All categories

2 years ago

How do human population growth trends differ between developed nations and developing nations?

Physics

1 answer:

ololo11 [35]2 years ago

4 0

Answer:

Replacement-Level Fertility

Another important population characteristic that differ btw develop nation and developing nations is relates to births is replacement-level fertility. Replacement-level fertility is the fertility rate that will result in the replacement of the parents in the population. Again, in an ideal world, the human replacement-level fertility rate would be exactly two. This would mean that each couple would produce two offspring that would replace them in the population. If this occurred, then the human population would stay at a stable rate

You might be interested in

If an atom has 8 valence electrons, how many electrons does it need to gain to achieve a stable electron configuration?

krek1111 [17]

It needs 8 to be stable so it is already stable

7 0

4 years ago

Where is the centre of mass of a system of two particles is situated?

Sever21 [200]

Answer:

In a two particle system, the center of mass lies on the center of the line joining the two particles.

4 0

3 years ago

An insulated pipe carries steam at 300°C. The pipe is made of stainless steel (with k = 15 W/mK), has an inner diameter is 4 cm,

insens350 [35]

Answer:

The answers to the question are

(i) The rate of heat loss per-unit-length (W/m) from the pipe is 131.62 W

(ii) The temperature of the outer surface of the insulation is 49.89 °C

Explanation:

To solve the question, we note that the heat transferred is given by

$Q = \frac{2\pi L(t_{hf} - t_{cf}) }{\frac{1}{h_{hf}r_1}+\frac{ln(r_2/r_1)}{k_A} + \frac{ln(r_3/r_2)}{k_B} +\frac{1}{h_{cf}r_3}}$

Where

$t_{hf}$ = Temperature at the inside of the pipe = 300 °C

$t_{f}$ = Temperature at the outside of the pipe = 20 °C

r₁ =internal radius of pipe = 4.0 cm

r₂ = Outer radius of pipe = 4.5 cm

r₃ = Outer radius of the insulation = r₂ + 2.5 = 7.0 cm

$k_A$ = 15 W/m·K

$k_B$ = 0.038 W/m·K

$h_{hf}$ = 75 W/m²·K

$h_{cf}$ = 10 W/m²·K

Plugging in the values in the above equation where for a unit length L = 1 m, we have

Q = 131.32 W

From which we have, for the film of air at the pipe outer boundary layer

$Q = \frac{t_A-t_B}{R_T}$ Where $R_T$ for the air film on the pipe outer surface is given by

$R_T= \frac{1}{\alpha A}$

where A =area of the outside of the pipe

= $\frac{1}{10*2\pi*0.07*1 }$ = 0.227 K/W

Therefore

131.32 W = $\frac{t_A-20}{0.227}$ which gives

$t_A$ = 49.89 °C

Heat transferred by radiation = q' = ε×σ×(T₁⁴ - T₂⁴)

Where ε = 0.9, σ, = 5.67×10⁻⁸W/m²·(K⁴)

T₁ = Surface temperature of the pipe = 49.89 °C and

T₂ = Temperature of the surrounding = 20.00 °C

Plugging in the values gives, q' = 0.307 W per m²

Total heat lost per unit length = 131.32 + 0.307 =131.62 W

8 0

3 years ago

Sam drives her scooter 7 kilometres north. She stops for lunch and then drives

Talja [164]

Answer:

1. Distance travelled = 12 km.

2. Displacement = 8.6 km

Explanation:

From the question given above, the following data were obtained:

Distance 1 (d₁) = 7 km

Distance 2 (d₂) = 5 km

Total distance =?

Displacement =?

1. Determination of the distance travelled.

Distance 1 (d₁) = 7 km

Distance 2 (d₂) = 5 km

Total distance (dₜ) =?

dₜ = d₁ + d₂

dₜ = 7 + 5

dₜ = 12 km

2. Determination of the displacement.

In the attached photo, R is the displacement.

We can obtain the value of R by using the pythagoras theory as illustrated below:

R² = 7² + 5²

R² = 49 + 25

R² = 74

Take the square root of both side

R = √74

R = 8.6 km

8 0

3 years ago

The rate constants for the reactions of atomic chlorine and of hydroxyl radical with ozone are given by 3 × 10-11 e-250/T and 2

Vlada [557]

Answer:

Calculate the ratio of the rates of ozone destruction by these catalysts at 20 km, given that at this altitude the average concentration of OH is about 100 times that of Cl and that the temperature is about -50 °C

Knowing

Rate constants for the reactions of atomic chlorine and of hydroxyl radical with ozone are given by 3x $10^{-11} e^{-255/T}$ and 2x $10^{-12} e^{-940/T}$

T = -50 °C = 223 K

The reaction rate will be given by [Cl] [O3] 3x $10^{-11} e^{-255/223} = 9.78^{-12} [Cl] [O3]$

Than, the reaction rate of OH with O3 is

Rate = [OH] [O3] 2x $10^{-12} e^{-940/223} = 2.95^{-14} [OH] [O3]$

Considering these 2 rates we can realize the ratio of the reaction with Cl to the reaction with OH is 330 * [Cl] / [OH]

Than, the concentration of OH is approximately 100 times of Cl, and the result will be that the reaction with Cl is 3.3 times faster than the reaction with OH

Calculate the rate constant for ozone destruction by chlorine under conditions in the Antarctic ozone hole, when the temperature is about -80 °C and the concentration of atomic chlorine increases by a factor of one hundred to about 4 × 105 molecules cm-3

Knowing

Rate constants for the reactions of atomic chlorine and of hydroxyl radical with ozone are given by 3x $10^{-11} e^{-255/T}$ and 2x $10^{-12} e^{-940/T}$

T = -80 °C = 193 K

The reaction rate will be given by [Cl] [O3] 3x $10^{-11} e^{-255/193} = 8.21^{-12} [Cl] [O3]$

Than, the reaction rate of OH with O3 is

Rate = [OH] [O3] 2x $10^{-12} e^{-940/193} = 1.53^{-14} [OH] [O3]$

Considering these 2 rates we can realize the ratio of the reaction with Cl to the reaction with OH is 535 * [Cl] / [OH]

Than, considering the concentration of Cl increases by a factor of 100 to about 4 × $10^{5}$ molecules $cm^{-3}$ , the result will be that the reaction with OH will be 535 + (100 to about 4 × $10^{5}$ molecules $cm^{-3}$ ) times faster than the reaction with Cl

Explanation:

4 0

3 years ago