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elixir [45]
3 years ago
8

List one advantage and one disadvantage of the use of the commutator?

Engineering
1 answer:
Mariulka [41]3 years ago
3 0

Answer and  Explanation:

Commutator are used in DC machine commutator is mainly used for the reversing the direction of the current .It is connected to the armature of the DC generator or motor

ADVANTAGE OF COMMUTATOR The main advantage of the commutator in DC motor is to keep keep the direction of the toque always in the same direction by changing the current direction

DISADVANTAGE OF COMMUTATOR : The main disadvantage is due to the friction between the commutator and brushes there is a friction loss.

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Which of the following types of protection is an employer required to pay for?
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Hearing protection would be your answer!

Explanation:

This includes earplugs,muffs etc.

Hope it helps!

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If extension done in a stork position, during backward bending in a standing position of a special test for lumbar spine, produc
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How engineer can find problems and solutions <br><br><br> Give example
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2 years ago
The fracture toughness of a stainless steel is 137 MPa*m12. What is the tensile impact load sustainable before fracture that a r
Charra [1.4K]

Answer:

7.7 kN

Explanation:

The capacity of a material having a crack to withstand fracture is referred to as fracture toughness.

It can be expressed by using the formula:

K = \sigma Y \sqrt{\pi a}

where;

fracture toughness K = 137 MPam^{1/2}

geometry factor Y = 1

applied stress \sigma = ???

crack length a = 2mm = 0.002

∴

137 =\sigma \times 1  \sqrt{ \pi \times 0.002 }

137 =\sigma \times 0.07926

\dfrac{137}{0.07926} =\sigma

\sigma = 1728.489 MPa

Now, the tensile impact obtained is:

\sigma = \dfrac{P}{A}

P = A × σ

P = 1728.289 × 4.5

P = 7777.30 N

P = 7.7 kN

7 0
3 years ago
A cylindrical metal specimen of initial diameter d0 =14 mm, initial length L0=53 mm, strain hardening exponent n=0.31, strength
Marrrta [24]

Answer:

a) Ef = 0.755

b) length of specimen( Lf )= 72.26mm

  diameter at fracture = 9.598 mm

c) max load ( Fmax ) = 52223.24 N

d) Ft = 51874.67 N

Explanation:

a) Determine the true strain at maximum load and true strain at fracture

True strain at maximum load

Df = 9.598 mm

True strain at fracture

Ef = 0.755

b) determine the length of specimen at maximum load and diameter at fracture

Length of specimen at max load

Lf = 72.26 mm

Diameter at fracture

= 9.598 mm

c) Determine max load force

Fmax = 52223.24 N

d) Determine Load ( F ) on the specimen when a true strain et = 0.25 is applied during tension test

F = 51874.67 N

attached below is a detailed solution of the question above

3 0
2 years ago
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