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Marta_Voda [28]

3 years ago

8

Express the angular speed of the earth in rad/s and rpm.

1 answer:

spin [16.1K]3 years ago

7 0

Answer:

we might say the earth rotates at 7.272×10-5 rad/s and this tells us in its angular speed

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Sedimentary rock forms at or near the

Natalija [7]

I believe it’s A bit I. could be wrong

5 0

3 years ago

Read 2 more answers

Electric Resistance Heating. A house that is losing heat at a rate of 50,000 kJ/h when the outside temperature drops to 4 0C is

ryzh [129]

Answer:

a) $\dot W = 0.978\,kW$ , b) $I = \left(50000\,\frac{kJ}{h} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s}\right)\cdot \left(\frac{1}{COP_{real}} \right) - 0.978\,kW$

Explanation:

a) The ideal Coefficient of Performance for the heat pump is:

$COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}$

$COP_{HP} = \frac{298.15\,K}{298.15\,K - 277.15\,K}$

$COP_{HP} = 14.198$

The reversible work input is:

$\dot W = \frac{\dot Q_{H}}{COP_{HP}}$

$\dot W = \left(\frac{50000\,\frac{kJ}{h} }{14.198} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s} \right)$

$\dot W = 0.978\,kW$

b) The irreversibility is given by the difference between real work and ideal work inputs:

$I = \dot W_{real} - \dot W_{ideal}$

$I = \left(50000\,\frac{kJ}{h} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s}\right)\cdot \left(\frac{1}{COP_{real}} \right) - 0.978\,kW$

7 0

4 years ago

An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500

Assoli18 [71]

Answer: The exit temperature of the gas in deg C is $32^{o}C$ .

Explanation:

The given data is as follows.

$C_{p}$ = 1000 J/kg K, R = 500 J/kg K = 0.5 kJ/kg K (as 1 kJ = 1000 J)

$P_{1}$ = 100 kPa, $V_{1} = 15 m^{3}/s$

$T_{1} = 27^{o}C = (27 + 273) K = 300 K$

We know that for an ideal gas the mass flow rate will be calculated as follows.

$P_{1}V_{1} = mRT_{1}$

or, m = $\frac{P_{1}V_{1}}{RT_{1}}$

= $\frac{100 \times 15}{0.5 \times 300}$

= 10 kg/s

Now, according to the steady flow energy equation:

$mh_{1} + Q = mh_{2} + W$

$h_{1} + \frac{Q}{m} = h_{2} + \frac{W}{m}$

$C_{p}T_{1} - \frac{80}{10} = C_{p}T_{2} - \frac{130}{10}$

$(T_{2} - T_{1})C_{p} = \frac{130 - 80}{10}$

$(T_{2} - T_{1})$ = 5 K

$T_{2}$ = 5 K + 300 K

$T_{2}$ = 305 K

= (305 K - 273 K)

= $32^{o}C$

Therefore, we can conclude that the exit temperature of the gas in deg C is $32^{o}C$ .

8 0

4 years ago

An agricultural manager requires

sp2606 [1]

C, being able to maintain legal information on grant programs

3 0

3 years ago

A company specification calls for a steel component to have a minimum tensile strength of 1240 MPa. Tension tests are conducted

polet [3.4K]

Answer:

HV = 372.4 KGf/mm^2

$HR_C = 39.445$

Explanation:

given data:

minimum tensile strength = 1240 MPa

we knwo that $\sigma_{vrs}$ and brinell hardness number (HB} relation given as

$\sigma_{vrs} = 3.4 \times H.B$

$H.B = \frac{1240}{3.4} = 364.7$

1) relation betwen H.B and Rockwell is

$HR_C = 6.96 + 0.089 HB$

$HR_C = 6.96 + (0.089\times 365} = 39.445$

2) relation between HB and Vickers hardnes HV is given as

$HB \times 1.05 = HV + 10.5$

$HV = HB\times 1.05 - 10.5$

HV = 372.4 KGf/mm^2

4 0

4 years ago