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3 years ago

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Calculate the reactions at 4 ends (supports) of this bookshelf. Assume that the weight of each book is approximately 1 lb. The w

idth of shelf is 10inches and its length is 40 inches. (Hint: Model the shelf as a beam with distributed load).

1 answer:

hoa [83]3 years ago

8 0

Answer:

R = 4.75 lb (↑)

Explanation:

Number of books = n = 19

Weight of each book = W = 1 lb

Length of the bookshelf = L = 40 inches

We can get the value of the distributed load as follows

q = n*W/L = 19*1 lb/ 40 inches = 0.475 lb/in

then the reactions at 4 ends (supports) of the bookshelf are

R = (q/2)/2 = 4.75 lb

We can see the bookshelf in the pic.

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The pressure in a water line is 1500 kPa. What is the line pressure in (a) lb/ft2units and (b) lbf/in2(psi) units?

n200080 [17]

Answer:

Part A:

$1500\ KPa= 31328.145 \frac{lb}{ft^2}$

Part B:

$1500 KPa=217.55656 \frac{lb}{in^2}$ (Psi)

Explanation:

Part A:

Line Pressure is 1500 KPa

We need a conversion factor which converts KPa to lb/ft^2.

$20.88543 \frac{lb}{ft^2}= 1\ KPa$

In order to convert 1500 KPa to lb/ft^2, we proceed as:

$1\ KPa=20.88543 \frac{lb}{ft^2} \\1500\ KPa= 1500 KPa*20.88543 \frac{lb}{ft^2.KPa}\\1500\ KPa= 31328.145 \frac{lb}{ft^2}$

1500 KPa is 31328.145 lb/ft^2

Part B:

We will use the same procedure we did in Part A:

1 ft= 12 in

$(1\ ft)^2=(12\ in)^2\\1 ft^2=144 in^2$

Converting $1500 KPa\ into\ \frac{lb}{in^2}$

$1500\ KPa= 1500 KPa*20.88543 \frac{lb}{ft^2.KPa} * \frac{ft^2}{144\ in^2}$

$1500 KPa=217.55656 \frac{lb}{in^2}$ (Psi)

1500 KPa is 217.55656 lb/in^2 (psi)

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3 years ago

The disk of radius 0.4 m is originally rotating at ωo=4 rad/sec. If it is subjected to a constant angular acceleration of α=5 ra

Lina20 [59]

Answer:32.4m/ $s^2$

Explanation:

Given data

$radius\left ( r\right )$ =0.4m

Intial angular velocity $\left ( \omega_0\right )$ =4rad/s

angular acceleration $\left ( \alpha\right )$ =5rad/ $s^2$

angular velocity after 1 sec

$\omega$ = $\omega_0$ + $\alpha\times\t$

$\omega$ =4+5 $\left ( 1\right )$

$\omega$ =9rad/s

Velocity of point on the outer surface of disc $\left ( v\right )$ = $\omega_0\timesr$

v= $9\times0.4$ m/s=3.6m/s

Normal component of acceleration $\left ( a_c\right )$ = $\frac{v^2}{r}$

$a_c$ = $\frac{3.6\times3.6}{0.4}$ =32.4m/ $s^2$

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4 years ago

What does product integration refer to in an advanced manufacturing setting?

creativ13 [48]

Answer:

C!!

Explanation:

Combining all manufacturing processes to provide higher efficiency and fulfilling the requeriments.

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4 years ago

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The diesel fuel cooler is being discussed. Technician A says that high-pressure fuel systems generate a lot of heat. Technician

GuDViN [60]

Answer:

Both the technicians are correct.

Explanation:

As the pressure increases the temperature will also increase in accordance with the Boyle's law hence greater amount of heat is formed.

$Pressure\propto Temperature$

When the temperature increases the intermolecular spaces increase as the molecules of the fuel gain energy and their kinetic energy increases. This is limited to temperatures below dissociation/combustion temperature of diesel .

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3 years ago

One kilogram of water fills a 0.140 m^3 rigid container at an initial pressure of 1.8 MPa. The container is then cooled to 40°C.

Pavel [41]

Answer:

T1 = 299.18 °C

P2 = 0.00738443 MPa

Explanation:

From the data, we can get two properties for the initial condition. These are pressure and specific volume.

The pressure is 1.8 MPa and the specific volume, we can get it with the mass and volume of the container, since it’s filled this is also the volume of the water in it.

$v=\frac{vol (m^{3})}{mass (kg)} = \frac{0.140 m^{3}}{1 kg} = 0.140 \frac{ m^{3}}{kg}$

When we check in the thermodynamic tables, the conditions for saturation at 1.8 MPa we found the following:

$P^{sat} = 1.8 MPa$

$T^{sat} = 207.12 C$

$v_{g} = 0.1103\frac {m^{3}}{kg}$ specific volume for the saturated vapor

$v_{l} = 0.001167 \frac{m^{3}}{kg}$ specific volume for the saturated liquid

Since the specific volume in our condition is higher that the specific volume for the saturated vapor, we have a superheated steam.

Looking in the thermodynamic tables for superheated steam we found that the temperature where the steam has a specific volume of $0.140 \frac{ m^{3}}{kg}$ at 1.8 MPa is 299.18 °C. This is the initial temperature in the container.

Since the only information that we have about the final condition is that the container was cooled. We can assume that it was cooled until a condition of saturation. So, the final pressure for the water will be the pressure of saturation for a temperature of 40°C. From thermodynamic tables we get:

$P^{sat} at 40C = 0.00738443 MPa$

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3 years ago