<span>There is six horizen.
1. O Horizon - The top, organic layer of soil,
2. A Horizon - The layer called topsoil;
3. E Horizon - This layer is beneath the A Horizon and above the
B Horizon. It is made up mostly of sand.
4. B Horizon - Also called the subsoil - this layer is beneath the E
Horizon and above the C Horizon.
5. C Horizon - it's called regolith: the layer beneath the B Horizon
and above the R Horizon.
6 R Horizon - this is last and the unweathered rock layer that is
beneath all the other layers.</span>
Answer:
The weight lifter would not get past this sticking point.
Explanation:
Generally torque applied on the weight is mathematically represented as
T = F z
To obtain Elbow torque we substitute 4000 N for F (the force ) and 2cm 
for z the perpendicular distance
So Elbow Torque is 
To obtain the torque required we substitute 300 N for F and 30cm 
So the Required Torque is 
Now since 
it mean that the weight lifter would not get past this sticking point
The total work <em>W</em> done by the spring on the object as it pushes the object from 6 cm from equilibrium to 1.9 cm from equilibrium is
<em>W</em> = 1/2 (19.3 N/m) ((0.060 m)² - (0.019 m)²) ≈ 0.031 J
That is,
• the spring would perform 1/2 (19.3 N/m) (0.060 m)² ≈ 0.035 J by pushing the object from the 6 cm position to the equilibrium point
• the spring would perform 1/2 (19.3 N/m) (0.019 m)² ≈ 0.0035 J by pushing the object from the 1.9 cm position to equilbrium
so the work done in pushing the object from the 6 cm position to the 1.9 cm position is the difference between these.
By the work-energy theorem,
<em>W</em> = ∆<em>K</em> = <em>K</em>
where <em>K</em> is the kinetic energy of the object at the 1.9 cm position. Initial kinetic energy is zero because the object starts at rest. So
<em>W</em> = 1/2 <em>mv</em> ²
where <em>m</em> is the mass of the object and <em>v</em> is the speed you want to find. Solving for <em>v</em>, you get
<em>v</em> = √(2<em>W</em>/<em>m</em>) ≈ 0.46 m/s
That is true imo not false
