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3 years ago

6

A city bus travels 6 blocks east and 8 blocks north. Each block is 100 m long. If the bus travels this distance in 15 minutes, w

hat is the average speed of the bus (m/s)?

1 answer:

nexus9112 [7]3 years ago

6 0

Answer:

The average speed of the bus, v = 1.55 m/s

Explanation:

Given that,

Number of blocks traveled by bus towards east = 6

Number of blocks traveled by bus towards north = 8

Length of each block = 100 m

Distance traveled by bus towards east 6 x 100 = 600 m

Distance traveled by bus towards north 8 x 100 = 800 m

The total distance traveled, d = 600 + 800 = 1400

Time taken by the bus to travel is, t = 15 minutes

The velocity is given by the formula

v = d/t m

Substituting the values in the above equation

v = 1400 m /(15 x 60) s

= 1.55 m/s

Hence, the average speed of the bus, v = 1.55 m/s

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

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Where

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$T_2 =$ Temperature at exit of turbine

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

$m_i = m_0 = m$

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

Where,

m = mass

$m_i$ = mass at inlet

$m_0$ = Mass at outlet

$h_i$ = Enthalpy at inlet

$h_0$ = Enthalpy at outlet

W = Work done

Q = Heat transferred

$V_i$ = Velocity at inlet

$V_0$ = Velocity at outlet

$Z_i$ = Height at inlet

$Z_0$ = Height at outlet

For the insulated system with neglecting kinetic and potential energy effects

$h_i = h_0 + W$

$W = h_i -h_0$

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

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

$\frac{T_2}{1400K} = (\frac{0.8bar}{8nar})^{(\frac{1.4-1}{1.4})}$

$T_2 = 725.126K$

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

So:

$W = h_i -h_0$

$W = C_p (T_1-T_2)$

$W = 1.005(1400-725.126)$

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Therefore the maximum theoretical work that could be developed by the turbine is 678.248kJ/kg

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Answer:

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Unknown:

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Speed = $\frac{distance}{time}$

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