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3 years ago

100 POINTS

Engineering

2 answers:

goldfiish [28.3K]3 years ago

8 0

Answer:

Primary succession begins in barren areas, such as on bare rock exposed by a retreating glacier. The first inhabitants are lichens or plants—those that can survive in such an environment. Over hundreds of years these “pioneer species” convert the rock into soil that can support simple plants such as grasses. These grasses further modify the soil, which is then colonized by other types of plants. Each successive stage modifies the habitat by altering the amount of shade and the composition of the soil. The final stage of succession is a climax community, which is a very stable stage that can endure for hundreds of years.

Explanation:

this is the correct answer for your question

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jenyasd209 [6]3 years ago

3 0

Answer:

Depending on the size of the shrubs and the tress, will give you a general idea on how long they take to grow. Unless there are small plants that take forever to grow and or large plants that grow quickly. Hope this helps....

Explanation:

You might be interested in

Compressed Air In a piston-cylinder device, 10 gr of air is compressed isentropically. The air is initially at 27 °C and 110 kPa

Helen [10]

Answer:

(a) 2.39 MPa (b) 3.03 kJ (c) 3.035 kJ

Explanation:

Solution

Recall that:

A 10 gr of air is compressed isentropically

The initial air is at = 27 °C, 110 kPa

After compression air is at = a450 °C

For air, R=287 J/kg.K

cv = 716.5 J/kg.K

y = 1.4

Now,

(a) W efind the pressure on [MPa]

Thus,

T₂/T₁ = (p₂/p₁)^r-1/r

=(450 + 273)/27 + 273) =

=(p₂/110) ^0.4/1.4

p₂ becomes 2390.3 kPa

So, p₂ = 2.39 MPa

(b) For the increase in total internal energy, is given below:

ΔU = mCv (T₂ - T₁)

=(10/100) (716.5) (450 -27)

ΔU =3030 J

ΔU =3.03 kJ

ΔW = mR ( (T₂ - T₁) / k- 1

(10/100) (287) (450 -27)/1.4 -1

ΔW = 3035 J

Hence, the total work required is = 3.035 kJ

4 0

3 years ago

An engineer is testing the shear strength of spot welds used on a construction site. The engineer's null hypothesis at a 5% leve

lilavasa [31]

Answer:

b) The null hypothesis should be rejected.

Explanation:

The null hypothesis is that the mean shear strength of spot welds is at least

3.1 MPa

H0: u ≥3.1 MPa against the claim Ha: u< 3.1 MPa

The alternate hypothesis is that the mean shear strength of spot welds is less than 3.1 MPa.

This is one tailed test

The critical region Z(0.05) < ± 1.645

The Sample mean= x`= 3.07

The number of welds= n= 15

Standard Deviation= s= 0.069

Applying z test

z= x`-u/s/√n

z= 3.07-3.1/0.069/√15

z= -0.03/0.0178

z= -1.68

As the calculated z= -1.68 falls in the critical region Z(0.05) < ± 1.645 the null hypothesis is rejected and the alternate hypothesis is accepted that the mean shear strength of spot welds is less than 3.1 MPa

8 0

2 years ago

An insulated piston-cylinder device initially contains 0.16 m2 of CO2 at 150 kPa and 41 °C. Electric resistance heater supplied

lyudmila [28]

Answer:

I=0.3636

Explanation:

See the attached picture for explanation.

4 0

3 years ago

You want a potof water to boil at 105 celcius. How heavy a

ankoles [38]

Answer:

35.7 kg lid we put

Explanation:

given data

temperature = 105 celcius

diameter = 15 cm

Patm = 101 kPa

to find out

How heavy a lid should you put

solution

we know Psaturated from table for temperature is 105 celcius is

Psat = 120.8 kPa

area will be here

area = $\frac{\pi }{4} d^2$ ..................1

here d is diameter

put the value in equation 1

area = $\frac{\pi }{4} 0.15^2$

area = 0.01767 m²

so net force is

Fnet = ( Psat - Patm ) × area

Fnet = ( 120.8 - 101 ) × 0.01767

Fnet = 0.3498 KN = 350 N

we know

Fnet = mg

mass = $\frac{Fnet}{g}$

mass = $\frac{350}{9.8}$

mass = 35.7 kg

so 35.7 kg lid we put

3 0

3 years ago

Plane wall of material A with internal heat generation is insulated on one side and bounded by a second wall of material B, whic

viktelen [127]

Sorry❤

Have a nice day ✨

8 0

3 years ago

*100 POINTS*

100 POINTS