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In some places timber companies remove all the trees from entire hillside when they are harvesting logs and farmers till the soi

laila [671]

Answer:

The timber companies removing all the trees from entire hillside when they are harvesting logs is a practice that could cause the following when it is time to plant in spring:

1. It could affect the quality of plant needed nutrients in the soil and beneficial microorganism population in the soil which could impact negatively the planting season.

2. Trees serves as protection of soil nutrients against wind erosions, so the soil nutrients would be affected.

The farmers tilling the soil means preparing the soil for a farming season as the following effect:

1. Helps control weed for the planting season.

2. This practice could further encourage soil erosion if not done well.

3. Tilling the soil could make leftover plants from the felling of trees mix well with the soil and also add nutrients to the soil when they decay.

8 0

3 years ago

A horizontal cylinder equipped with a frictionless piston contains 785 cm3 of steam at 400 K and 125 kPa pressure. A total of 83

guapka [62]

Answer:

a. 478.69 K

b. 939.43 $cm^{3}$

c. 19.30 J

d. 64.5J

Explanation:

From the question, we can identify the following;

$V_{o}$ = 785 $cm^{3}$ = 0.000785 $m^{3}$

$T_{o}$ = 400K

$P_{o}$ = 125 Kpa = 125 000 Pa

Using the ideal gas equation,

PV = nRT

where R is the molar gas constant = 8.314 $m^{3}$ ⋅Pa⋅ $K^{-1}$ ⋅ $mol^{-1}$

Thus, n = PV/RT = (125000 × 0.000785)/(8.314 × 400) = 0.03 mol

a. Steam temperature in K

To calculate this, we use the constant pressure process;

q = nΔH

Where q is 83.8J according to the question

Thus;

83.8 = 0.03 × [34980 + 35.5 $T_{1}$ - (34980 + 35.5 $T_{o}$ )]

83.8 = (0.03 × 35.5) ( $T_{1}$ - 400K)

83.8 = 1.065 ( $T_{1}$ - 400)

78.69 = ( $T_{1}$ - 400)

$T_{1}$ = 400 + 78.69

$T_{1}$ = 478.69 K

b. Final cylinder volume

To calculate this, we make use of the Charles' law(Temperature and pressure are directly proportional)

$V_{1}$ / $T_{1}$ = $V_{o}$ / $T_{o}$

$V_{1}$ = $V_{o}$ $T_{1}$ / $T_{o}$

$V_{1}$ = (785 × 478.69)/400

$V_{1}$ = 939.43 $cm^{3}$

c. Work done by the system

Mathematically, the work done by the system is calculated as follows;

w = P( $V_{1}$ - $V_{o}$ ) = 125 KPa ( 939.43 - 785) = 19.30 J

d. Change in internal energy of the steam in J

ΔU = q - w = 83.8 - 19.3 = 64.5J

6 0

3 years ago

A forklift raises a crate weighing 8.35 × 10Æ newtons to a height of 6.0 meters. What amount of work does the forklift do?

katrin2010 [14]

13.4998JOULES it does not put anymore work lol.

3 0

3 years ago

Read 2 more answers

Why do warm currents begin near the equator?

r-ruslan [8.4K]

Answer:

I would go with A

Explanation:

Because the earths equator is warmed by most direct rays of the sun, air a the equator is hotter than air further north or the south. The hotter air rises up at the equator and as colder air moves in to take its place, the wind begins to blow and push the ocean into waves and currents

5 0

3 years ago

What are all four Quantum numbers of 11th electron of magnesium

ruslelena [56]

Answer:n=0

l=0

ml=0

ms = =1/2 or +1/2

Explanation:

3 0

3 years ago