All categories

3 years ago

The sum ofall microscopic forms of energy of a system is quantified as flow energy. a)True b) False

Engineering

1 answer:

Sliva [168]3 years ago

3 0

Answer: b) False

Explanation: Microscopic energy is the the energy that is based on the molecular level in a particular energy system. Microscopic energy basically comprise with tiny particles like atoms and molecules .The sum of all microscopic form of energy e together make the internal energy .Therefore, the statement given is false because the sum of all the microscopic forms of energy of a system is quantified as internal energy not flow energy.

You might be interested in

I’m doing a project on renewable energy. There are 6 energy sources. Solar, wind, geothermal, hydroelectric, tidal, and biomass.

nalin [4]

Answer:

"Biofuels"

Explanation:

I don't know if this counts but I guess it's not one of those.

6 0

2 years ago

Read 2 more answers

Following are several z-transforms. For each one, determine inverse z-transform using both the method based on the partial-fract

tigry1 [53]

Answer:

For now the answer to this question is only for partial fraction. Find attached.

6 0

3 years ago

Exhaust gas from a furnace is used to preheat the combustion air supplied to the furnace burners. The gas, which has a flow rate

Monica [59]

Answer:

The total tube surface area in m² required to achieve an air outlet temperature of 850 K is 192.3 m²

Explanation:

Here we have the heat Q given as follows;

Q = 15 × 1075 × (1100 - $t_{A2}$ ) = 10 × 1075 × (850 - 300) = 5912500 J

∴ 1100 - $t_{A2}$ = 1100/3

$t_{A2}$ = 733.33 K

$\Delta \bar{t}_{a} =\frac{t_{A_{1}}+t_{A_{2}}}{2} - \frac{t_{B_{1}}+t_{B_{2}}}{2}$

Where

$\Delta \bar{t}_{a}$ = Arithmetic mean temperature difference

$t_{A_{1}$ = Inlet temperature of the gas = 1100 K

$t_{A_{2}$ = Outlet temperature of the gas = 733.33 K

$t_{B_{1}$ = Inlet temperature of the air = 300 K

$t_{B_{2}$ = Outlet temperature of the air = 850 K

Hence, plugging in the values, we have;

$\Delta \bar{t}_{a} =\frac{1100+733.33}{2} - \frac{300+850}{2} = 341\tfrac{2}{3} \, K = 341.67 \, K$

Hence, from;

$\dot{Q} = UA\Delta \bar{t}_{a}$ , we have

5912500 = 90 × A × 341.67

$A = \frac{5912500 }{90 \times 341.67} = 192.3 \, m^2$

Hence, the total tube surface area in m² required to achieve an air outlet temperature of 850 K = 192.3 m².

4 0

3 years ago

In the LC-3 data path, the output of the address adder goes to both the MARMUX and the PCMUX, potentially causing two very diffe

dangina [55]

Answer:

no need for that

Explanation:

they are not the same at all

3 0

3 years ago

A loss in value caused by an undesirable or hazardous influence offsite is which type of depreciation?

Lubov Fominskaja [6]

External depreciation may be defined as a loss in value caused by an undesirable or hazardous influence offsite.

<h3>What is depreciation?</h3>

Depreciation may be defined as a situation when the financial value of an acquisition declines over time due to exploitation, fray, and incision, or obsolescence.

External depreciation may also be referred to as "economic obsolescence". It causes a negative influence on the financial value gradually.

Therefore, it is well described above.

To learn more about Depreciation, refer to the link:

brainly.com/question/1203926

#SPJ1

5 0

2 years ago