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4 years ago

What is the equivalent resistance for a series circuit with three resistors: 5.0 ohms, 2.0 ohms, and 12.0 ohms?

2 answers:

7 0

There are several information's already given in the question. Based on those given information's, the answer can be easily deduced.
Equivalent resistance = 5 + 2 + 12
= 19 ohms

From the above deduction, it can be deduced that the correct option among all the options that are given in the question is the first option or option "A". I hope that this is the answer that has actually come to your desired help.

frosja888 [35]4 years ago

3 0

For Plato users, this answer is correct! I just took the test. 5/5 :3

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When unpolarized light is incident on a sheet of polarizing material with a transmission axis oriented vertically, what percenta

yulyashka [42]

Answer:

If the light were incident upon two polarizers at right angles, no light would get thru - thus each polarizer must block 50% of the light.

One polarizer would allow 50% of the light to pass.

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2 years ago

How many regions does the spine have and what are them ?s

Viktor [21]

the spine is divided into four main regions: cervical, thoracic, lumbar and sacral.

Have an amazing day! <3

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3 years ago

An ideal monatomic gas at a pressure of 2.0×105N/m2 and a temperature of 300 K undergoes a quasi-static isobaric expansion from

liraira [26]

Answer:

a) $W= 400 J$

b) $T_{2}=600 K$

c) $n=0.16 mol$

d) $\Delta E_{int}=598 J$

e) $Q=998 J$

Explanation:

a) Let's recall the definition of work.

$W=\int^{Vf}_{V0}PdV$

Because the system is a quasi-static isobaric expansion, P is constant here, therefore:

$W=P\int^{Vf}_{V0}dV=P(V_{f}-V_{0})$

$W=2.0\cdot 10^{5}(4.0\cdot 10^{-3}-2.0\cdot 10^{-3})= 400 [Nm^{2}]$

b) Using the ideal gas equation we have:

$\frac{P_{1}V_{1}}{T_{1}}=nR$ (1)

and $\frac{P_{2}V_{2}}{T_{2}}=nR$ (2)

We can note that n times R is a constant in (1) and (2), so we can equal those equations.

$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$ (3)

Let's solve T₂ for (3), let's recall that P₁ = P₂, so they canceled out

$T_{2}=T_{1}\cdot V_{2}/V_{1}$

$T_{2}=300\cdot 4x10^{3}/2x10^{3}=600 K$

c) Using the equation of ideal gas we have:

$\frac{P_{1}V_{1}}{RT_{1}}=n$

$n=\frac{P_{1}V_{1}}{RT_{1}}=0.16 mol$

d) We can write the internal energy as a function of Cv, and as we know the Cv is 1.5R for a monoatomic gas.

$\Delta E_{int}=n1.5R\Delta T$

$\Delta E_{int}=0.16\cdot 1.5\cdot 8.3 \cdot (600-300)$

$\Delta E_{int}=598 J$

e) Using the first law of thermodynamic, we have:

$\Delta E_{int}=Q-W$

Finally,

$Q=\Delta E_{int}+W=598+400=998 J$

Have a nice day!

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3 years ago

Problem: At the local swimming hole, a favorite trick is torun

denis23 [38]

Answer:

2 revolutions

Explanation:

Assume that when she runs off the edge of the 8.3m high cliff, her vertical speed is 0. So gravitational acceleration g = 9.8m/s2 is the only thing that makes her fall down. So we can use the following equation of motion to calculate the time it takes for her to fall down:

$s = gt^2/2$