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Galina-37 [17]
3 years ago
6

2. Which biome has many grasses and few trees due to fires, drought, and grazing?

Physics
2 answers:
inessss [21]3 years ago
8 0

"He who lays up treasures on earth spends his life backing away from his treasures. To him, death is loss. He who lays up treasures in heaven looks forward to eternity; he’s moving daily toward his treasures. To him, death is gain." - Randy Alcorn

maksim [4K]3 years ago
3 0

Answer:

C. Temperate grasslands

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A gas‑forming reaction produces 1.90 m 3 1.90 m3 of gas against a constant pressure of 179.0 kPa. 179.0 kPa. Calculate the work
goldfiish [28.3K]

Answer:

The work done is 3.4 × 10⁵ J.

Explanation:

Given:

Pressure of the gas produced (P) = 179 kPa

Volume of the gas produced (ΔV) = 1.90 m³

We need to find the work done in joules. For that, we don't need any conversion as the units are already in SI units which will give the result in Joules only.

Now, let us verify our results by using conversion factors and without using them.

Using conversion factors:

1 m³ = 1000 L

So, 1.90 m³ = 1.90 m³ × 1000 \frac{L}{m^3} = 1900 L

Also, 1 atm = 101.325 kPa

So, 179 kPa = 179 kPa × \frac{1\ atm}{101.325\ kPa} = 1.767 atm

Now, work done in a constant pressure process is given as:

Work = Pressure × Volume change

Work = P × ΔV

Work = 1.767 atm × 1900 L

Work = 3.36 × 10³ atm-L

Now, again using the energy conversion for work.

1 atm-L = 101.325 J

So, 3.36 × 10³ atm-L = 3.36 × 10³ atm-L × \frac{101.325\ J}{1\ atm-L} = 3.4 × 10⁵ J

Therefore, the work done is 3.4 × 10⁵ J.

Now, let us verify the above result without any conversion.

Work = P × ΔV = 179 × 1000 × 1.90 = 3.4 × 10⁵ J.

Therefore, the work done is same by both ways.

Hence the work done is 3.4 × 10⁵ J.

8 0
3 years ago
If the mass of the block is 5 kg and the speed 7 m/s, what is the work done <br> on the block?
Kamila [148]

Answer:

A block of mass M = 5 kg is resting on a rough horizontal surface for which the coefficient of friction is 0.2. When a force F = 40N is applied, the acceleration of the block will be then (g=10ms

2 ).

Mass of the block=5kg

Coeffecient of friction=0.2

external applied force, F=40N

The angle at which the force is applied=30degree

So the horizontal component of force=Fcos30=40×

23 =20 3 N

While the uertical component of the force acting in upward direction=Fsin30=40× 21

​ =20N

The normal reaction from the surface (N)=mg−Fsin30=50−20=30N

So the ualue of limiting friction=μN=0.2×30=6N

Hence the net horizontal force on the block=Fcos30=μN=20

3

​

N−6N=28.64N

The horizontal acceleration of the block=

m

Fcos30−μN = 528.64

​  =5.73m/s 2

8 0
3 years ago
A rocket engine has a chamber pressure 4 MPa and a chamber temperature of 2000 K. Assuming isentropic expansion through the nozz
gladu [14]

This question is incomplete, the complete question is;

A rocket engine has a chamber pressure 4 MPa and a chamber temperature of 2000 K. Assuming isentropic expansion through the nozzle, and an exit Mach number of 3.2, what are the stagnation pressure and temperature in the exit plane of the nozzle?  Assume the specific heat ratio is 1.2.

Answer:

- stagnation pressure is 274.993 Mpa

- the stagnation temperature Tt is 4048 K

Explanation:

Given the data in the question;

To determine the stagnation pressure and temperature in the exit plane of the nozzle;

we us the expression;

Pt/P = (1 + (γ-1 / 2) M²)^(γ/γ -1) = ( Tt/T )^(γ/γ -1)

where Pt is stagnant pressure = ?

P is static pressure = 4 MPa = 4 × 10⁶ Pa  

Tt is stagnation temperature = ?

T is the static temperature  = 2000 K

γ is ratio of specific heats = 1.2

M is Mach number M = 3.2

we substitute

Pt/P = (1 + (γ-1 / 2) M²)^(γ/γ -1)

Pt = P(1 + (γ-1 / 2) M²)^(γ/γ -1)

Pt = 4 × 10⁶(1 + (1.2-1 / 2) 3.2²)^(1.2/1.2 -1)

Pt = 4 × 10⁶ × 68.7484

Pt = 274.993 × 10⁶ Pa

Pt = 274.993 Mpa

Therefore stagnation pressure is 274.993 Mpa

Now, to get our stagnation Temperature

Pt/P = ( Tt/T )^(γ/γ -1)

we substitute

274.993 × 10⁶ Pa / 4 × 10⁶ Pa =  ( Tt / 2000 )^(1.2/1.2 -1)

68.7484 =  Tt⁶ / 6.4 × 10¹⁹

Tt⁶ = 68.7484 × 6.4 × 10¹⁹

Tt⁶ = 4.3998976 × 10²¹

Tt = ⁶√(4.3998976 × 10²¹)

Tt = 4047.999 ≈ 4048 K

Therefore, the stagnation temperature Tt is 4048 K

6 0
3 years ago
Definition of synthesis
Finger [1]
Combination or composition, in particular.
6 0
4 years ago
What’s an example of contact friction
kotegsom [21]

Answer:

a nightstand on a lamp table

Explanation:

5 0
3 years ago
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