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Lubov Fominskaja [6]
3 years ago
14

Consider the aquatic phosphorus cycle. On land most phosphorus is found in rocks and minerals. In the oceans, phosphorus is depo

sited from decomposition, excretion from animals, run-off from the land, and finally ______________ of rocks and minerals.
A) deposition
B) erosion
C) reformation
D) weathering

Physics
2 answers:
Paraphin [41]3 years ago
8 0

Answer: D) weathering

Phosphorus is a chemical organic element that is found in numerous compounds of the Earth typically found in water, sediments and soils. The phosphorus in the soil is being added up by the chemical fertilizers. In the ocean or marine water the phosphorus is deposited as animal waste and by the weathering of the rocks and minerals.

The weathering is a process by which the rocks are broken down into small parts and sediments by the action of the physical agents such as water, wind and chemicals. In the given situation the weathering is the process by which the element phosphorus will be added to the bottom bed of the ocean.

dmitriy555 [2]3 years ago
4 0
D) weathering  im 100% sure i took the test and i pasted so i know 
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A professor designing a class demonstration connects a parallel-plate capacitor to a battery, so that the potential difference b
Lesechka [4]

Answer:

a)  Q = 397.57 pC , Q = 3.18 104 pC , b) C = 1.157 10⁻¹⁰ F ,  V = 3.4375 V ,

c)  U = 54.7 nJ ,  d) ΔU = 54 nJ,

Explanation:

a) The capacity of a capacitor is defined

        C = Q / V

        Q = C V

         

can also be calculated using geometry consideration

        C = e or A / d

         

we reduce to the SI system

       A = 25.0 cm² (1 m / 10² cm) 2 = 25.0 10⁻⁴ m²

       d = 1.53 cm = 1.53 10⁻² m

we substitute

         Q = eo A / d V

         Q = 8.85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻² 275

         Q = 3.9757 10⁻¹⁰ C

         

let's reduce to pC

         Q = 3.9757 10⁻¹⁰ C (10¹² pC / 1 C)

          Q = 397.57 pC

when the capacitor is introduced into the water the dielectric constant is different

           Q = k Q₀

           Q = 80 397.57

           Q = 3.18 104 pC

b) Find capacitance and voltage after submerged in water

           C = k C₀

           C = 80 8.85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻²

           C = 1.157 10⁻¹⁰ F

           V = Vo / k

            V = 275/80

            V = 3.4375 V

c) The stored energy is

             U = ½ C V²

              U = ½, 85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻²     275²

             U = 5.47 10⁻⁸ J

let's reduce to nJ

              109 nJ = 1 J

               U = 54.7 nJ

d) energy after submerging

             U = ½ (kCo) (Vo / k) 2

             U = ½ Co Vo2 / k

             U = U₀ / k

             U = 54.7 / 80 nJ

              U = 0.68375 nJ

the energy change is

         ΔU = U₀ -U

          ΔU = 54.7 - 0.687375

           

6 0
3 years ago
A 1.8 kg uniform rod with a length of 90 cm is attached at one end to a frictionless pivot. It is free to rotate about the pivot
Leokris [45]

Answer:

a. 32.67 rad/s²  b. 29.4 m/s²

Explanation:

a. The initial angular acceleration of the rod

Since torque τ = Iα = WL (since the weight of the rod W is the only force acting on the rod , so it gives it a torque, τ at distance L from the pivot )where I = rotational inertia of uniform rod about pivot = mL²/3 (moment of inertia about an axis through one end of the rod), α = initial angular acceleration, W = weight of rod = mg where m = mass of rod = 1.8 kg and g = acceleration due to gravity = 9.8 m/s² and L = length of rod = 90 cm = 0.9 m.

So, Iα = WL

mL²α/3 = mgL

dividing through by mL, we have

Lα/3 = g

multiplying both sides by 3, we have

Lα = 3g

dividing both sides by L, we have

α = 3g/L

Substituting the values of the variables, we have

α = 3g/L

= 3 × 9.8 m/s²/0.9 m

= 29.4/0.9 rad/s²

= 32.67 rad/s²

b. The initial linear acceleration of the right end of the rod?

The linear acceleration at the initial point is tangential, so a = Lα = 0.9 m × 32.67 rad/s² = 29.4 m/s²

5 0
3 years ago
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