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1 year ago

What are the general rules for press fit allowances

Engineering

1 answer:

Keith_Richards [23]1 year ago

4 0

Explanation:

As a general rule of thumb, the large the diameter of a bearing, bushing or pin, the larger the tolerance range,” Brieschke points out. “The inverse is true for smaller-diameter pieces.”

Mike Brieschke, vice president of sales at Aries Engineering, says a 0.25-inch-diameter metal dowel that is press-fit into a mild steel hole usually has an interference of ±0.0015 inch. Parts in noncritical assemblies tend to have looser tolerances

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Answer:

1. Equatorial Evergreen or Rainforest

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3. Mediterranean forest

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A system consists of a disk rotating on a frictionless axle

kakasveta [241]

The system includes a disk rotating on a frictionless axle and a bit of clay transferring towards it, as proven withinside the determine above.

<h3>What is the angular momentum?</h3>

The angular momentum of the device earlier than and after the clay sticks can be the same.

Conservation of angular momentum the precept of conservation of angular momentum states that the whole angular momentum is usually conserved.

Li = Lf where;
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Thus, the angular momentum of the device earlier than and after the clay sticks can be the same.

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2 years ago

A motorcycle starts from rest with an initial acceleration of 3 m/s^2, and the acceleration then changes with the distance s as

katrin2010 [14]

Answer:

Follows are the solution to this question:

Explanation:

Calculating the area under the curve:

A = as

$=\frac{1}{2}(3 +6 \frac{m}{s^2})(100 \ m)+ \frac{1}{2}(6+4 \frac{m}{s^2})(100 m) \\\\=\frac{1}{2}(9 \frac{m}{s^2})(100 \ m)+ \frac{1}{2}(10\frac{m}{s^2})(100 m) \\\\=\frac{1}{2}(900 \frac{m^2}{s^2})+ \frac{1}{2}(1,000\frac{m^2}{s^2}) \\\\=(450 \frac{m^2}{s^2})+ (500\frac{m^2}{s^2}) \\\\= 950 \ \frac{m^2}{s^2}$

Calculating the kinematics equation:

$\to v^2 = v^2_{o} + 2as\\\\$

$=0+ \sqrt{2as}\\\\ = \sqrt{2(A)}\\\\= \sqrt{2(950 \frac{m^2}{s^2})}\\\\= 43.59 \frac{m}{s}$

Calculating the value of acceleration:

$\to a= \frac{dv}{dt}$

$=\frac{dv}{ds}(\frac{ds}{dt}) \\\\=v\frac{dv}{ds}\\\\\to \frac{dv}{ds}=\frac{a}{v}$

$\to \frac{dv}{ds} =\frac{4 \frac{m}{s^2}}{43.59 \frac{m}{s}} \\\\$

$=\frac{0.092}{s}$

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2 years ago

In the engineering design process, testing is the least important step.<br> True<br> False

Wittaler [7]

Answer:

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Explanation:

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lorasvet [3.4K]

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