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Naya [18.7K]
1 year ago
5

The 15-Mg tank car A and 25-Mg freight car B travel towards each other with the velocities shown. If the coefficient of restitut

ion between the bumpers is e = 0.8, determine the speed of each car just after the collision. Part A. Determine the speed of the car A just after the collision. Part B. Determine the speed of the car B just after the collision.
Engineering
1 answer:
irga5000 [103]1 year ago
6 0

The speed of the car A is 6.05 m/s and  the speed of the car B just after the collision 7.65 m/s.

Ma=15 mg , Mb=25mg

Vai=5 m/s   vbi=7 m/s

We know coeffecient of restitution

e=|Vaf-Vbf/Vai-Cbi|

0.8=|Vaf-Vbf/5-7|

Vaf-Vbf=1.6

MaVa+mbVb=MaVaf+MbVbf

15*5*25*7=15Vaf+25Vbf

3Vaf+5Vbf=50

sovleving eq 1 and 2

Vbf =6.05 m/s

Vaf=7.65 m/s

The speed of a change in an object's location in any direction. The distance traveled divided by the time required to travel that distance is the definition of speed.Due to its lack of magnitude and merely having a direction, speed is a scalar number. The average speed of an object can be determined if you know the distance traveled and the time it took. Distance times speed is how speed is calculated.

Learn more about speed here:

brainly.com/question/28224010

#SPJ4

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attached below

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3 years ago
A furnace wall composed of 200 mm, of fire brick. 120 mm common brick 50mm 80% magnesia and 3mm of steel plate on the outside. I
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Answer:

  • fire brick / common brick : 1218 °C
  • common brick / magnesia : 1019 °C
  • magnesia / steel : 90.06 °C
  • heat loss: 4644 kJ/m^2/h

Explanation:

The thermal resistance (R) of a layer of thickness d given in °C·m²·h/kJ is ...

  R = d/k

so the thermal resistances of the layers of furnace wall are ...

  R₁ = 0.200/4 = 0.05 °C·m²·h/kJ

  R₂ = 0.120 2.8 = 3/70 °C·m²·h/kJ

  R₃ = 0.05/0.25 = 0.2 °C·m²·h/kJ

  R₄ = 0.003/240 = 1.25×10⁻⁵ °C·m²·h/kJ

So, the total thermal resistance is ...

  R₁ +R₂ +R₃ +R₄ = R ≈ 0.29286 °C·m²·h/kJ

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The rate of heat loss is ΔT/R = (1450 -90)/0.29286 = 4643.70 kJ/(m²·h)

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The temperature drops across the various layers will be found by multiplying this heat rate by the thermal resistance for the layer:

  fire brick: (4543.79 kJ/(m²·h))(0.05 °C·m²·h/kJ) = 232 °C

so, the fire brick interface temperature at the common brick is ...

  1450 -232 = 1218 °C

For the next layers, the interface temperatures are ...

  common brick to magnesia = 1218 °C - (3/70)(4643.7) = 1019 °C

  magnesia to steel = 1019 °C -0.2(4643.7) = 90.06 °C

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<em>Comment on temperatures</em>

Most temperatures are rounded to the nearest degree. We wanted to show the small temperature drop across the steel plate, so we showed the inside boundary temperature to enough digits to give the idea of the magnitude of that.

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Pouring molten aluminum into a mold and allowing it to cool forms?
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Answer:

Assumption:

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2. The cylinder is well insulated and thus heat transfer is negligible.

3. The thermal energy stored in the cylinder itself is negligible.

4. The process is stated to be reversible

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a. This is reversible adiabatic(i.e isentropic) process and thus s_{1} =s_{2}

From the refrigerant table A11-A13

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b.) We take the content of the cylinder as the sysytem.

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Hence, the energy balance for adiabatic closed system can be expressed as:

E_{in} - E_{out}  =ΔE

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w_{b, out} =m([tex]u_{1} -u_{2)

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=5.8491(246.82-219.9)

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The additives is one of the type of chemical polymer which basically include polymer matrix for improving the ability of processing of the polymer. It also helps to enhance the production and requirement of the polymer products in the environment.

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3 years ago
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