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Serhud [2]
3 years ago
14

A disk is rotating around an axis located at its center. The angular velocity is 0.5 rad/s. The radius of the disk is 0.4 m. Wha

t is the magnitude of the velocity at a point located on the outer edge of the disk, in units of m/s?
Engineering
2 answers:
dimaraw [331]3 years ago
4 0

Answer:

0.2 m/s

Explanation:

The velocity of a point on the edge of a disk rotating disk can be calculated as:

v=\omega*r

Where \omega is the angular velocity and r the radius of the disk. This leads to:

v=0.5\,rad/s\,*\,0.4\,m=0.2\,m/s

Anna007 [38]3 years ago
4 0

Answer:

0.2 m/s

Explanation:

<u>Step 1: </u>identify the given parameters

angular velocity, ω = 0.5 rad/s

radius of the disk, r = 0.4m

<u>Note</u>: the inner part of the disk and outer edge spin at the same rate. This means that the velocity at inner part of the disk is the same as the outer part.

<u>Step 2:</u> calculate the velocity of the disk at the outer edge in m/s

Velocity = Angular velocity (rad/s) X radius (m)

Velocity = 0.5 rad/s X 0.4 m

Velocity = 0.2 m/s

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Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320
lesya [120]

Answer:

The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.

Explanation:

The complete statement of this question is "Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320 MPa, and a factor of safety of 4.2 is required with respect to fracture. Determine the minimum allowable bolt diameter required to support an applied load of P = 450 kN"

Each bolt is subjected to shear forces. In this case, safety factor is the ratio of the ultimate shear strength to maximum allowable shear stress. That is to say:

n = \frac{S_{uts}}{\tau_{max}}

Where:

n - Safety factor, dimensionless.

S_{uts} - Ultimate shear strength, measured in pascals.

\tau_{max} - Maximum allowable shear stress, measured in pascals.

The maximum allowable shear stress is consequently cleared and computed: (n = 4.2, S_{uts} = 320\times 10^{6}\,Pa)

\tau_{max} = \frac{S_{uts}}{n}

\tau_{max} = \frac{320\times 10^{6}\,Pa}{4.2}

\tau_{max} = 76.190\times 10^{6}\,Pa

Since each bolt has a circular cross section area and assuming the shear stress is not distributed uniformly, shear stress is calculated by:

\tau_{max} = \frac{4}{3} \cdot \frac{V}{A}

Where:

\tau_{max} - Maximum allowable shear stress, measured in pascals.

V - Shear force, measured in kilonewtons.

A - Cross section area, measured in square meters.

As connection consist on five bolts, shear force is equal to a fifth of the applied load. That is:

V = \frac{P}{5}

V = \frac{450\,kN}{5}

V = 90\,kN

The minimum allowable cross section area is cleared in the shearing stress equation:

A = \frac{4}{3}\cdot \frac{V}{\tau_{max}}

If V = 90\,kN and \tau_{max} = 76.190\times 10^{3}\,kPa, the minimum allowable cross section area is:

A = \frac{4}{3} \cdot \frac{90\,kN}{76.190\times 10^{3}\,kPa}

A = 1.640\times 10^{-3}\,m^{2}

The minimum allowable cross section area can be determined in terms of minimum allowable bolt diameter by means of this expression:

A = \frac{\pi}{4}\cdot D^{2}

The diameter is now cleared and computed:

D = \sqrt{\frac{4}{\pi}\cdot A}

D =\sqrt{\frac{4}{\pi}\cdot (1.640\times 10^{-3}\,m^{2})

D = 0.0457\,m

D = 45.7\,mm

The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.

5 0
2 years ago
thier only motto and goal is to work for society and not make any profits A.small business entreprenuership B.scalable start up
sweet-ann [11.9K]

Answer:

social entrepreneurship

8 0
3 years ago
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What are some quality assurance systems
Aloiza [94]
Examples of quality assurance activities include process checklists, process standards, process documentation and project audit. Examples of quality control activities include inspection, deliverable peer reviews and the software testing process. You may like to read more about the quality assurance vs quality control.
7 0
2 years ago
Why was the lack of preparedness of the Federal Emergency Management Agency in the Hurricane Katrina disaster so damaging for th
kolbaska11 [484]

Answer:

B

Explanation:

The Bush administration had prided itself on its unique focus for homeland security and combating terrorism, without adequate preparation for the disaster.

8 0
3 years ago
Write the chemical equation of the polymerization reaction to synthesize PS
marin [14]

Explanation:

Styrene is  a vinyl monomer in which there is a carbon carbon double bond.

The polymerization of the styrene, which is initiated by using a free radical which reacts with the styrene and the compound thus forms react again and again to form polystyrene (PS).

The equation is shown below as:

\begin{matrix}& C_6H_5 \\&|\\n H_2C & =CH\end{matrix}        ⇒                \begin{matrix}&C_6H_5 \\&|\\ -[-H_2C & -CH-]-_n\end{matrix}

6 0
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