On the 74ls74 d flip-flop, the pr (preset) and clr (clear) inputs have a circle. what do these symbols mean?

Think for a moment about the potential problems with big data that the speaker mentioned:

Natalka [10]

Answer: 1. sadly yes, some people are treated unfairly for crimes people yet have commited. 2. no 3. yes

Explanation:

i did this last year

5 0

3 years ago

Read 2 more answers

Which term defines the amount of mechanical work an engine can do per unit of heat energy it uses?

skad [1K]

Answer:

d

Explanation:

is the because that's the amount of work in making machine can do producing heat

7 0

3 years ago

In the final stages of production, a pharmaceutical is sterilized by heating it from 25 to 75°C as it moves at 0.2 m/s through a

stepan [7]

Answer:

The required heat flux = 12682.268 W/m²

Explanation:

From the given information:

The initial = 25°C

The final = 75°C

The volume of the fluid = 0.2 m/s

The diameter of the steel tube = 12.7 mm = 0.0127 m

The fluid properties for density $\rho$ = 1000 kg/m³

The mass flow rate of the fluid can be calculated as:

$m = pAV$

$m = \rho \dfrac{\pi}{4}D^2V$

$m = 1000 \times \dfrac{\pi}{4} \times ( 0.0127)^2 \times 0.2$

$m = 0.0253 \ kg/s$

To estimate the amount of the heat by using the expression:

$q = mc_p(T_{final}-T_{initial})$

q = 0.0253 × 4000(75-25)

q = 101.2 (50)

q = 5060 W

Finally, the required heat of the flux is determined by using the formula:

$q" = \dfrac{q}{A_s}$

$q" = \dfrac{q}{\pi D L}$

$q" = \dfrac{5060}{\pi \times 0.0127 \times 10}$

q" = 12682.268 W/m²

The required heat flux = 12682.268 W/m²

3 0

3 years ago

A thin rim with a mean diameter of 1.2 m cross-section of 15 mm x 200 mm is subjected to an internal pressure of 10 MPa and rota

Soloha48 [4]

Answer:

The net centrifugal force over the rim is 30000N, the radial stress is 397887 Pa and the total change in diameter is 4.98 mm.

Explanation:

Lets first calculate the force in the rim due to the centrifugal force. For doing this we may assume that the centrifugal force is constant along with thick because of the thin thick.

Fc = m.ω^2/R

Where m is the mass, w the angular speed and R the mean radius. The mass is computing by the rim density and its volume:

m=p.V

m=p*(A*R)

Where A is the cross-sectional area in meters:

m=((0.015m*0.200m)*0.6m)*(7800 kg/m^3)=28.08 kg

The angular speed in rad/s is:

ω = 800r/m . 1m/60s = 133.33 r/s

Thus the centrifugal force is:

Fc = (28.08 kg)*(133.33 rad/s)^2*(0.6m) = 299505N = 30000N

Note that the calculating value is the net contribution to the whole rim but the centrifugal force is distributed along the rim's external area:

fc = Fc / (2π .R .b)

Where b is rim's with equal to 200mm :

fc = 300000 N / (2π*0.6m*0.2m) = 397887 N/m^2

The centrifugal force can be taken as internal pressure:

Pfc = 397887 N/m^2 = 3978787 Pa

As both pressures act expanding the rim it can be summed:

Pt=Pi+Pfc

Pt = 10MPa+397887Pa= 10000000Pa+397887Pa= 10397887Pa

Then for a thinner thick the stress is calculated by:

Pt*d =2σr*t

Take into account that the stress σr is over the radial direction. Then solving for o and by replacing the total pressure:

σr = Pt.d/(2*t)

σr = 10397887 Pa / (2*0.015m*0.2m) = 415915480Pa = 415MPa

We know that the radial specific deformation ε is:

σr = E / εr

εr = σr / E

For a young modulus of 200GPa:

εr = 415MPa / 200GPa

εr = 415MPa / 200000MPa=0.002075

By definition the specific deformation is written in terms of the total change in the radius:

εr = Δr / R

Δr = R / εr =0.002075 * 1.2 m = 0.00249m

As we need the change in diameter:

Δd = 2Δr =0.00498m= 4.98mm [/tex]

5 0

3 years ago

Steam enters a turbine operating at steady state with a mass flow of 10 kg/min, a specific enthalpy of 3100 kJ/kg, and a velocit

Natali [406]

Answer:

$\dot W_{out} = 133.327\,kW$

Explanation:

The model for the turbine can be derived by means of the First Law of Thermodynamics:

$-\dot Q_{out}-\dot W_{out} +\dot m \cdot \left[(h_{in}-h_{out})+\frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2}) + g\cdot (z_{in}-z_{out})\right] =0$

The work produced by the turbine is:

$\dot W_{out}=-\dot Q_{out} +\dot m \cdot \left[(h_{in}-h_{out})+\frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2}) + g\cdot (z_{in}-z_{out})\right]$

The mass flow and heat transfer rates are, respectively:

$\dot m = (10\frac{kg}{min})\cdot (\frac{1\,min}{60\,s} )$

$\dot m = 0.167\,\frac{kg}{s}$

$\dot Q_{out} = (0.167\,\frac{kg}{s} )\cdot (1.1\times 10^{3}\,\frac{J}{kg} )$

$\dot Q_{out} = 183.7\,W$

Finally:

$\dot W_{out} = -183.7\,W + (0.167\,\frac{kg}{s} )\cdot \left(8\times 10^{5}\,\frac{J}{kg} -562,5\,\frac{J}{kg} +29.43\,\frac{J}{kg} \right)$

$\dot W_{out} = 133.327\,kW$

3 0

3 years ago