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AleksAgata [21]
2 years ago
7

A block attached to a spring undergoes simple harmonic motion on a horizontal frictionless surface. Its total energy is 50 J. Wh

en the displacement is half the amplitude, the kinetic energy is
Physics
1 answer:
Nina [5.8K]2 years ago
3 0

Answer:

The kinetic energy at a displacement of half the amplitude is 37.5 J

Explanation:

Given;

total energy on the spring, E = 50 J

When the displacement is half the amplitude, the total energy in the spring is sum of the kinetic energy and elastic potential energy.

E = K + U

Where;

K is the kinetic energy

U is the elastic potential energy

K = E - U

K = E - ¹/₂KA²

When the displacement is half = ¹/₂(A) = A/₂

K = E - ¹/₂K(A/₂)²

K = E - ¹/₂K(A²/₄)

K = E - ¹₄(¹/₂KA²)

Recall, E = ¹/₂KA²

K = ¹/₂KA² - ¹₄(¹/₂KA²)     (recall from simple arithmetic, 1 - ¹/₄ = ³/₄)

K = 1(¹/₂KA²) - ¹₄(¹/₂KA²)  = ³/₄(¹/₂KA²)

K = ³/₄(¹/₂KA²)

But E = ¹/₂KA² = 50J

K = ³/₄ (50J)

K = 37.5 J

Therefore, the kinetic energy at a displacement of half the amplitude is 37.5 J

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Here is the missing part of the question

To Determine the heat transfer, in kJ  if the final temperature in the tank is 110 deg C

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Explanation:

The image attached below shows the process on T - v diagram

<u>At State 1:</u>

The first step is to find the vapor pressure

P_{v1} = \rho_1 P_g_1

= \phi_1 P_{x  \ at \ 125^0C}

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<u>At State 1:</u>

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710000= m_a \times 114220.642

m_a = \dfrac{710000}{114220.642}

m_a = 6.216 \ kg

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T_{sat} = 100 + (110 -100) \bigg(\dfrac{1.584-1.673}{1.210 - 1.673}\bigg)

T_{sat} =101.92 ^0\ C

Thus, at T_{sat} =101.92 ^0\ C, condensation needs to begin.

However, since the exit temperature tends to be higher than the saturation temperature, then there will be an absence of condensation during the process.

Heat can now be determined by using the formula

Q = ΔU + W

Recall that: For a rigid tank, W = 0

Q = ΔU + 0

Q = ΔU

Q = U₂ - U₁

Also, the mass will remain constant given that there will not be any condensation during the process from state 1 and state 2.

<u>At State 1;</u>

The internal energy is calculated as:

U_1 = (m_a u_a \ _{ at \ 125^0 C})+ ( m_{v1} u_v \ _{ at \ 125^0 C} )

At T_1 = 125° C, we obtain the specific internal energy of air

SO;

U_{a \ at \ 125 ^0C } = 278.93 + ( 286.16 -278.93) (\dfrac{398-390}{400-390}   )

=278.93 + ( 7.23) (\dfrac{8}{10}   )

= 284.714 \ kJ/kg\\

At T_1 = 125° C, we obtain the specific internal energy of  water vapor

U_{v1 \ at \ 125^0C} = u_g = 2534.5 \ kJ/kg

U_1 = (m_a u_a \ at \ _{  125 ^0C }) + ( m_{v1} u_v  \ at \ _{125^0C} )

= 6.216 × 284.714 + 1.578 × 2534.5

= 5768.716 kJ

<u>At State 2:</u>

The internal energy is calculated as:

U_2 = (m_a u_a \ _{ at \ 110^0 C})+ ( m_{v1} u_v \ _{ at \ 110^0 C} )

At temperature 110° C, we obtain the specific internal energy of air

SO;

U_{a \ at \ 110^0C } = 271.69+ ( 278.93-271.69) (\dfrac{383-380}{390-380}   )

271.69+ (7.24) (0.3)

= 273.862 \ kJ/kg\\

At temperature 110° C, we obtain the specific internal energy of  water vapor

U_{v1 \ at \ 110^0C}= 2517.9 \ kJ/kg

U_2 = (m_a u_a \ at \ _{  110 ^0C }) + ( m_{v1} u_v  \ at \ _{110^0C} )

= 6.216 × 273.862 + 1.578 × 2517.9

= 5675.57 kJ

Finally, the heat transfer during the process is

Q = U₂ - U₁

Q = (5675.57 - 5768.716 ) kJ

Q = -93.146 kJ

with the negative sign, this indicates that heat is lost from the system.

6 0
2 years ago
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