Which of the following is an activity of daily living? jogging cleaning weightlifting all of the above

What is CAD/CAM (Computer-Aided Designing and Manufacturing) used for in school?<br> Plz help

joja [24]

It used for designing homes and drafting products before production

8 0

3 years ago

Steam enters a turbine operating at steady state at 2 MPa, 323 °C with a velocity of 65 m/s. Saturated vapor exits at 0.1 MPa an

Lera25 [3.4K]

Answer:

$\dot Q_{out} = 13369.104\,kW$

Explanation:

The turbine is modelled after the First Law of Thermodynamics:

$-\dot Q_{out} - \dot W_{out} + \dot H_{in} - \dot H_{out} + \dot K_{in} - \dot K_{out} + \dot U_{in} - \dot U_{out} = 0$

The rate of heat transfer between the turbine and its surroundings is:

$\dot Q_{out} = \dot H_{in}-\dot H_{out} + \dot K_{in} - \dot K_{out} - \dot W_{out} + \dot U_{in} - \dot U_{out}$

The specific enthalpies at inlet and outlet are, respectively:

$h_{in} = 3076.41\,\frac{kJ}{kg}$

$h_{out} = 2675.0\,\frac{kJ}{kg}$

The required output is:

$\dot Q_{out} = \left(8\,\frac{kg}{s} \right)\cdot \left\{3076.41\,\frac{kJ}{kg}-2675.0\,\frac{kJ}{kg}+\frac{1}{2}\cdot \left[\left(65\,\frac{m}{s} \right)^{2}-\left(42\,\frac{m}{s} \right)^{2}\right] + \left(9.807\,\frac{m}{s^{2}} \right)\cdot (4\,m) \right\} - 8000\,kW$ $\dot Q_{out} = 13369.104\,kW$

4 0

3 years ago

A sphere is assumed to have the properties of water and has an initial heat generation 46480 W/m^3 How much should the heat gene

Verdich [7]

Answer:

Resulting heat generation, Q = 77.638 kcal/h

Given:

Initial heat generation of the sphere, $Q_{Gi} = 46480 W/m^{3}$

Maximum temperature, $T_{m} = 360 K$

Radius of the sphere, r = 0.1 m

Ambient air temperature, $T = 25^{\circ}C$ = 298 K

Solution:

Now, maximum heat generation, $Q_{m}$ is given by:

$T_{m} = \frac{Q_{m}r^{2}}{6K} + T$ (1)

where

K = Thermal conductivity of water at $T_{m} = 360 K = 0.67 W/m^{\circ}C$

Now, using eqn (1):

$360 = \frac{Q_{m}\times 0.1^{2}}{6\times 0.67} + 298$

$Q_{m} = 24924 W/m^{3}$

max. heat generation at maintained max. temperature of 360 K is 24924 $W/m^{3}$

For excess heat generation, Q:

$Q = (Q_{Gi} - Q_{m})\times volume of sphere, V$

where

$V = \frac{4}{3}\pi r^{3}$

$Q = (46480 - 24924)\times \frac{4}{3}\pi\0.1^{3} = 21556\times \frac{4}{3}\pi\0.1^{3} W/m^{3}$

$Q = 90.294 W$

Now, 1 kcal/h = 1.163 W

Therefore,

$Q = \frac{90.294}{1.163} = 77.638 kcal/h$

3 0

3 years ago

Suppose the measured background level is 5.1 mV. A signal of 20.7 mV is measured at a distance of 29 mm and 15.8 mV is measured

Paladinen [302]

Answer: The normalized corrected value at 32.5 mm = 0.69 mV

Explanation:

Signal value V1 = 20.7 mV at

distance value = 29 mm and

V2 = 15.8 mV is measured at 32.5 mm.

It is necessary, therefore, to subtract off this background level from the data to obtain a valid measurement

V1 = 20.7 - 5.1 = 15.6 mV

V2 = 15.8 - 5.1 = 10.7mV

The normalized corrected value at 32.5 mm will be

Vn = V2/V1 since V1 is the maximum value

Vn = 10.7/15.6 = 0.69 mV

3 0

3 years ago

An inductor is connected to a voltage source and it is found that it has a time constant, t. When a 10-ohm resistor is placed in

prisoha [69]

Answer:

A) 30 mH

B ) 10-ohm

Explanation:

resistor = 10-ohm

Inductor = 30mH ( l )

L = inductance

R = resistance

r = internal resistance

values of the original Inductors

Note : inductor = constant time (t) case 1

inductor + 10-ohm resistor connected in series = constant time ( t/2) case2

inductor + 10-ohm resistor + 30 mH inductance in series = constant time (t) case3

<em>From the above cases</em>

case 1 = time constant ( t ) = L / R

case 2 = Req = R + r hence time constant t / 2 = L / R + r therefore

t = $\frac{2L}{R+r}$

case 3 = Leq = L + l , Req = R + r . constant time ( t )

hence Z = $\frac{L + l}{R + r}$ = t

A) Inductance

To calculate inductance equate case 1 to case 3

$\frac{L}{R}$ = $\frac{L + l}{R + r}$ = L / 10 = (L + 30) / ( 10 + 10 )

= 20 L = 10 L + 300 mH

10L = 300 m H

therefore L = 30 mH

B ) The internal resistance

equate case 1 to case 2

$\frac{L}{R}$ = $\frac{2L}{R + r}$

R + r = 2 R therefore ( r = R ) therefore internal resistance = 10-ohm

6 0

3 years ago