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2 years ago

7

A monkey weighs 6.00 x 102 N and swings from vine to vine. As the monkey grabs a new vine, both vines make an angle of 35.0° wit

h the vertical. In such a condition of static equilibrium, what is the net force on the new vine? ANS: 366 N

1 answer:

zmey [24]2 years ago

4 0

Answer:

$T=366.23\ N$

Explanation:

Given:

mass of monkey, $w=600\ N$

angle of vine from the vertical, $\theta=35^{\circ}$

Now follow the schematic to understand the symmetry and solution via Lami's theorem.

<u>The weight of the monkey will be balanced equally by the tension in both the vines:</u>

Using Lami's Theorem:

$\frac{w}{sin\ 70^{\circ}} =\frac{T}{sin\ 145^{\circ}}$

$\frac{600}{sin\ 70^{\circ}} =\frac{T}{sin\ 145^{\circ} }$

$T=366.23\ N$

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Answer:

Final temperature of the steam = 304.29 K = 31.14°C

Rate Entropy generation of the process should be 0 because no heat is transferred into or out of the system. And ΔS = Q/T = 0 J/K.

But 0.01 J/k.s was obtained though, which is approximately 0.

Explanation:

For an adiabatic system, the Pressure and temperature are related thus

P¹⁻ʸ Tʸ = constant

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P₁¹⁻ʸ T₁ʸ = P₂¹⁻ʸ T₂ʸ

P₁ = 700 bar

P₂ = 10 bar

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T₂ = 304.29 K = 31.14°C

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To prove this

Entropy of the process

dQ - dW = dU

dQ = dU + dW

dU = mCv dT

dW = PdV

dQ = TdS

TdS = mCv dT + PdV

dS = (mCv dT/T) + (PdV/T) =

PV = mRT; P/T = mR/V

dS = (mCv dT/T) + (mRdV/V)

On integrating, we obtain

ΔS = mCv In (T₂/T₁) + mR In (V₂/V₁)

To obtain V₁ and V₂, we use PV = mRT

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Pv = RT

R for steam = 461.52 J/kg.K

For V₁

P = 700 bar = 700 × 10⁵ Pa, T = 873.15 K

v₁ = (461.52 × 873.15)/(700 × 10⁵) = 0.00570 m³/kg

For V₂

P = 10 bar = 10 × 10⁵ Pa, T = 304.29 K

v₂ = (461.52 × 304.29)/(10 × 10⁵) = 0.143 m³/kg

ΔS = mCv In (T₂/T₁) + mR In (V₂/V₁)

m = 2 kg/min = 0.0333 kg/s, Cv = 1410.8 J/kg.K

ΔS = [0.03333 × 1410.8 × In (304.29/873.15)] + (0.03333 × 461.52 × In (0.143/0.00570)

ΔS = - 49.56 + 49.57 = 0.01 J/K.s ≈ 0 J/K.s

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Answer:

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Explanation:

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A 1 530-kg automobile has a wheel base (the distance between the axles) of 2.70 m. The automobile's center of mass is on the cen

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As we know that the weight of the car is balanced by the normal force on the front wheel and rear wheels

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Determine the line-to-line sending end voltage which is the sending end voltage.

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