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romanna [79]
3 years ago
9

What is the normal balance side of an asset?

Engineering
1 answer:
Mademuasel [1]3 years ago
8 0
The normal balance for asset and expense accounts is the debit side, while for income, equity, and liability accounts it is the credit side. An account's assigned normal balance is on the side where increases go because the increases in any account are usually greater than the decreases
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Determine the number of flipflops required to build a binary counter that count from 0 to 2043
Pie

Answer:

10 flip -flops are required to build a binary counter circuit to count to from 0 to 1023 .

Explanation:

3 0
3 years ago
A square loop of wire surrounds a solenoid. The side of the square is 0.1 m, while the radius of the solenoid is 0.025 m. The sq
Semmy [17]

Answer:

I=9.6×e^{-8}  A

Explanation:

The magnetic field inside the solenoid.

B=I*500*muy0/0.3=2.1×e ^-3×I.

so the total flux go through the square loop.

B×π×r^2=I×2.1×e^-3π×0.025^2

=4.11×e^-6×I

we have that

(flux)'= -U

so differentiating flux we get

so the inducted emf in the loop.

U=4.11×e^{-6}×dI/dt=4.11×e^-6×0.7=2.9×e^-6 (V)

so, I=2.9×e^{-6}÷30

I=9.6×e^{-8}  A

6 0
3 years ago
A binary geothermal power plant uses geothermal water at 160°C as the heat source. The cycle operates on the simple Rankine cycl
bogdanovich [222]

A binary geothermal power operates on the simple Rankine cycle with isobutane as the working fluid. The isentropic efficiency of the turbine, the net power output, and the thermal efficiency of the cycle are to be determined

Assumptions :

1.  Steady operating conditions exist.

2.  Kinetic and potential energy changes are negligible.

Properties:  The specific heat of geothermal water ( c_{geo}[) is taken to be 4.18 kJ/kg.ºC.  

Analysis (a) We need properties of isobutane, we can obtain the properties from EES.

a. Turbine

PP_{3} = 3.25mPa = (3.25*1000) kPa\\= 3250kPa\\from the EES TABLE\\h_{3} = 761.54 kJ/kg\\s_{3} = 2.5457 kJ/kg\\P_{4} = 410kPa\\\\s_{4} = s_{3} \\h_{4s} = 470.40kJ/kg\\\\T_{4} = 179.5^{0} C\\\\h_{4} = 689.74 kJ/KG\\\\ The  isentropic  efficiency, n_{T} = \frac{h_{3}-h_{4}  }{h_{3}- h_{4s} }

==\frac{761.54-689.74}{761.54-670.40} \\=\frac{71.8}{91.14} \\=0.788

b. Pump

h_{1} = h_{f} @ 410kPa = 273.01kJ/kg\\v_{1} = v_{f} @ 410kPa = 0.001842 m^{3}/kgw_{p,in} =  \frac{v_{1}(P_{2}-P_{1})   }{n_{p} } \\\\= \frac{0.01842(3250-410)}{0.9} \\\\ =5.81kJ/kg\\h_{2} =h_{1} + w_{p,in}\\          = 273.01+5.81\\           = 278.82 kJ/kg\\\\w_{T,out} = m^{.}  (h_{3} -h_{4} )\\=(305.6)(761.54-689.74)\\=305.6(71.8)\\=21,942kW\\\\

W^{.} _ {P,in} = m^{.} (h_{2} -h_{1}) \\=m^{.}  w_{p,in \\=305.6(5.81)\\\\=1,777kW\\W^{.}  _{net} = W^{.} _{T, out} - W^{.}  _{P,in} \\= 21,942-1,777\\=20,166 kW\\\\HEAT EXCHANGER\\\\Q_{in} = m^{.} _{geo} c_{geo} (T_{in-T_{out} } )\\=555.9(4.18)(160-90)\\=162.656kW\\

c. The thermal efficiency of the cycle  n_{th}  =\frac{W^{.} _{net} }{Q^{._{in} } } \\\\= \frac{20,166}{162,656} \\=0.124\\=12.4%

7 0
3 years ago
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Imagine you compare the effectiveness of four different types of stimulant to keep you awake while revising statistics using a o
Arlecino [84]

Answer:

The correct answer is At least two of the stimulants will have different effects on the mean time spent awake.

Explanation:

The null hypothesis exposed leads to consider an alternative hypothesis that differs from the behavior of the 4 stimulants. In this case it is related to the effects on patients, since in practice at least one will present differences compared to the others, influencing the sleep time of consumers.

4 0
3 years ago
PLEASE HELP!! Its easy!!!
Rina8888 [55]

Answer:

C is tire

F is cassette

D is hub

4 0
3 years ago
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