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zavuch27 [327]
3 years ago
11

In electric heaters, electrical energy is converted to potential energy. a)-True b)-false?

Engineering
1 answer:
son4ous [18]3 years ago
8 0

Answer:

False

Explanation:

In electric heater electric energy is converted into heat energy. In heater wires are present which have resistance and current is flow in heater when we connect the heater to supply.

And we know that whenever current is flow in any resistance then heat is produced so in electric heaters electric energy is converted into heat energy

So this is a false statement

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Engineering is a broad category that includes a variety of occupations and attempts to solve problems using math and
nadezda [96]

Answer:

Engineering is a broad category that includes a variety of occupations and attempts to solve problems using math and  systematic and scientific processes.

Explanation:

Engineering is the scientific discipline that deals with the implementation of science on materials, constructions, machines, systems and processes to realize a specific goal. It is an area of ​​technical activity that includes a number of specialized areas and disciplines aimed at the practical application of scientific, economic, social and practical knowledge in order to turn natural resources for the benefit of humans.

The goals of engineering activities are the invention, development, creation, implementation, repair, maintenance and / or improvement of equipment, materials or processes.  Engineering is closely intertwined with science, relying on the postulates of fundamental science and the results of applied research. In this sense, it is a branch of scientific and technical activity.

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Determine (a) the principal stresses and (b) the maximum in-plane shear stress and average normal stress at the point. Specify t
raketka [301]

Answer:

a) 53 MPa,  14.87 degree

b) 60.5 MPa  

Average shear = -7.5 MPa

Explanation:

Given

A = 45

B = -60

C = 30

a) stress P1 = (A+B)/2 + Sqrt ({(A-B)/2}^2 + C)

Substituting the given values, we get -

P1 = (45-60)/2 + Sqrt ({(45-(-60))/2}^2 + 30)

P1 = 53 MPa

Likewise P2 = (A+B)/2 - Sqrt ({(A-B)/2}^2 + C)

Substituting the given values, we get -

P1 = (45-60)/2 - Sqrt ({(45-(-60))/2}^2 + 30)

P1 = -68 MPa

Tan 2a = C/{(A-B)/2}

Tan 2a = 30/(45+60)/2

a = 14.87 degree

Principal stress

p1 = (45+60)/2 + (45-60)/2 cos 2a + 30 sin2a = 53 MPa

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Average = (45-(-60))/2 = -7.5 MPa

5 0
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