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2 years ago

ECONOMICS GRADE 10 CASE STUDY TOPIC: South African growth and development: Mining and industry Manufacturing and services.

Physics

1 answer:

bearhunter [10]2 years ago

6 0

The correct answe is South Africa - Economic Growth and Development:

After the formal end of the previous apartheid system two decades ago, South Africa can now boast having one of the richest economies in Africa and a well-functioning democracy. It is the largest economy in Africa, but it also has deeply ingrained structural issues that limit its ability to expand and flourish.

One of the largest economies on the African continent is that of South Africa. However, despite a period of rapid development from 2003 to 2007, its real GDP average yearly growth rate between 2001 and 2010 has been very weak and unquestionably significantly below the African average. A number of African nations have had substantially faster growth rates, which has improved a number of development-related indices.

By the standards of the recent growth records of several Euro Zone nations, South Africa's development is hardly sluggish! One crucial aspect of the economy is that South Africa has achieved relatively modest progress in meeting a number of important development targets and some of the Millennium Development Goals, but her economy does not appear to have achieved the "take-off" required to kick start significant development progress, especially against the backdrop of her deep social problems.

To learn more about South Africa's development refer the link:

brainly.com/question/24400106

#SPJ9

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A rock that has deformed ____ under stress keeps its new shape when the stress is released.

Anna007 [38]

Answer:

Elastically

Explanation:

A rock that has deformed Elastically under stress keeps its new shape when the stress is released.

In elastic deformation the original shape of the object is regained when the stress is removed. Whereas in plastic deformation the original shape is parmanently deformed with the application of stress.

8 0

3 years ago

A photon of wavelength 7.33 pm scatters at an angle of 157° from an initially stationary, unbound electron. What is the de Brogl

Ann [662]

Answer:

4.63 p.m.

Explanation:

The problem given here can be solved by the Compton effect which is expressed as

$\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)$

here, $\lambda$ is the initial photon wavelength, $\lambda^{'}$ is the scattered photon wavelength, h is he Planck's constant, $m_e$ is the free electron mass, c is the velocity of light, $\theta$ is the angle of scattering.

Given that, the scattering angle is, $\theta=157^{\circ}$

Putting the respective values, we get

$\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42(1-cos157^\circ ) p.m.$

Therfore,

$\lambda^{'}-\lambda=4.64 p.m.$

Here, the photon's incident wavelength is $\lamda=7.33pm$

So,

$\lambda^{'}=7.33+4.64=11.97 p.m$

From the conservation of momentum,

$\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}$

here, $\vec{P_\lambda}$ is the initial photon momentum, $\vec{P_{\lambda^{'}}}$ is the final photon momentum and $\vec{P_e}$ is the scattered electron momentum.

Expanding the vector sum, we get

$P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta$

Now expressing the momentum in terms of De-Broglie wavelength

$P=h/\lambda$ and putting it in the above equation we get,

$\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}$

Therfore,

$\lambda_{e}=\frac{7.33\times 11.97}{\sqrt{7.33^{2}+11.97^{2}-2\times 7.33\times 11.97\times cos157^\circ }} p.m.\\\lambda_{e}=\frac{87.7401}{18.935} = 4.63 p.m.$

This is the de Broglie wavelength of the electron after scattering.

8 0

3 years ago

A driver must always stop within 50 ft but not less than ____________ ft from the nearest rail when the signal is flashing and t

k0ka [10]

Answer:

20 ft

Explanation:

5 0

3 years ago

Read 2 more answers

A long, hollow, cylindrical conductor (inner radius 3.4 mm, outer radius 7.3 mm) carries a current of 36 A distributed uniformly

Elden [556K]

Answer:

a. $B= 9.45 \times10^{-3} T$

b. $B= 0.820 T$

c. $B= 0.0584 T$

Explanation:

First, look at the picture to understand the problem before to solve it.

a. d1 = 1.1 mm

Here, the point is located inside the cilinder, just between the wire and the inner layer of the conductor. Therefore, we only consider the wire's current to calculate the magnetic field as follows:

To solve the equations we have to convert all units to those of the international system. (mm→m)

$B=\frac{u_{0}I_{w}}{2\pi d_{1}} =\frac{52 \times4\pi \times10^{-7} }{2\pi 1.1 \times 10^{-3}} =9.45 \times10^{-3} T\\$

μ0 is the constant of proportionality

μ0=4πX10^-7 N*s2/c^2

b. d2=3.6 mm

Here, the point is located in the surface of the cilinder. Therefore, we have to consider the current density of the conductor to calculate the magnetic field as follows:

J: current density

c: outer radius

b: inner radius

The cilinder's current is negative, as it goes on opposite direction than the wire's current.

$J= \frac {-I_{c}}{\pi(c^{2}-b^{2} ) }}$

$J=\frac{-36}{\pi(5.33\times10^{-5}-1.16\times10^{-5}) } =-274.80\times10^{3} A/m^{2}$

$B=\frac{u_{0}(I_{w}-JA_{s})}{2\pi d_{2} } \\A_{s}=\pi (d_{2}^{2}-b^2)=4.40\times10^{-6} m^2\\$

$B=\frac{6.68\times10^{-5}}{8.14\times10^{-5}} =0.820 T$

c. d3=7.4 mm

Here, the point is located out of the cilinder. Therefore, we have to consider both, the conductor's current and the wire's current as follows:

$B=\frac{u_{0}(I_w-I_c)}{2\pi d_3 } =\frac{2.011\times10^-5}{3.441\times10^{-4}} =0.0584 T$

As we see, the magnitud of the magnetic field is greater inside the conductor, because of the density of current and the material's nature.

3 0

3 years ago

0. 85 kg of lead, specific heat 128 J/kgC, is heated from 50 C to 95 C. How much heat energy did the sample absorb?

Zanzabum

4896

0.85 x 45 x 128 = 4896

Change in energy = specific heat capacity x mass x change in temperature

5 0

2 years ago

ECONOMICS GRADE 10 CASE STUDY TOPIC: South African growth and development: Mining and industry Manufacturing and services.​

ECONOMICS GRADE 10 CASE STUDY TOPIC: South African growth and development: Mining and industry Manufacturing and services.