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iVinArrow [24]
2 years ago
12

A hot piece of iron is thrown into the ocean and its temperature eventually stabilizes. Which of the following statements concer

ning this process is correct? (There may be more than one correct choice.)
A. The entropy gained by the iron is equal to the entropy lost by the ocean.
B. The ocean gains less entropy than the iron loses.
C. The change in the entropy of the iron-ocean system is zero.
D. The entropy lost by the iron is equal to the entropy gained by the ocean.
E. The ocean gains more entropy than the iron loses.
Physics
1 answer:
777dan777 [17]2 years ago
8 0

Answer:

E. The ocean gains more entropy than the iron loses.

Explanation:

When there is a spontaneous process , entropy of the system increases . Here hot iron is losing entropy and ocean is gaining entropy . Net effect will be gain of entropy . That means entropy gained by ocean is more than entropy lost by iron .

Hence option E is correct .

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A spring with a spring constant of 2500 n/m. is stretched 4.00 cm. what is the work required to stretch the spring?
Yuri [45]
W = 1/2k*x^2.

k = spring constant = 2500 n/m.
x = distance = 4 cm = 0.04m (convert to same units).

W = 1/2(2500)(0.04)^2 = 2J.
5 0
2 years ago
Read 2 more answers
Charge q1 is distance r from a positive point charge q. charge q2=q1/3 is distance 2r from q. what is the ratio u1/u2 of their p
makvit [3.9K]
We need the power law for the change in potential energy (due to the Coulomb force) in bringing a charge q from infinity to distance r from charge Q. We are only interested in the ratio U₁/U₂, so I'm not going to bother with constants (like the permittivity of space). 

<span>The potential energy of charge q is proportional to </span>
<span>∫[s=r to ∞] qQs⁻²ds = -qQs⁻¹|[s=r to ∞] = qQr⁻¹, </span>

<span>so if r₂ = 3r₁ and q₂ = q₁/4, then </span>
<span>U₁/U₂ = q₁Qr₂/(r₁q₂Q) = (q₁/q₂)(r₂/r₁) </span>
<span>= 4•3 = 12.</span>
5 0
2 years ago
Two parallel co-axial disks are floating in deep space (far from sun and planets). Each disk is 1 meter in diameter and the disk
HACTEHA [7]

Answer:

T₂ = 5646 K

Explanation:

Let's start by finding the power received by the first disc, for this we use Stefan's law

          P = σ. A e T⁴

Where next is the Stefam-Bolztmann constant with value 5,670 10-8 W / m² K⁴, A is the area of ​​the disk, T the absolute temperature and e the emissivity that for a black body is  1

The intensity is defined as the amount of radiation that arrives per unit area. For this we assume that the radiation expands uniformly in all directions, the intensity is

           I = P / A

Writing this expression for both discs

          I₁ A₁ = I₂ A₂

          I₂ = I₁ A₁ / A₂

The area of ​​a sphere is

          A = 4π r²

           I₂ = I₁ (r₁ / r₂)²

          r₂ = r₁ ± 5

          I₁ = I₂ ( (r₁ ± 5)/r₁)²

.

        Let's write the Stefan equation

         P / A = σ e T⁴

          I = σ e T⁴

This is the intensity that affects the disk, substitute in the intensity equation

         σ e₁ T₁⁴ = σ e₂ T₂⁴ (r₂ / r₁)²

The first disc indicates that it is a black body whereby e₁ = 1, the second disc, as it is painted white, the emissivity is less than 1, the emissivity values ​​of the white paint change between 0.90 and 0.95, for this calculation let's use 0.90 matt white

        e₁ T₁⁴ = T₂⁴   (r1 + 5)²/r₁²

       T₁ = T₂  {(e₂/e₁)}^{1/4}  √(1 ± 1/ r₁)  

If we assume that r₁ is large, which is possible since the disks are in deep space, we can expand the last term

           (1 ±x) n = 1 ± n x

Where x = 5 / r₁ << 1

We replace

          T₁ = T₂ {(e₂/e₁)}^{1/4}  (1 ± ½   5/r1)

           T₁ = T₂ {(e₂)}^{1/4}   (1 ± 5/2 1/r1)

If the discs are far from the star, they indicate that they are in deep space, the distance r₁ from being grade by which we can approximate; this is a very strong approach

              T₁ = T₂  {(e₂)}^{1/4} ¼

              T<u>₁</u> = T₂  0.90.9^{1/4}

               5500 = T₂  0.974

               T₂ = 5646 K

3 0
2 years ago
Two rocks are tied to massless strings and whirled in nearly horizontal circles so that the time to travel around the circle onc
Fynjy0 [20]

Answer:m_1=m_2

Explanation:

Given

Time period for both string is same

\frac{2\pi r}{v_1}=\frac{2\pi 2r}{v_2}

2v_1=v_2

and tension in string 2 is  twice the first string

2T_1=T_2

Tension will provide centripetal acceleration

2\frac{m_1v_1^2}{r}=\frac{m_2v_2^2}{2r}

2\frac{m_1v_1^2}{r}=\frac{m_2\times 4v_1}{2r}

thus m_1=m_2

3 0
2 years ago
An inventor claims to have developed a power cycle having a thermal efficiency of 40%, while operating between hot and cold rese
lesya [120]

From Carnot's theorem, for any engine working between these two temperatures:


efficiency <= (1-tc/th) * 100


Given: tc = 300k (from question assuming it is not 5300 as it seems)

For a, th = 900k, efficiency = (1-300/900) = 70%

For b, th = 500k, efficiency = (1-300/500) = 40% 

For c, th = 375k, efficiency = (1-300/375) = 20% 


Hence in case of a and b, efficiency claimed is lesser than efficiency calculated, which is valid case and in case of c, however efficiency claimed is greater which is invalid. 

7 0
2 years ago
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