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3 years ago

WHICH OF THE FOLLOWING IS A HAZARD OF GLOBALIZATION???

Physics

2 answers:

cricket20 [7]3 years ago

7 0

Answer:

correct option will be option B)

Explanation:

the correct option will be option B)

the hazard of globalization will be that the companies will start violating human right.

As due to increase in globalization the job insecurity will increase and the fluctuation of the price will also increase.

This will lead company to not following the human right not paying proper wages , job insecurity etc .

PolarNik [594]3 years ago

4 0

The correct option is B.
There are many hazards that may occur as a result of globalization. One of this is the violation of human right by companies. This can occur in several ways, for instance, there are some companies from developed countries which introduce hazardous technological devices to less developing countries.

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For a solution to be a physical solution, it must satisfy several criteria. First, it must be continuous everywhere. Second, it

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Answer:

Explanation:

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3 years ago

A fuel pump sends gasoline from a car's fuel tank to the engine at a rate of 6.55x10-2 kg/s. The density of the gasoline is 740

klasskru [66]

Answer:

speed = 3.95 m/s

Explanation:

area = π x radius^2

area = π x (2.67 x 10^-3)^2

volume flow rate = area x speed

volume / time = area x speed

density = mass / volume

volume = mass / density

<u>mass / (density x time) = area *speed</u>

mass flow rate = mass / time

<u>mass flow rate / density = area x speed</u>

6.55 x 10^-2 / 740 = pi * (2.67 x 10^-3)^2 * speed

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3 years ago

Air enters a nozzle steadily at 2.21 kg/m3 and 20 m/s and leaves at 0.762 kg/m3 and 150 m/s. If the inlet area of the nozzle is

saveliy_v [14]

Answer:

a) The mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle is 23.6 cm².

Explanation:

a) The mass flow rate through the nozzle can be calculated with the following equation:

$\dot{m_{i}} = \rho_{i} v_{i}A_{i}$

Where:

$v_{i}$ : is the initial velocity = 20 m/s

$A_{i}$ : is the inlet area of the nozzle = 60 cm²

$\rho_{i}$ : is the density of entrance = 2.21 kg/m³

$\dot{m} = \rho_{i} v_{i}A_{i} = 2.21 \frac{kg}{m^{3}}*20 \frac{m}{s}*60 cm^{2}*\frac{1 m^{2}}{(100 cm)^{2}} = 0.27 kg/s$

Hence, the mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle can be found with the Continuity equation:

$\rho_{i} v_{i}A_{i} = \rho_{f} v_{f}A_{f}$

$0.27 kg/s = 0.762 kg/m^{3}*150 m/s*A_{f}$

$A_{f} = \frac{0.27 kg/s}{0.762 kg/m^{3}*150 m/s} = 0.00236 m^{2}*\frac{(100 cm)^{2}}{1 m^{2}} = 23.6 cm^{2}$

Therefore, the exit area of the nozzle is 23.6 cm².

I hope it helps you!

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3 years ago

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AysviL [449]

Answer:

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2 years ago

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Bogdan [553]

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3 years ago

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