Answer:

Explanation:
<u>Work and Kinetic Energy
</u>
The work an object does due to its motion is equal to the change of its kinetic energy. Being ko and k1 the initial and final kinetic energy respectively and m the mass of the object, then

Since

We have

The truck has a mass of 60,000 kg and is moving at 27 m/s. The runaway truck ramp must stop the truck, so the final speed is 0. Thus



The compressor constricts the refrigerant vapor, raising its pressure, and pushes it into the coils on the outside of the refrigerator. 2. When the hot gas in the coils meets the cooler air temperature of the kitchen, it becomes a liquid. ... The refrigerant absorbs the heat inside the fridge, cooling down the air.
To solve this problem it is necessary to apply the continuity equations in the fluid and the kinematic equation for the description of the displacement, velocity and acceleration.
By definition the movement of the Fluid under the terms of Speed, acceleration and displacement is,

Where,
Velocity in each state
g= Gravity
h = Height
Our values are given as,



Replacing at the kinetic equation to find
we have,



Applying the concepts of continuity,

We need to find A_2 then,

So the cross sectional area of the water stream at a point 0.11 m below the faucet is



Therefore the cross-sectional area of the water stream at a point 0.11 m below the faucet is 
Answer:
Alveoli
The bronchioles end in tiny air sacs called alveoli, where oxygen is transferred from the inhaled air to the blood.
Hope this helps :)