A circuit is a pathway for electron flow.

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

A laboratory in the Y building keep a vacuum pressure of 0.1 kPa abs. What is the net force acting on the door considering the a

seropon [69]

Answer:

net force acting on the floor is 100 kN

Explanation:

Given data:

$P_{vaccum} = 0.1 kPa$

$P_{atm} = 101.325 kPa$

dimension of floor = 2 m \times 0.5 m

we know that

Net force can be calculated as follow

$f_{net} = P_{vaccum} \times area$

$f_{net} = 0.1\times 10^3 \times 2\times 0.5$

$f_{net} = 0.1\times 10^3 \times 1$

$f_{net} = 100 kN$

Therefore net force acting on the floor is 100 kN

7 0

4 years ago

An important fluid property is the kinematic viscosity which determines the viscous, or frictional, forces acting in a flow. The

Aleksandr [31]

Answer:- 400 K - 26.665

600 K - 53.67

Explanation: y−y1=(y2−y1)/(x2−x1)×(x−x1)

Basic usage of the interpolation formula to get the values of the required answers kinematic viscosities at 400 and 600 K

5 0

4 years ago

Explain orthographic<br>and multi-view<br>projection

Elina [12.6K]

Answer:

Orthographic projection is a projection method to present three-dimensional shapes in two-dimensional format in which the projection lines are drawn to be orthogonal to the plane of projection, such that each view of the three-dimensional object is translated to a view of the orthographic projection and orthogonal to the view

A multiview projection is the representation of a three-dimensional projection by two or more two-dimensional views

Explanation:

3 0

4 years ago

Suppose there are 76 packets entering a queue at the same time. Each packet is of size 5 MiB. The link transmission rate is 2.1

tia_tia [17]

Answer:

938.7 milliseconds

Explanation:

Since the transmission rate is in bits, we will need to convert the packet size to Bits.

1 bytes = 8 bits

1 MiB = 2^20 bytes = 8 × 2^20 bits

5 MiB = 5 × 8 × 2^20 bits.

The formula for queueing delay of <em>n-th</em> packet is : (n - 1) × L/R

where L : packet size = 5 × 8 × 2^20 bits, n: packet number = 48 and R : transmission rate = 2.1 Gbps = 2.1 × 10^9 bits per second.

Therefore queueing delay for 48th packet = ( (48-1) ×5 × 8 × 2^20)/2.1 × 10^9

queueing delay for 48th packet = (47 ×40× 2^20)/2.1 × 10^9

queueing delay for 48th packet = 0.938725181 seconds

queueing delay for 48th packet = 938.725181 milliseconds = 938.7 milliseconds

4 0

3 years ago