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Rashid [163]
3 years ago
8

In 2002, a gargantuan iceberg broke away from the Ross Ice Sheet in Antarctica. It was approximately a rectangle 218 km long, 25

.0 km wide, and 250.0 m thick.
A) What is the mass of this iceberg, given that the density of ice is 917 kg/m3?
B) How much heat transfer (in joules) is needed to melt the iceberg?
C) How many years would it take sunlight alone to melt ice this thick, if the ice absorbs an average of 109 W/m2, 12.0 hours per day?
Physics
1 answer:
Musya8 [376]3 years ago
5 0

Answer:

a) 1.25e15 kg

b) 4.17e20 J

c) 44.55 years

Explanation:

To find the volume you need to multiply 218 km * 25 km * 250 m (be careful with units), so the volume is 1.3625e12 m^3, if you multiply this value by the density you will obtain the mass, that is 1.25e15 kg.

To find the energy needed to melt the ice, you use the latent heat, in this case, it is 3.34e5 J/kg. Now you multiply this value by the mass, so you need 4.17e20 J to melt the iceberg.

The surface area of the iceberg is 545e7 m^2, so the ice absorbs 594e9 W, one W is one J/s, so in 12 hours the iceberg absorbs 2.56e16 J, so in 365 days absorbs 9.36e18 J. Now you just divide 4.17e20 J by the amount f energy per year, and obtain 44.55 years.

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Answer:

0.124 m

Explanation:

the period of a simple pendulum with a small amplitude is given as

T = 2π [√(I/mgd)]

From the above stated formula,

I = moment of inertia

m = mass of the pendulum

g = acceleration due to gravity, 9.8 m/s²

d = distance from rotation axis due to center of gravity

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T = 2π [√(2mr²/mgd)]

If we assume that our r = d, then we have

T = 2π [√(2r/g)]

If we make r the subject of the formula in the above equation, we get

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A gas‑forming reaction produces 1.90 m 3 1.90 m3 of gas against a constant pressure of 179.0 kPa. 179.0 kPa. Calculate the work
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Answer:

The work done is 3.4 × 10⁵ J.

Explanation:

Given:

Pressure of the gas produced (P) = 179 kPa

Volume of the gas produced (ΔV) = 1.90 m³

We need to find the work done in joules. For that, we don't need any conversion as the units are already in SI units which will give the result in Joules only.

Now, let us verify our results by using conversion factors and without using them.

Using conversion factors:

1 m³ = 1000 L

So, 1.90 m³ = 1.90 m³ × 1000 \frac{L}{m^3} = 1900 L

Also, 1 atm = 101.325 kPa

So, 179 kPa = 179 kPa × \frac{1\ atm}{101.325\ kPa} = 1.767 atm

Now, work done in a constant pressure process is given as:

Work = Pressure × Volume change

Work = P × ΔV

Work = 1.767 atm × 1900 L

Work = 3.36 × 10³ atm-L

Now, again using the energy conversion for work.

1 atm-L = 101.325 J

So, 3.36 × 10³ atm-L = 3.36 × 10³ atm-L × \frac{101.325\ J}{1\ atm-L} = 3.4 × 10⁵ J

Therefore, the work done is 3.4 × 10⁵ J.

Now, let us verify the above result without any conversion.

Work = P × ΔV = 179 × 1000 × 1.90 = 3.4 × 10⁵ J.

Therefore, the work done is same by both ways.

Hence the work done is 3.4 × 10⁵ J.

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