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Wewaii [24]
4 years ago
13

Convert 25 mm into in.

Engineering
1 answer:
mihalych1998 [28]4 years ago
4 0

Answer:

25 mm = 0.984252 inches

Explanation:

Millimeter and inches are both units of distance. The conversion of millimeter into inches is shown below:

<u>1 mm = 1/25.4 inches</u>

From the question, we have to convert 25 mm into inches

Thus,

<u>25 mm = (1/25.4)*25 inches</u>

So,

25 mm=\frac{25}{25.4} inches

Thus, solving we get:

<u>25 mm = 0.984252 inches</u>

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An aluminum part will be subjected to cyclic loading where the maximum stress will be 300 MPa and the minimum stress will be-100
Dominik [7]

Answer:

a) The mean stress experimented by the aluminium part is 100 megapascals, b) The stress amplitude of the aluminium part is 400 megapascals, c) The stress ratio of the aluminium part is 4.

Explanation:

a) The mean stress is determined by this expression:

\sigma_{m} = \frac{\sigma_{min}+\sigma_{max}}{2}

Where:

\sigma_{m} - Mean stress, measured in megapascals.

\sigma_{min} - Minimum stress, measured in megapascals.

\sigma_{max} - Maximum stress, measured in megapascals.

If we know that \sigma_{min} = -100\,MPa and \sigma_{max} = 300\,MPa, the mean stress is:

\sigma_{m} = \frac{-100\,MPa+300\,MPa}{2}

\sigma_{m} = 100\,MPa

The mean stress experimented by the aluminium part is 100 megapascals.

b) The stress amplitude is given by the following difference:

\sigma_{a} = |\sigma_{max}-\sigma_{min}|

Where \sigma_{a} is the stress amplitude, measured in megapascals.

If we know that \sigma_{min} = -100\,MPa and \sigma_{max} = 300\,MPa, the stress amplitude is:

\sigma_{a} = |300\,MPa-(-100\,MPa)|

\sigma_{a} = 400\,MPa

The stress amplitude of the aluminium part is 400 megapascals.

c) The stress ratio (R) is the ratio of the stress amplitude to mean stress. That is:

R = \frac{\sigma_{a}}{\sigma_{m}}

If we know that \sigma_{m} = 100\,MPa and \sigma_{a} = 400\,MPa, the stress ratio is:

R = \frac{400\,MPa}{100\,MPa}

R = 4

The stress ratio of the aluminium part is 4.

3 0
3 years ago
determine the components of each reaction at the ball and socket joint A and the tension in each cable necessary for equilibrium
Kitty [74]

The components of the reaction force in a ball and socket joint include the horizontal and vertical reaction forces.

<h3>What is a reaction force?</h3>

A reaction force is an equal but oppositely directed force which forms a pair which the applied or action force.

Tension is a type of force which is present in a string or a rope supporting a body or object.

A ball and socket jont is a type of joint in the body which is able to rotate freely about an axis.

The components of the reaction force in a ball and socket joint include the horizontal and vertical reaction forces.

Learn more about reaction forces at: brainly.com/question/1013858

#SPJ1

7 0
2 years ago
Question 7 of 10
exis [7]

Answer: Define the problem

Explanation: I just took the test

8 0
3 years ago
What is the weight density of a 2.24 in diameter titanium sphere that weights 0.82 lb?
nasty-shy [4]

Answer:

0.14\ lb/in^{3}

Explanation:

Density is defined as mass ler unit volume, expressed as

\rho=\frac {m}{v}

Where m is mass, \rho is density and v is the volume. For a sphere, volume is given as

v=\frac {4\pi r^{3}}{3}

Replacing this into the formula of density then

\rho=\frac {m}{\frac {4\pi r^{3}}{3}}

Given diameter of 2.24 in then the radius is 1.12 in. Substituting 0.82 lb for m then

\rho=\frac {0.82}{\frac {4\pi 1.12^{3}}{3}}=0.13932044952427\approx 0.14 lb/in^{3}

4 0
4 years ago
What is differences Between hard shoulder &amp; soft shoulder in civil Engineerin?
r-ruslan [8.4K]

Answer:

<em><u>The 'shoulder' of a road is the land to the edge of the road. On most roads without pavements, the shoulder is a strip of grass or a hedgerow. This is known as a 'soft shoulder'. On a motorway, this strip of land is hardstanding, hence the name 'hard shoulder.'</u></em>

<em><u>Mark</u></em><em><u> </u></em><em><u>as</u></em><em><u> brilliant</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u> </u></em>

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4 years ago
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