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Free_Kalibri [48]
3 years ago
10

The two boxcars A and B have a weight of 20 000 Ib and 30 000 Ib, respectively. If they coast freely down the incline when the b

rakes are applied to all the wheels of car A causing it to skid, determine the force in the coupling C between the two cars. The coefficient of kinetic friction between the wheels of A and the tracks is μk=0.5. The wheels of car B are free to roll. Neglect their mass in calculation.

Engineering
2 answers:
harkovskaia [24]3 years ago
8 0

Answer:

a = 3.608 ft/s²

T = 5.977 kip

Explanation:

Angle of inclined track is not given. It is 5 degree as shown in attached image 1.

Given data:

Weight of boxcar A = Wα = 20,000 lb

Weight of boxcar B = Wb = 30,000 lb

Coefficient of kinetic friction = υk = 0.5

angle of inclined track = 5 degree

Value of g = 32.2 ft/s²

Solution:

These boxcars are moving therefore,

∑Fₓ = ma

∑Fy = 0

Car A

Use ∑Fy = 0

Forces acting in Y-axis are weight and its counter N (as shown in Free body diagram). Weight is Negative (acting downward) while N is positive (acting upward)

∑Fy = Nα - Wα*Cos(∅) = 0

∑Fy = Nα - 20,000*Cos(5°) = 0

∑Fy = Nα - 20,000*0.9962 = 0

∑Fy = Nα - 19,924 = 0                

Nα = 19,924

Use ∑Fₓ = maₓ

Forces acting in x-axis are Fa, T and horizontal component of weight and force having acceleration a (as shown in Free body diagram). Horizontal component of weight and Tension (T) are Negative (acting in negative X-axis direction) while Fa is positive (acting in positive X-axis direction)

∑Fₓ = maₓ

(0.5*19924) - 20,000*Sin(5°) - T = (20,000/32.2)*a

9962 - 1744 - T = 621.12*a

8218 - T = 621.12*a  .................. Eq(1)

Both cars

∑Fₓ = maₓ

Forces acting are; horizontal component of weights, Fa and force having acceleration a (as shown in Free body diagram).

(0.5*19924) - 20,000*Sin(5°) - 30,000*Sin(5°) = (20,000/32.2)/a + (30,000/32.2)/a

9962 - 1744 - 2616 = 621.12/a + 931.68/a

5602 = 1552.8 / a

⇒ a = 3.608 ft/s²

put a = 3.608 ft/s² in equation (1)

⇒ T = 8218 - 621.12*3.608

    T = 8218 - 2240.79

    T = 5977.21 lb

    T = 5.977 kip

Tpy6a [65]3 years ago
3 0

Answer:

Answer for the question :

"the two boxcars A and B have a weight of 20 000 Ib and 30 000 Ib, respectively. If they coast freely down the incline when the brakes are applied to all the wheels of car A causing it to skid, determine the force in the coupling C between the two cars. The coefficient of kinetic friction between the wheels of A and the tracks is μk=0.5. The wheels of car B are free to roll. Neglect their mass in calculation."

is explained in the attachment.

Explanation:

Download pdf
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sleet_krkn [62]

Answer:

Q=15.7Kw

Explanation:

From the question we are told that:

Initial Pressure P_1=4bar

Initial Temperature T_1=20 C

Final Pressure  P_2=12 bar

Final Temperature T_2=80C

Work Output W= 60 kJ/kg

Generally Specific Energy from table is

At initial state

 P_1=4bar \& T_1=20 C

 E_1=262.96KJ/Kg

With

Specific Volume V'=0.05397m^3/kg

At Final state

 P_2=12 bar \& P_2=80C

 E_1=310.24KJ/Kg

Generally the equation for The Process is mathematically given by

 m_1E_1+w=m_2E_2+Q

Assuming Mass to be Equal

 m_1=m_1

Where

 m=\frac{V}{V'}

 m=frac{0.06666}{V'=0.05397m^3/kg}

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4 0
3 years ago
For each of the following combinations of parameters, determine if the material is a low-loss dielectric, a quasi-conductor, or
Alborosie

Answer:

Glass: Low-Loss dielectric

  α = 8.42*10^-11 Np/m

  β = 468.3 rad/m

  λ = 1.34 cm

  up = 1.34*10^8 m/s

  ηc = 168.5 Ω

Tissue: Quasi-Conductor

  α = 9.75 Np/m

  β = 12.16 rad/m

  λ = 51.69 cm

  up = 0.52*10^8 m/s

  ηc = 39.54 + j 31.72 Ω        

Wood: Good conductor

  α = 6.3*10^-4 Np/m

  β = 6.3*10^-4 Np/m

  λ = 10 km

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Explanation:

Given:

Glass with µr = 1, εr = 5, and σ = 10−12 S/m at 10 GHz

Animal tissue with µr = 1, εr = 12, and σ = 0.3 S/m at 100 MHz.

Wood with µr = 1, εr = 3, and σ = 10−4 S/m at 1 kHz

Find:

Determine if  the material is a low-loss dielectric, a quasi-conductor, or a good conductor, and then  calculate α, β, λ, up, and ηc:

Solution:

- We need to determine the loss tangent to determine category of the medium as follows:

                                σ / w*εr*εo

Where, w is the angular speed of wave

            εo is the permittivity of free space = 10^-9 / 36*pi

- Now we classify as follows:

    Glass = \frac{10^-^1^2 }{2*\pi * 10*10^9 * \frac{5*10^-^9}{36\pi } } = 3.6*10^-^1^3\\\\Tissue = \frac{0.3 }{2*\pi * 100*10^6 * \frac{12*10^-^9}{36\pi } } = 4.5\\\\Wood = \frac{10^-^4 }{2*\pi * 1*10^3 * \frac{3*10^-^9}{36\pi } } = 600\\  

- For σ / w*εr*εo < 0.01 --- Low-Loss dielectric and σ / w*εr*εo > 100 --- Good conducting material.

    Glass: Low-Loss dielectric

    Tissue: Quasi-Conductor

    Wood: Good conductor

- Now we will use categorized material base equations from Table 17-1 as follows:

     Glass: Low-Loss dielectric

          α = (σ / 2)*sqrt(u / εr*εo) = (10^-12 / 2)*sqrt( 4*pi*10^-7/5*8.85*10^-12)

          α = 8.42*10^-11 Np/m

          β = w*sqrt (u*εr*εo) = 2pi*10^10*sqrt (4*pi*10^-7*5*8.85*10^-12)

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          λ = 2*pi / β = 2*pi / 468.3

          λ = 1.34 cm

          up = λ*f = 0.0134*10^10

          up = 1.34*10^8 m/s

          ηc = sqrt ( u / εr*εo ) = sqrt( 4*pi*10^-7/12*8.85*10^-12)

          ηc = 168.5 Ω

     Tissue: Quasi-Conductor

          α = (σ / 2)*sqrt(u / εr*εo) = (0.3 / 2)*sqrt( 4*pi*10^-7/12*8.85*10^-12)

          α = 9.75 Np/m

          β = w*sqrt (u*εr*εo) = 2pi*100*10^6*sqrt (4*pi*10^-7*12*8.85*10^-12)

          β = 12.16 rad/m

          λ = 2*pi / β = 2*pi / 12.16

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          up = λ*f = 0.5169*100*10^6

          up = 0.52*10^8 m/s

          ηc = sqrt ( u / εr*εo )*( 1 - j (σ / w*εr*εo))^-0.5

          ηc = sqrt (4*pi*10^-7*12*8.85*10^-12)*( 1 - j 4.5)^-0.5

          ηc = 39.54 + j 31.72 Ω

     Wood: Good conductor

          α = sqrt (pi*f*σ u) = sqrt( pi* 10^3 *4*pi* 10^-7 * 10^-4 )

          β = α = 6.3*10^-4 Np/m

          λ = 2*pi / β = 2*pi / 6.3*10^-4

          λ = 10 km

          up = λ*f = 10,000*1*10^3

          up = 0.1*10^8 m/s

          ηc = α*( 1 + j ) / б = 6.3*10^-4*( 1 + j ) / 10^-4

          ηc = 6.28*( 1 + j )

         

           

         

8 0
3 years ago
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Vanyuwa [196]

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3 years ago
The exhaust steam from a power station turbine is condensed in a condenser operating at 0.0738 bar(abs). The surface of the heat
lozanna [386]

Answer:

Percentage change 5.75 %.

Explanation:Given ;

Given

 Pressure of condenser =0.0738 bar

Surface temperature=20°C

Now from steam table

Properties of steam at 0.0738 bar  

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Total heat =4.2(40-20)+2406

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h_1=120.3\frac{KJ}{kg-m^2K}

Now film coefficient after inclusion of sensible heat

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