cancel("m/s"))$
$= ulcolor(blue)(1.1 xx 10^(-38) "m")$
And this number is justifiably puny.
A macroscopic particle like a sports utility vehicle has practically no wave characteristics to speak of. Only really light particles moving at very fast speeds are quantum mechanical.
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