A landscape in Antarctica has an area of 500,000 square miles. It is covered with ice, giving the landscape an average elevation of 7520 feet. Without the ice, it is estimated the landscape would have an average elevation of 1500 ft. The average density of the ice is 0.917 g/mL. How many tonnes of ice cover that landscape?

m = 2.18x10¹⁵ tons of ice

Explanation:

To solve this exercise, we need to work with the same units of conversion. In this case let's see the conversion of every unit here:

1 mi = 5280 ft

1 ton = 1000 kg

1 kg = 1000 g

1 mL = 1 cm³

1 mi = 1609.34 m or 160,934 cm

Now that we know the units, let's analyze the data.

The landscape has a 500,000 mi² of area, this is however without including the ice. The ice is apart, and has a 7520 ft of height. And if the landscape wouldn't have ice, then the height will be 1500 ft only.

With that data, we can know the actual height of the ice by a simple difference:

Hi = Hli - Hl

Where:

Hi: Height of ice

Hli: Height of ice + landscape

Hl: Height of landscape

Applying this expression, the height of ice is:

Hi = 7520 - 1500 = 6020 ft of ice

Now that we know the height of the ice, we can calculate the volume of whole landscape. But first, let's do the conversion of foot to mile:

Hi = 6020 ft * (1 mi/5280 ft) = 1.14 mi

And now, let's calculate the volume of the landscape in mile, and then, to centimeter:

V = 1.14 mi * 500,000 mi² = 570,000 mi³

To convert the cubic mile to cubic centimeter we just do the following:

V = 570,000 mi³ * (160934 cm / 1 mi)³ = 2.38x10²¹ cm³

Now, we can get the mass of the ice using the given density of ice (d = 0.917 g/cm³)

If d = m/V  

then m = d * V

Replacing the data:

m = 0.917 g/cm³ * 2.38x10²¹ cm³

m = 2.18x10²¹ g

Finally, the tons of ice will be:

m = 2.18x10²¹ g * (1 kg / 1000 g) * (1 ton / 1000 kg)

Hope this helps

just did the assignment and the answer was #1 +2 n

може ли да зареждате повече информация, моля