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Elena-2011 [213]
3 years ago
9

A rigid tank contains 1 kg of air (ideal gas) at 15 °C and 210 kPa. A paddle wheel supplies work input to the air such that fina

l temperature is 97 "C. The specific heats are Cp 1.005 kJ(kg.K) and C-0.718 kJ/(kg.K). Determine the total work done in kJ:
Physics
1 answer:
lisov135 [29]3 years ago
5 0

Answer:

-58.876 kJ

Explanation:

m = mass of air = 1 kg

T₁ = Initial temperature = 15°C

T₂ = Final temperature = 97°C

Cp = Specific heat at constant pressure = 1.005 kJ/kgk

Cv = Specific heat at constant volume = 0.718 kJ/kgk

W = Work done

Q = Heat = 0 (since it is not mentioned we are considering adiabatic condition)

ΔU = Change in internal energy

Q = W+ΔU

⇒Q = W+mCvΔT

⇒0 = W+mCvΔT

⇒W = -mCvΔT

⇒Q = -1×0.718×(97-15)

⇒Q = -58.716 kJ

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adoni [48]

Answer:

<u> </u><u>»</u><u> </u><u>Image</u><u> </u><u>distance</u><u> </u><u>:</u>

{ \tt{ \frac{1}{v}  +  \frac{1}{u} =  \frac{1}{f}  }} \\

  • v is image distance
  • u is object distance, u is 10 cm
  • f is focal length, f is 5 cm

{ \tt{ \frac{1}{v} +  \frac{1}{10} =  \frac{1}{5}   }} \\  \\  { \tt{ \frac{1}{v}  =  \frac{1}{10} }} \\  \\ { \tt{v = 10}} \\  \\ { \underline{ \underline{ \pmb{ \red{ \: image \: distance \: is \: 10 \: cm \:  \: }}}}}

<u> </u><u>»</u><u> </u><u>Magnification</u><u> </u><u>:</u>

• Let's derive this formula from the lens formula:

{ \tt{ \frac{1}{v}  +  \frac{1}{u} =  \frac{1}{f}  }} \\

» Multiply throughout by fv

{ \tt{fv( \frac{1}{v} +  \frac{1}{u} ) = fv( \frac{1}{f}  )}} \\   \\ { \tt{ \frac{fv}{v}  +  \frac{fv}{u}  =  \frac{fv}{f} }} \\  \\  { \tt{f + f( \frac{v}{u} ) = v}}

• But we know that, v/u is M

{ \tt{f + fM = v}} \\  { \tt{f(1 +M) = v }} \\ { \tt{1 +M =  \frac{v}{f}  }} \\  \\ { \boxed{ \mathfrak{formular :  } \: { \tt{ M =  \frac{v}{f}  - 1 }}}}

  • v is image distance, v is 10 cm
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{ \tt{M =  \frac{10}{5} - 1 }} \\  \\ { \tt{M = 5 - 1}} \\  \\ { \underline{ \underline{ \pmb{ \red{ \: magnification \: is \: 4}}}}}

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4 0
2 years ago
1.A body of certain mass is kept at a height of 15m from the ground, taking the value of g as 10m/s find out the mass of the bod
ehidna [41]

Answer:

1.#potential energy = PE, m = mass in kg, g = force of gravity, h= vertical height above the ground.  ** means to the power of ie exponent. * means multiply.

PE = mgh

300 = m(10)(15)

m = 300/(10)(15)

m= 2kg

2. KE = 1/2 mv**2

     = 1/2(50)(50)**2

     = 2500 joules

Explanation

Is as in solution

4 0
3 years ago
(b) The distance of mass from mass A if there is no gravitational force acted on C
shepuryov [24]

Answer:

(a) The force, acting on object 'C' is approximately 2.66972 × 10⁻¹⁰ Newtons

(b) The distance of 'C' from 'A', in the direction particle 'B' if there is no  meters gravitational force acting on 'C' is appromimately 0.829 meters or 1.877 meters

Explanation:

The given parameters are;

The mass of particle, A, m₁ = 2 kg

The mass of particle, B, m₂ = 0.3 kg

The mass of particle, C, m₃ = 0.05 kg

The distance between particle 'A' and particle 'B', r₁ = 0.15 m

The distance between particle 'B' and particle 'C', r₂ = 0.05 m

(a) The gravitational force, 'F', is given as follows;

F =G \times \dfrac{m_{1} \times m_{2}}{r^{2}}

Where;

F = The force between the two masses

G = The gravitation constant = 6.67430 × 10⁻¹¹ N·m²/kg²

m₁ = The mass of object 1

m₂ = The mass of object 2

If 'C' is placed at 0.05 m from 'B', we have;

F₂₃ =  6.67430 × 10⁻¹¹ × 0.05 × 0.3/(0.05²) ≈ 4.00458 × 10⁻¹⁰

The gravitational force between force between particle 'B' and particle 'C', F₂₃ = 4.00458 × 10⁻¹⁰ N (towards the right)

F₁₃ =  6.67430 × 10⁻¹¹ × 0.05 × 2/(0.1²) ≈ × 10⁻¹⁰

The gravitational force between force between particle 'A' and particle 'B', F₁₃ = 6.6743 × 10⁻¹⁰ N (towards the left)

The force, 'F', acting on object 'C' = F₁₃ - F₂₃

F = (6.6743 - 4.00458) × 10⁻¹⁰ = 2.66972 × 10⁻¹⁰ N

The force, acting on object 'C' ≈ 2.66972 × 10⁻¹⁰ N

(b), When there is no gravitational force acting on 'C', let the distance of 'C' from 'A' = x

We have;

F₂₃ = F₁₂

F_{23} =G \times \dfrac{m_{1} \times m_{2}}{r_1^{2}} = F_{13} =G \times \dfrac{m_{1} \times m_{3}}{r_2^{2}}

By plugging in the values and removing like terms, we get;

\dfrac{0.3 \times 0.05}{(1.15 - x)^{2}}  = \dfrac{2 \times 0.05}{x^2}

(1.15 - x)² × 2 × 0.05 = 0.3 × 0.05 × x²

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0.1·x² - 0.23·x + 1.3225 - 0.015·x² = 0

0.085·x² - 0.23·x + 0.13225= 0

x = (0.23± √((-0.23)² - 4 × 0.085 × ( 0.13225)))/(2 × 0.085))

x ≈ 0.829, or x ≈ 1.877

Therefore, the distance of 'C' from 'A', if there is no gravitational force acting on 'C', x ≈ 0.829 m, or x = 1.877 m, in the direction of 'B'

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3 years ago
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