Answer:
%Reduction in area = 73.41%
%Reduction in elongation = 42.20%
Explanation:
Given
Original diameter = 12.8 mm
Gauge length = 50.80mm
Diameter at the point of fracture = 6.60 mm (0.260 in.)
Fractured gauge length = 72.14 mm.
%Reduction in Area is given as:
((do/2)² - (d1/2)²)/(do/2)²
Calculating percent reduction in area
do = 12.8mm, d1 = 6.6mm
So,
%RA = ((12.8/2)² - 6.6/2)²)/(12.8/2)²
%RA = 0.734130859375
%RA = 73.41%
Calculating percent reduction in elongation
%Reduction in elongation is given as:
((do) - (d1))/(d1)
do = 72.14mm, d1 = 50.80mm
So,
%RA = ((72.24) - (50.80))/(50.80)
%RA = 0.422047244094488
%RA = 42.20%
Answer:
Could ask a family member to help
Explanation:
Relay contacts that are defined as being normally open (n.o.) have contacts that are open only if the relay coil is known to have de-energized.
<h3>What is meant by normally open contacts?</h3>
Normally open (NO) are known to be open if there is no measure of current that is flowing through a given coil but it often close as soon as the coil is said to be energized.
Note that Normally closed (NO) contacts are said to be closed only if the coil is said to be de-energized and open only if the coil is said to carry current or is known to have energized.
The role of relay contact is wide. The Relays are tools that are often used in the work of switching of control circuits and it is one that a person cannot used for power switching that has relatively bigger ampacity.
Therefore, Relay contacts that are defined as being normally open (n.o.) have contacts that are open only if the relay coil is known to have de-energized.
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Answer:
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Answer:
250 N
Explanation:
Drag opposes the direction of motion, so it points in the opposite direction of thrust. Therefore, the net force is:
∑F = 450 N − 200 N
∑F = 250 N
