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2 years ago

Find the height or length of these natural wonders in kilometers, meters, and centimeters.

Physics

1 answer:

cluponka [151]2 years ago

5 0

A cave system with a mapped length of 369 miles:

594 km, 594,000 m, 59,400,000 cm

A waterfall that drops 1,180.8 ft :

0.35989 km, 359.89 m, 35,989 cm

A 20,250 ft tall mountain:

6.1738 km, 6,173.8 m , 617,380 cm

A canyon with a depth of 73,800 ft:

22.493 km, 22,493 m , 2,249,300 cm

Explanation:

Given:

The height or length of these natural wonders

To find:

The height or length of these natural wonders in kilometers, meters, and centimeters.

Solution:

(a) a cave system with a mapped length of 369 miles

Length of the cave system = 369 miles

$1 mile = 1.609 km\\369 miles=369\times 1.609 km=594 km\\1 km = 1000 m\\594 km = 594\times 1000 m= 594,000 m\\1 m = 100 cm\\594,000 m=594,000 \times 100 cm=59,400,00 cm\\$

(b) a waterfall that drops 1,180.8 ft

The height of the waterfall = 1,180.8 ft

$1 km= 3281 ft\\1ft=\frac{1}{3281} km\\1,189.8 ft=1,180.8 \times \frac{1}{3281}km=0.35989 km\\1 km = 1000 m\\0.35989 km=0.35989 \times 1000=359.89 m\\1 m = 100 cm\\359.89 m=359.89 \times 100 cm=35,989 cm\\$

(c) A 20,250 ft tall mountain.

The height of the mountain = 20.250 ft

$1 km= 3281 ft\\1ft=\frac{1}{3281} km\\20,250 ft=20,250 \times \frac{1}{3281}km=6.1738km\\1 km = 1000 m\\6.1738km=6.1738\times 1000=6,173.8 m\\1 m = 100 cm\\6,173.8 m=6,173.8 m\times 100 cm=617,380 cm\\$

(d) A canyon with a depth of 73,800 ft

The height of the canyon = 73,800 ft

$1 km= 3281 ft\\1ft=\frac{1}{3281} km\\73,800 ft=73,800\times \frac{1}{3281}km=22.493km\\1 km = 1000 m\\22.493km=22.493\times 1000=22,493m\\1 m = 100 cm\\22493 m=22493m\times 100 cm=2,249,300cm\\$

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Answer:

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v = 1/3(√k/m)x₀ = 1/3(√145/2.2)0.2 = 0.54 m/s

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a = kx₀/3m = 145 × 0.2/(2.2 × 3)= 4.39 m/s²

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