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3 years ago

12

The boy in the wagon begins throwing bricks out of the wagon to simulate rocket propulsion. The wagon begins at rest, and the bo

y throws three bricks. The boy’s weight is 80lbs, and the weight of the wago 20 lbs. The bricks weigh 10 lbs each, and he throws them with a horizontal velocity of 10 ft/s relative to the wagon. Neglect horizontal forces on the wagon's wheels.
A) What velocity does the boy attain if he throws the bricks one at a time?
B) What velocity does the boy attain if he throws all three bricks at once?
C) Compare the results of parts (a) and (b). What does this suggest about rocket propulsion? Why are these results different?

1 answer:

Digiron [165]3 years ago

5 0

Q:What velocity does the boy attain if he throws the bricks one at a time?

Answer:Linear velocity since it moves back and firth and does not rotate like angular velocity.

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3 years ago

Air at 400kPa, 970 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occu

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Answer:

a

The rate of work developed is $\frac{\r W}{\r m}= 300kJ/kg$

b

The rate of entropy produced within the turbine is $\frac{\sigma}{\r m}= 0.0861kJ/kg \cdot K$

Explanation:

From the question we are told

The rate at which heat is transferred is $\frac{\r Q}{\r m } = - 30KJ/kg$

the negative sign because the heat is transferred from the turbine

The specific heat capacity of air is $c_p = 1.1KJ/kg \cdot K$

The inlet temperature is $T_1 = 970K$

The outlet temperature is $T_2 = 670K$

The pressure at the inlet of the turbine is $p_1 = 400 kPa$

The pressure at the exist of the turbine is $p_2 = 100kPa$

The temperature at outer surface is $T_s = 315K$

The individual gas constant of air R with a constant value $R = 0.287kJ/kg \cdot K$

The general equation for the turbine operating at steady state is \

$\r Q - \r W + \r m (h_1 - h_2) = 0$

h is the enthalpy of the turbine and it is mathematically represented as

$h = c_p T$

The above equation becomes

$\r Q - \r W + \r m c_p(T_1 - T_2) = 0$

$\frac{\r W}{\r m} = \frac{\r Q}{\r m} + c_p (T_1 -T_2)$

Where $\r Q$ is the heat transfer from the turbine

$\r W$ is the work output from the turbine

$\r m$ is the mass flow rate of air

$\frac{\r W}{\r m}$ is the rate of work developed

Substituting values

$\frac{\r W}{\r m} = (-30)+1.1(970-670)$

$\frac{\r W}{\r m}= 300kJ/kg$

The general balance equation for an entropy rate is represented mathematically as

$\frac{\r Q}{T_s} + \r m (s_1 -s_2) + \sigma = 0$

=> $\frac{\sigma}{\r m} = - \frac{\r Q}{\r m T_s} + (s_1 -s_2)$

generally $(s_1 -s_2) = \Delta s = c_p\ ln[\frac{T_2}{T_1} ] + R \ ln[\frac{v_2}{v_1} ]$

substituting for $(s_1 -s_2)$

$\frac{\sigma}{\r m} = \frac{-\r Q}{\r m} * \frac{1}{T_s} + c_p\ ln[\frac{T_2}{T_1} ] - R \ ln[\frac{p_2}{p_1} ]$

Where $\frac{\sigma}{\r m}$ is the rate of entropy produced within the turbine

substituting values

$\frac{\sigma}{\r m} = - (-30) * \frac{1}{315} + 1.1 * ln\frac{670}{970} - 0.287 * ln [\frac{100kPa}{400kPa} ]$

$\frac{\sigma}{\r m}= 0.0861kJ/kg \cdot K$

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4 years ago

In order to keep an automobile operating, it is necessary to keep adding fuel as it is used up. Explain why this doesn't contrad

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Answer:

There is conversion of energy from one form to another and this justifies the law of conservation of energy, which states that energy can neither be created nor destroyed but can be converted from one form to another.

Explanation:

The fuel added to automobile is a chemical energy, which provides thermal energy of the combustion engine of the automobile, which is then converted to mechanical energy of the moving parts of the automobile. Thus, there is conversion of energy from one form to another. This justifies the law of conservation of energy, which states that energy can neither be created nor destroyed but can be converted from one form to another.

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Varvara68 [4.7K]

The question is not complete. We are supposed to find the average value of v_o.

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So, 12/T = 0.7/t1

So, t1 = 0.7T/12

t1 = 0.0583 T

Also, from symmetry of triangles,

t2 = T/2 - t1

So, t2 = T/2 - 0.0583 T

t2 = 0.4417T

Average of voltage output is;

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v_o,avg = (1/T) x 2.3 x 0.1917T

T will cancel out to give;

v_o,avg = 0.441V

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