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g100num [7]
4 years ago
6

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

Physics
2 answers:
cricket20 [7]4 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]4 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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Romeo traveled by a 900 kg horse from Manitua to Verona accelerating at the rate of 20 km/hr. With what force is Romeo moving at
Nookie1986 [14]

Answer:

Force(Romeo moving) = 5,000 N

Explanation:

Given:

Mass of horse = 900 kg

Acceleration = 20 km/hr

Find:

Force(Romeo moving)

Computation:

Acceleration = 20 km/hr

Acceleration in m/s = 20 / 3.6 = 5.555556 m/s²

Force = m x a

Force(Romeo moving) = 900 x 5.555556

Force(Romeo moving) = 5,000 N

3 0
3 years ago
A train travels 77 kilometers in 4 hours, and then 76 kilometers in 2 hours. What is its average speed?
Masja [62]

( (77/4) + 76/2 )/2 = 28.625 km/h is what i got

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4 years ago
The difference between experimental technique and procedure
kaheart [24]
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4 0
4 years ago
Two automobiles traveling at right angles to each other collide and stick together. Car A has a mass of 1200 kg and had a speed
sergij07 [2.7K]

Answer:

v_{B0}=15.73 m/s

Explanation:

We can use the conservation of momentum. The initial momentum is equal to the final momentum:

x-coordinate

p_{0x}=p_{fx}

m_{A}v_{A0}=(m_{A}+m_{B})v_{cx}  

m_{A}v_{A0}=(m_{A}+m_{B})v_{c}cos(40) (1)

y-coordinate

p_{0y}=p_{fy}

m_{B}v_{B0}=(m_{A}+m_{B})v_{cy}  

m_{B}v_{B0}=(m_{A}+m_{B})v_{c}sin(40) (2)

We can divide equations (2) and (1):

\frac{m_{B}v_{B0}}{m_{A}v_{A0}}=\frac{sin(40)}{cos(40)}

\frac{m_{B}v_{B0}}{m_{A}v_{A0}}=tan(40)

v_{B0}=\frac{m_{A}v_{A0}}{m_{B}}*tan(40)

v_{B0}=\frac{1200*25}{1600}*tan(40)

v_{B0}=15.73 m/s

I hope it helps you!

           

4 0
4 years ago
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Answer:

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