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murzikaleks [220]
3 years ago
8

Sam constructs a circuit, connects a lead acid battery of 2 V to a lamp of resistance 3 Ω and places an ammeter across it. What

must be the reading of the ammeter?
A.
0.66 A
B.
0.5 A
C.
0.54 A
D.
0.61 A
Engineering
2 answers:
Anon25 [30]3 years ago
8 0
It’s A because voltage equals current times resistance
frosja888 [35]3 years ago
4 0

Answer:

A. 0.66 Amps

Explanation:

Using ohms law, we can say that Voltage is equivalent to Current times Resistance.  We are given the voltage and the resistance of the circuit, so we simply need to find the current.

V = IR

Solve for I, where V = 2volts and R = 3ohms.

V = IR

V * 1/R = I * R * 1/R

I = V/R

I = 2/3 Amps

Hence, we should choose option A, 0.66 Amps for the current in this simple circuit.

Cheers.

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skin is the largest organ

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What is the composition of the two phases that form when a stream of 40% A, 39% B, and 21% C separates into two phases? Label th
irga5000 [103]

Answer:

vapor fraction = 0.4 and 0.08

Explanation:

At reasonably high temperatures, a mixture will exist in the form of a sub cooled liquid. Between these extremes, the mixture exists in a two phrase region where it is a vapor liquid equilibrium. From a vapor-liquid phase diagram, a mixture of 40% A, 39% B, and 21% C separates to give the vapor compositions of 0.4 and 0.08.

8 0
3 years ago
A gasoline engine has a piston/cylinder with 0.1 kg air at 4 MPa, 1527◦C after combustion, and this is expanded in a polytropic
Roman55 [17]

Answer:

The expansion work is 71.24 kJ and heat transfer is -16.89 kJ

Explanation:

From ideal gas law,

Initial volume (V1) = nRT/P

n is the number of moles of air in the cylinder = mass/MW = 0.1/29 = 0.00345 kgmol

R is gas constant = 8314.34 J/kgmol.K

T is initial temperature = 1527 °C = 1527+273 = 1800 K

P is initial pressure = 4 MPa = 4×10^6 Pa

V1 = 0.00345×8314.34×1800/(4×10^6) = 0.013 m^3

V2 = 10×V1 = 10×0.013 = 0.13 m^3

The process is a polytropic expansion process

polytropic exponent (n) = 1.5

P2 = P1(V1/V2)^n = 4×10^6(0.013/0.13)^1.5 = 1.26×10^5 Pa

Expansion work = (P1V1 - P2V2) ÷ (n - 1) = (4×10^6 × 0.013 - 1.26×10^5 × 0.13) ÷ (1.5 - 1) = 35620 ÷ 0.5 = 71240 J = 71240/1000 = 71.24 kJ

Heat transfer = change in internal energy + expansion work

change in internal energy (∆U) = Cv(T2 - T1)

T2 = PV/nR = 1.26×10^5 × 0.13/0.00345×8314.34 = 571 K

Cv = 20.785 kJ/kgmol.K

∆U = 20.785(571 - 1800) = -25544.765 kJ/kgmol × 0.00345 kgmol = -88.13 kJ

Heat transfer = -88.13 + 71.24 = -16.89 kJ

5 0
3 years ago
Air within a piston cylinder assembly executes a Carnot refrigeration cycle between hot and cold reservoirs at TH=600 K and TC=3
Nataly [62]

Answer:

See explaination

Explanation:

for a reverse carnot cycle T-S diagram is a rectangle which i have shown

net work for a complete cycle must be equal to net heat interaction.

Kindly check attachment for the step by step solution of the given problem.

5 0
3 years ago
A large heat pump should upgrade 5 MW of heat at 85°C to be delivered as heat at 150°C. Suppose the actual heat pump has a COP o
AysviL [449]

Answer:

W=2 MW

Explanation:

Given that

COP= 2.5

Heat extracted from 85°C  

Qa= 5 MW

Lets heat supplied at 150°C   = Qr

The power input to heat pump = W

From first law of thermodynamics

Qr= Qa+ W

We know that COP of heat pump given as

COP=\dfrac{Qr}{W}

2.5=\dfrac{5}{W}

2.5=\dfrac{5}{W}

W=2 MW

For Carnot heat pump

COP=\dfrac{T_2}{T_2-T_1}

2.5=\dfrac{T_2}{T_2-(273+85)}

2.5 T₂ -  895= T₂

T₂=596.66 K

T₂=323.6 °C

7 0
3 years ago
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