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4 years ago

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Find the height or length of these natural wonders in km, m, and cm.A. A cave system with a mapped length of 354 miles.B. A wate

rfall that drops 1,235.2 ft.C. A 21320 ft tall mountain.
D. A canyon with a depth of 6630 ft.

1 answer:

cricket20 [7]4 years ago

4 0

Explanation:

Some standard unit conversion are 1 mile = 1.609344 km, 1 ft.= 30.48 cm,

1 km= 1000 m or 1 m = 0.001 km and 1 m= 100 cm or 1 cm=0.01 m.

Now, use these values to convert the given lengths.

A. length of cave = 5 miles (given)

From standard value 1 mile = 1.609344 km

$\Rightarrow$ 5 miles = $\times$ 1.609344 km= 8.04672 km,

$\Rightarrow$ 5 miles =8.04672 $\times$ 1000 m= 8046.72 m [ as 1 km = 1000 m]

$\Rightarrow$ 5 miles = 8046.72 $\times$ 100 cm=804672 cm

B. Height of the waterfall = 1235 ft. (given)

1 ft.= 30.48 cm [ from standard value]

$\Rightarrow$ 1235.2 ft.= 1235.2 $\times$ 30.48 cm=37648.896 cm,

$\Rightarrow$ 1235.2 ft.=37648.896 $\times$ 0.01 m =376.48896 m [ as 1 cm = 0.01 m]

$\Rightarrow$ 1235.2 ft.=376.48896 $\times$ 0.001 km =0.37648896 km [ as 1 m = 0.001 km]

C. Height of the mountain= 21320 ft. (given)

From standard value: 1 ft.= 30.48 cm

$\Rightarrow$ 21320 ft.= 21320 $\times$ 30.48 cm=649833.6 cm,

$\Rightarrow$ 21320 ft.=649833.6 $\times$ 0.01 m =6498.336 m [ as 1 cm = 0.01 m]

$\Rightarrow$ 21320 ft.=6498.336 $\times$ 0.001 km =6.498336 km [ as 1 m = 0.001 km].

D. Depth of canyon =6630 ft.

From standard value: 1 ft.= 30.48 cm

$\Rightarrow$ 6630 ft.= 6630 $\times$ 30.48 cm=202082.4 cm,

$\Rightarrow$ 6630 ft.=202082.4 $\times$ 0.01 m =2020.824 m [ as 1 cm = 0.01 m]

$\Rightarrow$ 6630 ft.=2020.824 $\times$ 0.001 km =2.020824 km [ as 1 m = 0.001 km].

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Using the following relation to determine the amount of calories burned per minute while walking; we have:

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For an object thrown upward and subjected to free fall, the height of the object at any time t is given by the suvat equation:

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