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Anastaziya [24]
4 years ago
15

A light, inextensible cord passes over a light, frictionless pulley with a radius of 15 cm. It has a(n) 14 kg mass on the left a

nd a(n) 5.2 kg mass on the right, both hanging freely. Initially their center of masses are a vertical distance 2.5 m apart. The acceleration of gravity is 9.8 m/s 2 .
Physics
2 answers:
jeyben [28]4 years ago
4 0

Answer:

The rate of acceleration is 7.441 m/s/s.

(19 - 2.6) * 9.8 / (19 + 2.6) = 7.441

The tension in the cord is 44.826 Newtons.

2.6 * (9.8 + 7.441) = 44.826

or

19 * (9.8 - 7.441) = 44.826

Explanation:

Yuki888 [10]4 years ago
3 0

Answer with Explanation:

We are given that

m_1=14 kg

m_2=5.2 lg

Vertical distance,d=2.5 m

Acceleration due to gravity=g=9.8 m/s^2

a.Acceleration=a=\frac{(m_1-m_2)}{m_1+m_2}g

Substitute the values

Acceleration, a=\frac{14-5.2}{14+5.2}\times 9.8

Acceleration, a=4.49 m/s^2

b.Tension in the cord, T=m_2(a+g)

Tension in the cord,T=5.2(9.8+4.49)=74.3 N

Hence, the tension in the cord=74.3 N

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Answer:

the maximum theoretical work that could be developed by the turbine is 775.140kJ/kg

Explanation:

To solve this problem it is necessary to apply the concepts related to the adiabatic process that relate the temperature and pressure variables

Mathematically this can be determined as

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^{(\frac{\gamma-1}{\gamma})}

Where

Temperature at inlet of turbine

Temperature at exit of turbine

Pressure at exit of turbine

Pressure at exit of turbine

The steady flow Energy equation for an open system is given as follows:

m_i = m_0 = mm(h_i+\frac{V_i^2}{2}+gZ_i)+Q = m(h_0+\frac{V_0^2}{2}+gZ_0)+W

Where,

m = mass

m(i) = mass at inlet

m(o)= Mass at outlet

h(i)= Enthalpy at inlet

h(o)= Enthalpy at outlet

W = Work done

Q = Heat transferred

v(i) = Velocity at inlet

v(o)= Velocity at outlet

Z(i)= Height at inlet

Z(o)= Height at outlet

For the insulated system with neglecting kinetic and potential energy effects

h_i = h_0 + WW = h_i -h_0

Using the relation T-P we can find the final temperature:

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^{(\frac{\gamma-1}{\gamma})}\\

\frac{T_2}{1600K} = (\frac{0.8bar}{8nar})^{(\frac{1.4-1}{1.4})}\\ = 828.716K

From this point we can find the work done using the value of the specific heat of the air that is 1,005kJ / kgK

W = h_i -h_0W = C_p (T_1-T_2)W = 1.005(1600 - 828.716)W = 775.140kJ/Kg

the maximum theoretical work that could be developed by the turbine is 775.140kJ/kg

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3 years ago
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