Buying shop supplies from the shop owner to work on your own car at home is an ethical practice

A circular ceramic plate that can be modeled as a blackbody is being heated by an electrical heater. The plate is 30 cm in diame

denis23 [38]

Answer:

Q = 125.538 W

Explanation:

Given data:

D = 30 cm

Temperature $T_\infity = 15$ degree celcius

$T_S = 220 + 273 = 473 K$

Heat coefficient = 12 W/m^2 K

Efficiency 80% = 0.8

$Q = hA(T_S - T_{\infty}) \eta$

$= 12(\frac{\pi}{4} 0.3^2) (473 - 288) 0.8$

Q = 125.538 W

5 0

2 years ago

Consider a system with two tasks, Task1 and Task2. Task1 has a period of 200 ms, and Task2 has a period of 300 ms. All tasks ini

Murrr4er [49]

<u>Explanation:</u>

Task 1 time period = 200ms, Task 2 time period = 300ms

Task ticked = $\frac{1000ms}{200ms}= 5$ → 5 times

Task 2 ticked = $\frac{1000ms}{300ms} = 3.33$ → 3 times

At 600 ms → 200ms 200ms 200ms

300ms → $\frac{30ms}{60ms}$

Largest time period = H.C.M of (200ms, 300ms)

= 600ms

4 0

2 years ago

Water at 15°C is to be discharged from a reservoir at a rate of 18 L/s using two horizontal cast iron pipes connected in series

love history [14]

Answer:

The required pumping head is 1344.55 m and the pumping power is 236.96 kW

Explanation:

The energy equation is equal to:

$\frac{P_{1} }{\gamma } +\frac{V_{1}^{2} }{2g} +z_{1} =\frac{P_{2} }{\gamma } +\frac{V_{2}^{2} }{2g} +z_{2}+h_{i} -h_{pump} , if V_{1} =0,z_{2} =0\\h_{pump} =\frac{V_{2}^{2}}{2} +h_{i}-z_{1}$

For the pipe 1, the flow velocity is:

$V_{1} =\frac{Q}{\frac{\pi D^{2} }{4} }$

Q = 18 L/s = 0.018 m³/s

D = 6 cm = 0.06 m

$V_{1} =\frac{0.018}{\frac{\pi *0.06^{2} }{4} } =6.366m/s$

The Reynold´s number is:

$Re=\frac{\rho *V*D}{u} =\frac{999.1*6.366*0.06}{1.138x10^{-3} } =335339.4$

$\frac{\epsilon }{D} =\frac{0.00026}{0.06} =0.0043$

Using the graph of Moody, I will select the f value at 0.0043 and 335339.4, as 0.02941

The head of pipe 1 is:

$h_{1} =\frac{V_{1}^{2} }{2g} (k_{L}+\frac{fL}{D} )=\frac{6.366^{2} }{2*9.8} *(0.5+\frac{0.0294*20}{0.06} )=21.3m$

For the pipe 2, the flow velocity is:

$V_{2} =\frac{0.018}{\frac{\pi *0.03^{2} }{4} } =25.46m/s$

The Reynold´s number is:

$Re=\frac{\rho *V*D}{u} =\frac{999.1*25.46*0.03}{1.138x10^{-3} } =670573.4$

$\frac{\epsilon }{D} =\frac{0.00026}{0.03} =0.0087$

The head of pipe 1 is:

$h_{2} =\frac{V_{2}^{2} }{2g} (k_{L}+\frac{fL}{D} )=\frac{25.46^{2} }{2*9.8} *(0.5+\frac{0.033*36}{0.03} )=1326.18m$

The total head is:

hi = 1326.18 + 21.3 = 1347.48 m

The required pump head is:

$h_{pump} =\frac{25.46^{2} }{2*9.8} +1347.48-36=1344.55m$

The required pumping power is:

$P=Q\rho *g*h_{pump} =0.018*999.1*9.8*1344.55=236965.16W=236.96kW$

8 0

2 years ago

the increase of current when 15 V is applied to 10000ohm rheostat which is adjusted to 1000ohm value

Anastasy [175]

Given data:
•) applied voltage = 15 V
•). Resistance = 1000 ohm

Required:
•). The magnitude of current= ?

•••••••••••••SOLUTION•••••••••••••

We can find the relation ship between current, voltage and resistance with the help of Ohms law.

According to ohms law;

V= IR.

Rearranging the above equation;

I= V/ R

Putt the values in the above equation; we get

I= 15V/ 1000ohm

I = 0.015 A( ampere)

••••••••••••••• CONCLUSION•••••••

The value of the current would be 0.15 ampere when Resistance is equal to 1000 and that of Voltage is equal to 15 V.

4 0

2 years ago

Multiply. Write the answer in simplest form. 1 3/10×1/8

kicyunya [14]

9514 1404 393

Answer:

13/80

Explanation:

The product is ...

(1 3/10)×(1/8) = (13/10)×(1/8) = (13×1)/(10×8) = 13/80

4 0

1 year ago