Answer:
Final length= 746.175 mm
Explanation:
Given that Length of aluminium at 223 C is 750 mm.As we know that when temperature of material increases or decreases then dimensions of material also increases or decreases respectively with temperature.
Here temperature of aluminium decreases so the final length of aluminium decreases .
As we know that

Now by putting the values

ΔL=3.82 mm
So final length =750-3.82 mm
Final length= 746.175 mm
Answer:
5.833
Explanation:
Coefficient of Perfomance (COP) is the ratio of refrigeration effect to power input.
where RE is refrigeration effect and P is power input
Here, the power input is given as 30 kW
We also know that 1 ton cooling is equivalent to 3.5 kW hence for 50 tons, RE=50*3.5=175 kW
Now the 
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.
Answer:
The principal stresses are σp1 = 27 ksi, σp2 = -37 ksi and the shear stress is zero
Explanation:
The expression for the maximum shear stress is given:

Where
σx = stress in vertical plane = 20 ksi
σy = stress in horizontal plane = -30 ksi
τM = 32 ksi
Replacing:

Solving for τxy:
τxy = ±19.98 ksi
The principal stress is:

Where
σp1 = 20 ksi
σp2 = -30 ksi
(equation 1)
equation 2
Solving both equations:
σp1 = 27 ksi
σp2 = -37 ksi
The shear stress on the vertical plane is zero