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2 years ago

Why is education the main basis of profession? Write in point

Social Studies

1 answer:

Irina-Kira [14]2 years ago

6 0

Answer:

education offers many-fold benefits to individuals and their professions. They are:

People acquire relevant knowledge and skills, which prepares them to polish and commercialize their profession.
This would ensure their employments and income for survival and reduces the problem of unemployment.
It enhances their capacity to start and run their own business and thus contributes to their overall career development.
Professional education prepares competent and skillful human resources.
This would enable people involved in production sectors to produce goods of high quality.
Those working under service sectors can better deal with their clients or service holders
supported by relevant training and exposure.

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3 years ago

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2 years ago

Read 2 more answers

A basketball is dropped from the edge of a cliff at the same time that second basketball is thrown upward from the ground. When

pentagon [3]

Answer:

The collision occurs at a height of 1/3 of the total height of cliff.

Explanation:

The situation is represented in the attached figure:

For the stone dropped from top of cliff we have

Initial velocity of drop= 0 m/s.

Now the time at which it attains a velocity of 2v is obtained from first equation of kinematics as

$v_{f}=u_{i}+gt\\\\2v=0+gt\\\\\therefore t=\frac{2v}{g}$

Thus the distance it covers in this time can be calculated using third equation of kinematics as

$v_{f}^{2}=v_{i}^{2}+2gh\\\\(2v)^{2}=2gh_{1}\\\\\therefore h_{1}=\frac{2v^{2}}{g}$

For the stone thrown upwards the velocity at collision is give as 'v' thus the velocity at which it is thrown upwards can be calculated by first equation of kinematics as

$v=v_{i}-gt\\\\\therefore v_{i}=v+gt\\\\v_{i}=v+g\times \frac{2v}{g}\\\\\therefore v_{i}=3v$

The height over which it changes it's velocity from '3v' to 'v' can be again obtained from third equation of kinematics as

$v^{2}=(3v)^{2}-2gh_{2}\\\\v^{2}=9v^{2}-2gh_{2}\\\\\therefore h_{2}=\frac{4v^{2}}{g}$

$\therefore h_{2}=2h_{1}$

Thus the collision occurs at at a fraction of 1/3 of height.

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3 years ago