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xz_007 [3.2K]
3 years ago
13

The small washer is sliding down the cord OA. When it is at the midpoint, its speed is 28 m/s and its acceleration is 7 m/s 2 .

Express the velocity and acceleration of the washer at this point in terms of its cylindrical components.

Engineering
1 answer:
Neporo4naja [7]3 years ago
3 0

Answer:

Velocity components

V_r = -16.28 m/s

V_z = -22.8 m/s

V_q = 0 m/s

For Acceleration components;

a_r = -4.07m/s^2

a_z = -5.70m/s^2

a_q = 0m/s^2

Explanation:

We are given:

Speed v_o = 28 m/s

Acceleration a_o= 7 m/s^2

We first need to find the radial position r of washer in x-y plane.

Therefore

r = \sqrt{300^2 + 400^2}

r = 500 mm

To find length along direction OA we have:

L = \sqrt{500^2 + 700^2}L = 860 mm

Therefore, the radial and vertical components of velocity will be given as:

V_r = V_o*cos(Q)

V_z = V_o*sin(Q)

Where Q is the angle between OA and vector r.

Therefore,

V_r = 28 * \frac{r}{L} = > 28 * \frac{500}{860}

V_r = -16.28 m/s

• V_z = 28 * \frac{700}{860} = -22.8

• V_q = 0 m/s

The radial and vertical components of acceleration will be:

a_r = a_o*cos(Q)

a_z = a_o*sin(Q)

Therefore we have:

• a_r = 7* \frac{500}{860} = -4.07m/s^2

• a_z = 7 * \frac{700}{860} = -5.70 m/s^2

• a_q = 0 m/s^2

Note : image is missing, so I attached it

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Answer:

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Explanation:

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The density of water is = 997 kg/m³

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since we are given the mass, therefore, the  volume will be mass/density

25/997 = 0.0251 m^3/s

volumetric flow rate = 0.0251 m^3/s

Average velocity calculations:

<em>Pipe section A:</em>

cross-sectional area =

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mass flow rate = density X cross-sectional area X velocity

velocity = mass flow rate /(density X cross-sectional area)

velocity = 25/(997 \times 3.85\times10^{-3}) = 6.513m/s

<em>Pipe section B:</em>

cross-sectional area =

\pi \times d^2\\=\pi \times 0.05^2= 1.96\times10^{-3}m^2

mass flow rate = density X cross-sectional area X velocity

velocity = mass flow rate /(density X cross-sectional area)

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7 0
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A slab-milling operation is performed on a 0.7 m long, 30 mm-wide cast-iron block with a feed of 0.25 mm/tooth and depth of cut
denis23 [38]

Answer:

a)  T_m=1.787min

b)  MRR=35259.7mm^3/min

Explanation:

From the question we are told that:

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Width w=30mm

FeedF=0.25mm/tooth

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Generally the equation for Approach is mathematically given by

x=\sqrt{Dd-d^2}

X=\sqrt{75*3-3^2}

X=14.69mm

Therefore

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L_e=700+14.69

L_e=714.69mm

a)

Generally the equation for Machine Time is mathematically given by

T_m=\frac{L_e}{F_m}

Where

F_m=F*n*N

F_m=0.25*8*200

F_m=400

Therefore

T_m=\frac{714.69}{400}

T_m=1.787min

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Can a 1½ " conduit, with a total area of 2.04 square inches, be filled with wires that total 0.93 square inches if the maximum f
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Answer:

it is not possible to place the wires in the condui

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to find out

Can it is possible place wire in conduit conduit

solution

we know maximum fill is 40%

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<u>Explanation:</u>

<u />

Given-

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Vacancies/cm³ = 1 / 200 X (3.5089 X 10⁻⁸cm)³

Vacancies/cm³ = 1.157 X 10²⁰

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3 years ago
A water-filled manometer is used to measure the pressure in an air-filled tank. One leg of the manometer is open to atmosphere.
ddd [48]

Answer:

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Explanation:

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P = 150.335\,kPa

4 0
3 years ago
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