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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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The amount of power change if less work is done in more time"then the amount of power will decrease".

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3 years ago
If a ball is thrown straight up into the air with an initial velocity of 65 ft/s, its height in feet after t seconds is given by
fgiga [73]

Answer:

a) v_{1}=\frac{(62.5-66)ft}{(2.5-2)s}=-7ft/s

v_{2}=\frac{(65.94-66)ft}{(2.1-2)s}=-0.6ft/s

v_{3}=\frac{(66.0084-66)ft}{(2.01-2)s}=0.84ft/s

v_{4}=\frac{(66.001-66)ft}{(2.001-2)s}=1ft/s

b) v=65-32(2)=1ft/s

Explanation:

From the exercise we got the ball's equation of position:

y=65t-16t^{2}

a) To find the average velocity at the given time we need to use the following formula:

v=\frac{y_{2}-y_{1}  }{t_{2}-t_{1}  }

Being said that, we need to find the ball's position at t=2, t=2.5, t=2.1, t=2.01, t=2.001

y_{t=2}=65(2)-16(2)^{2} =66ft

y_{t=2.5}=65(2.5)-16(2.5)^{2} =62.5ft

v_{1}=\frac{(62.5-66)ft}{(2.5-2)s}=-7ft/s

--

y_{t=2.1}=65(2.1)-16(2.1)^{2} =65.94ft

v_{2}=\frac{(65.94-66)ft}{(2.1-2)s}=-0.6ft/s

--

y_{t=2.01}=65(2.01)-16(2.01)^{2} =66.0084ft

v_{3}=\frac{(66.0084-66)ft}{(2.01-2)s}=0.84ft/s

--

y_{t=2.001}=65(2.001)-16(2.001)^{2} =66.001ft

v_{4}=\frac{(66.001-66)ft}{(2.001-2)s}=1ft/s

b) To find the instantaneous velocity we need to derivate the equation

v=\frac{df}{dt}=65-32t

v=65-32(2)=1ft/s

7 0
3 years ago
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