what is an example of an innovative solution to an engineering problem? Explain briefly why you chose this answer.

Answers

Answer 1

Answer:

robotic technology

Explanation:

Innovation is nothing but the use of various things such as ideas, products, people to build up a solution for the benefit of the human. It can be any product or any solution which is new and can solve people's problems.

Innovation solution makes use of technology to provide and dispatch new solutions or services which is a combination of both technology and ideas.

One such example of an innovative solution we can see is the use of "Robots" in medical science or in any military operations or rescue operation.

Sometimes it is difficult for humans to do everything or go to everywhere. Thus scientist and engineers have developed many advance robots or machines using new ideas and technology to find solutions to these problems.

Using innovations and technologies, one can find solutions to many problems which is difficult for the peoples. Robots can be used in any surveillance operation or in places of radioactive surrounding where there is a danger of humans to get exposed to such threats. They are also used in medical sciences to operate and support the patient.

Related Questions

An unknown impedance Z is connected across a 380 V, 60 Hz source. This causes a current of 5A to flow and 1500 W is consumed. Determine the following: a. Real Power (kW) b. Reactive Power (kvar) c. Apparent Power (kVA) d. Power Factor e. The impedance Z in polar and rectangular form

Answers

Answer:

a) Real Power (kW) = 1.5 kW

b) Reactive Power (kvar) is 1.1663 KVAR

c) Apparent Power (kVA) is 1.9 KVA

d) the Power Factor cos∅ is 0.7894

e) the impedance Z in polar and rectangular form is 76 ∠ 37.87° Ω

Explanation:

Given that;

V = 380v

i = 5A

P = 1500 W

determine;

a) Real Power (kW)

P = 1500W = 1.5 kW

therefore Real Power (kW) = 1.5 kW

b) Reactive Power (kvar)

p = V×i×cos∅

cos∅ = p / Vi

cos∅ = 1500 / ( 380 × 5 ) = 0.7894

∅ = cos⁻¹ (0.7894)

∅ = 37.87°

Q = VIsin∅

Q = 380 × 5 × sin( 37.87° )

Q = 1.1663 KVAR

Therefore Reactive Power (kvar) is 1.1663 KVAR

c) Apparent Power (kVA)

S = P + jQ

= ( 1500 + J 1166.3 ) VA

S = 1900 ∠ 37.87° VA

S = 1.9 KVA

Therefore Apparent Power (kVA) is 1.9 KVA

d) Power Factor

p = V×i×cos∅

cos∅ = p / Vi

cos∅ = 1500 / ( 380 × 5 ) = 0.7894

Therefore the Power Factor cos∅ is 0.7894

e) The impedance Z in polar and rectangular form

Z = 380 / ( S∠-37.87) = V/I

Z = ( 60 + j 46.647) Ω

Z = 76 ∠ 37.87° Ω

Therefore the impedance Z in polar and rectangular form is 76 ∠ 37.87° Ω

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