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2 years ago

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A car weighs 2200 lb and is travelling 100 mi/h on a race track that is on a 3% upgrade. The car is preparing to pass a slower c

ar and its torque/engine speed curve is given by (with Me in ftlb and ne in revolutions per second): Me = 8ne - 0.05n; Drivetrain efficiency is 90%, drive axle slippage is 2%, wheel radius is 15 inches, frontal area is 22 ft2, drag coefficient is 0.35, and air density is 0.0022 slugs/ft3. If the car is in a gear that produces maximum torque, what would the car's maximum acceleration be?

2 answers:

Lyrx [107]2 years ago

5 0

Answer:

Detailed solution is given below:

White raven [17]2 years ago

4 0

Answer:

- 0.3275 ft / s^2 cars maximum acceleration

Explanation:

The average drag force is calculated as

fo = 1/2 ∝ A $v^{2}$ α

α = drag coefficient = 0.35

∝ = 1.328

A = 22 ft^2 = 2.04387 m^2

S = 0.0022 slugs/ft3 = 1.1338 kg/m^3

V = 100 m/h = 44.704 m/s

fo = 1/2 * 1.328 * 2.04387 * $44.704^{2}$ * 0.35

= 81044 N

next calculate the torque/engine speed curve

Me = 8ne - 0.05 * $ne^{2}$

ne = 4800 rev /min = 80 rev /sec

Me = 8.80 - 0.05 * 80 ^2

= 320 Ib ft = 438.86 Nm

next calculate torque at the curve

= torque speed * drive train efficiency * 1 - drive axle slippage

= 433.86 * 0.9 * 0.98

= 382.66 N-m

force at the wheel ( F wheel) = 1004.368 N

radius = 15 in = 0.381 m

the opposing forces to this are drag force and weight along the upgrade

race track given = 0.03

which means the angle ∅ can be gotten from tan ∅ = 0.03 or ∅ = 1.71835

car mass = 2200 Ib = 997.9032 kg

to calculate the car's maximum acceleration we apply newton's second law

A max = F wheel - Fo - mg sin∅

1004.368 - 810.44 - 997.9032 * 9.8 * sin ∅

= 997.9032 Ra

= a = -0.09983 m/s^2

= - 0.3275 ft / s^2

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Answer:

long duct: 1.0 and 0.424

Hemisphere 1.0 ; 0.125; 0.5

Explanation:

For a long duct:

By inspection, $F_{12} = 1.0$

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Hemisphere:

By reciprocity gives = 0.125

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Answer:

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An excited electron in an Na atom emits radiation at a wavelength 589 nm and returns to the ground state. If the mean time for t

11Alexandr11 [23.1K]

Answer: Inherent width in the emission line: 9.20 × 10⁻¹⁵ m or 9.20 fm

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∆t∆E ≈ h ( Planck's Constant)

The transition time ∆t corresponds to the energy that is ∆E

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The corresponding uncertainty in the emitted frequency ∆v is:

∆v= ∆E/h = (5.273*10^-27 J)/(6.626*10^ J.s)= 7.958 × 10^6 s^-1

To find the corresponding spread in wavelength and hence the line width ∆λ, we can differentiate

λ = c/v

dλ/dv = -c/v² = -λ²/c

Therefore,

∆λ = (λ²/c)*(∆v) = {(589*10⁻⁹ m)²/(3.0*10⁸ m/s)} * (7.958*10⁶ s⁻¹)

= 9.20 × 10⁻¹⁵ m or 9.20 fm

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2 years ago

Poles are values of Laplace transform variable, s, that make denominator of transfer function zero. Zeros are values of Laplace

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Answer:

Zero 1 = -1

Zero 2 = -3

Pole 1 = 0

Pole 2 = -2

Pole 3 = -4

Pole 4 = -6

Gain = 4

Explanation:

For any given transfer function, the general form is given as

T.F = k [N(s)] ÷ [D(s)]

where k = gain of the transfer function

N(s) is the numerator polynomial of the transfer function whose roots are the zeros of the transfer function.

D(s) is the denominator polynomial of the transfer function whose roots are the poles of the transfer function.

k [N(s)] = 4s² + 16s + 12 = 4[s² + 4s + 3]

it is evident that

Gain = k = 4

N(s) = (s² + 4s + 3) = (s² + s + 3s + 3)

= s(s + 1) + 3 (s + 1) = (s + 1)(s + 3)

The zeros are -1 and -3

D(s) = s⁴ + 12s³ + 44s² + 48s

= s(s³ + 12s² + 44s + 48)

= s(s + 2)(s + 4)(s + 6)

The roots are then, 0, -2, -4 and -6.

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2 years ago

A thermal energy storage unit consists of a large rectangular channel, which is well insulated on its outer surface and encloses

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Answer:

the temperature of the aluminum at this time is 456.25° C

Explanation:

Given that:

width w of the aluminium slab = 0.05 m

the initial temperature $T_1$ = 25° C

$T{\infty} =600^0C$

h = 100 W/m²

The properties of Aluminium at temperature of 600° C by considering the conditions for which the storage unit is charged; we have ;

density ρ = 2702 kg/m³

thermal conductivity k = 231 W/m.K

Specific heat c = 1033 J/Kg.K

Let's first find the Biot Number Bi which can be expressed by the equation:

$Bi = \dfrac{hL_c}{k} \\ \\ Bi = \dfrac{h \dfrac{w}{2}}{k}$

$Bi = \dfrac{hL_c}{k} \\ \\ Bi = \dfrac{100 \times \dfrac{0.05}{2}}{231}$

$Bi = \dfrac{2.5}{231}$

Bi = 0.0108

The time constant value $\tau_t$ is :

$\tau_t = \dfrac{pL_cc}{h} \\ \\ \tau_t = \dfrac{p \dfrac{w}{2}c}{h}$

$\tau_t = \dfrac{2702* \dfrac{0.05}{2}*1033}{100}$

$\tau_t = \dfrac{2702* 0.025*1033}{100}$

$\tau_t = 697.79$

Considering Lumped capacitance analysis since value for Bi is less than 1

Then;

$Q= (pVc)\theta_1 [1-e^{\dfrac {-t}{ \tau_1}}]$

where;

$Q = -\Delta E _{st}$ which correlates with the change in the internal energy of the solid.

So;

$Q= (pVc)\theta_1 [1-e^{\dfrac {-t}{ \tau_1}}]= -\Delta E _{st}$

The maximum value for the change in the internal energy of the solid is :

$(pVc)\theta_1 = -\Delta E _{st}max$

By equating the two previous equation together ; we have:

$\dfrac{-\Delta E _{st}}{\Delta E _{st}{max}}= \dfrac{ (pVc)\theta_1 [1-e^{\dfrac {-t}{ \tau_1}}]} { (pVc)\theta_1}$

Similarly; we need to understand that the ratio of the energy storage to the maximum possible energy storage = 0.75

Thus;

$0.75= [1-e^{\dfrac {-t}{ \tau_1}}]}$

So;

$0.75= [1-e^{\dfrac {-t}{ 697.79}}]}$

$1-0.75= [e^{\dfrac {-t}{ 697.79}}]}$

$0.25 = e^{\dfrac {-t}{ 697.79}}$

$In(0.25) = {\dfrac {-t}{ 697.79}}$

$-1.386294361= \dfrac{-t}{697.79}$

t = 1.386294361 × 697.79

t = 967.34 s

Finally; the temperature of Aluminium is determined as follows;

$\dfrac{T - T _{\infty}}{T_1-T_{\infty}}= e ^ {\dfrac{-t}{\tau_t}}$

$\dfrac{T - 600}{25-600}= e ^ {\dfrac{-967.34}{697.79}$

$\dfrac{T - 600}{25-600}= 0.25$

$\dfrac{T - 600}{-575}= 0.25$

T - 600 = -575 × 0.25

T - 600 = -143.75

T = -143.75 + 600

T = 456.25° C

Hence; the temperature of the aluminum at this time is 456.25° C

3 0

2 years ago