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grin007 [14]
3 years ago
13

For the first challenge, Lex demands that Superman use the weight of a small stone to lift a boulder. Understanding the principl

es of fluids, Superman forms a hydraulic jack out of spare pipes, bits of metal, and river water. He places the small stone on the small piston, the boulder on the large piston, and, sure enough, the weight of the small stone raises the boulder sitting on the large piston. Explain how Superman uses Pascal’s Principle to succeed.
Physics
1 answer:
Nitella [24]3 years ago
7 0

Pascal's law states that the pressure exerted on an incompressible fluid and in equilibrium within a container of non-deformable walls is transmitted with equal intensity in all directions and at all points of the fluid. In other words, it could be summed up further, stating that any pressure exerted on a fluid will spread over the entire substance evenly.

When Superman puts the stone on one side of the hydraulic system, the weight of the stone creates a pressure on the fluid that ends up being transmitted to lift the piston attached to the large stone. In order for the law of conservation of energy to exist and be maintained, the small stone has to move a lot so that the large stone moves only slightly, but moves safely.

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3 years ago
An insulated thermos contains 106.0 cm3 of hot coffee at a temperature of 80.0 °C. You put in 11.0 g of ice cube at its melting
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Answer:

the final temperature is T f = 64.977 ° C≈ 65°C

Explanation:

Since the thermus is insulated, the heat absorbed by the ice is the heat released by the coffee. Thus:

Q coffee + Q ice = Q surroundings =0 (insulated)

We also know that the ice at its melting point , that is 0 °C ( assuming that the thermus is at atmospheric pressure= 1 atm , and has an insignificant amount of impurities ).

The heat released by coffee is sensible heat : Q = m * c * (T final - T initial)

The heat absorbed by ice is latent heat and sensible heat : Q = m * L + m * c * (T final - T initial)

therefore

m co * c co * (T fco - T ico) + m ice * L + m ice * c wat  * (T fwa - T iwa) = 0

assuming specific heat capacity of coffee is approximately the one of water c co = c wa = 4.186 J/g°C and the density of coffee is the same as water

d co = dw = 1 gr/cm³

therefore m co = d co * V co = 1 gr / cm³ * 106 cm³ = 106 gr

m co * c wat * (T f  - T ico) + m ice * L + m ice * c wat  * (T f - T iwa) = 0

m co * c wat * T f+ m ice * c wat  * T f  = m ice * c wat  * T iwa  + m co * c wat * Tico -m ice * L

T f  = (m ice * c wat  * T iwa  + m co * c wat * Tico -m ice * L ) /( m co * c wat * + m ice * c wat )

replacing values

T f = (11 g * 4.186 J/g°C * 0°C +  106 g * 4.186 J/g°C*80°C - 11 g * 334 J/gr) / ( 11 g * 4.186 J/g°C +  106 g * 4.186 J/g°C* ) = 64,977 ° C

T f = 64.977 ° C

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