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3 years ago

A driver is traveling at 90 km/h down a 3% grade on good, wet pavement. An accident

Engineering

1 answer:

Paul [167]3 years ago

4 0

Answer:

0.35

Explanation:

We resolve the component of the weight of the car along and perpendicular to the grade. We have mgsinФ and mgcosФ where Ф = angle of grade.

Now, the frictional force f = μN = μmgcosФ where μ = coefficient of friction

So, the net force along the grade is F = mgsinФ - μmgcosФ.

The work done by this force moving a distance, d along the grade is

W = (mgsinФ - μmgcosФ)d

This work equals the change in kinetic energy of the car. So ΔK = 1/2m(v₂² - v₁²) = W = (mgsinФ - μmgcosФ)d

1/2m(v₂² - v₁²) = (mgsinФ - μmgcosФ)d

1/2(v₂² - v₁²) = (gsinФ - μgcosФ)d

(v₂² - v₁²)/2d = (gsinФ - μgcosФ)

dividing through by gcosФ, we have

(v₂² - v₁²)/2dgcosФ = (gsinФ/gcosФ) - μgcosФ/gcosФ

(v₂² - v₁²)/2dgcosФ = tanФ - μ

μ = tanФ - (v₂² - v₁²)/2dgcosФ

given that tanФ = 3% = 3/100 and 1 + tan²Ф = 1/cos²Ф, cosФ = 1/(√1 + tan²Ф) = 1/(√1 + (3/100)²) = 1/(√1 + (9/10000)) = 1/(√10000 + 9/10000) = 1/√(10009/10000) = 100/√10009 = 100/100.05 = 0.9995.

Also, given that v₁ = 90 km/h = 90 × 1000/3600 m/s = 25 m/s and v₂ = 45 km/h = 45 × 1000/3600 m/s = 12.5 m/s, d = 75 m and g = 9.8 m/s².

So, substituting the values of the variables into the equation, we have

μ = tanФ - (v₂² - v₁²)/2dgcosФ

μ = 3/100 - ((12.5 m/s)² - (25 m/s)²)/(2 × 75 m × 9.8 m/s² × 0.9995)

μ = 3/100 - ((156.25 m/s)² - (625 m/s)²)/1,469.265 m²/s²

μ = 3/100 - (-468.75 m²/s²)/1,469.265 m²/s²

μ = 3/100 + 468.75 m²/s²/1,469.265 m²/s²

μ = 0.03 + 0.32

μ = 0.35

So, theoretical friction coefficient is 0.35

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Answer:

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Explanation:

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3 years ago

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2 years ago

2) The switch in the circuit below has been closed a long time. At t=0, it is opened.

saul85 [17]

Answer:

il(t) = e^(-100t)

Explanation:

The current from the source when the switch is closed is the current through an equivalent load of 15 + 50║50 = 15+25 = 40 ohms. That is, it is 80/40 = 2 amperes. That current is split evenly between the two parallel 50-ohm resistors, so the initial inductor current is 2/2 = 1 ampere.

The time constant is L/R = 0.20/20 = 0.01 seconds. Then the decaying current is described by ...

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2 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

Sonja [21]

Answer:

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

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

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

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

5 0

3 years ago