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3 years ago

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Two 125 kg bumper cars are moving toward each other in opposite directions. Car X is moving at 10 m/s and Car Z at −12 m/s when

they collide headon. If the resulting velocity of Car Z after the collision is 10 m/s, what is the velocity of Car X after the collision?

1 answer:

nignag [31]3 years ago

3 0

<h2>Given that,</h2>

Mass of two bumper cars, m₁ = m₂ = 125 kg

Initial speed of car X is, u₁ = 10 m/s

Initial speed of car Z is, u₂ = -12 m/s

Final speed of car Z, v₂ = 10 m/s

We need to find the final speed of car X after the collision. Let v₁ is its final speed. Using the conservation of momentum to find it as follows :

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

v₁ is the final speed of car X.

$m_1u_1+m_2u_2-m_1v_1=m_2v_2\\\\m_2v_2=m_1u_1+m_2u_2-m_2v_2\\\\m_1v_1=125\times 10+125\times (-12)-125\times 10\\\\v_1=\dfrac{-1500}{125}\\\\v_1=-12\ m/s$

So, car X will move with a velocity of -12 m/s.

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The correct option is d: T₁ > T₂ > T₃.

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Since W₃ = mg, where <em>m</em> is for mass and <em>g</em> is for the acceleration due to gravity, we have:

$T_{3} = W_{3} = mg$ (1) <u>Case T₂.</u>

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$T_{2} = T_{3} + W_{2}$ (2)

By entering W₂ = 2mg and equation (1) into eq (2) we have:

$T_{2} = T_{3} + W_{2} = mg + 2mg = 3mg$

<u>Case T₁.</u>

$T_{1} - (T_{2} + W_{1}) = 0$

$T_{1} = T_{2} + W_{1}$ (3)

Knowing that W₁ = 3mg and T₂ = 3mg, eq (3) is:

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I hope it helps you!

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An ideal Diesel cycle has a compression ratio of 18 and a cutoff ratio of 1.5. Determine the (1) maximum air temperature and (2)

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Answer:

(1) The maximum air temperature is 1383.002 K

(2) The rate of heat addition is 215.5 kW

Explanation:

T₁ = 17 + 273.15 = 290.15

$\frac{T_2}{T_1} =r_v^{k - 1} =18^{0.4} =3.17767$

T₂ = 290.15 × 3.17767 = 922.00139

$\frac{T_3}{T_2} =\frac{v_3}{v_2} = r_c = 1.5$

Therefore,

T₃ = T₂×1.5 = 922.00139 × 1.5 = 1383.002 K

The maximum air temperature = T₃ = 1383.002 K

(2)

$\frac{v_4}{v_3} =\frac{v_4}{v_2} \times \frac{v_2}{v_3} = \frac{v_1}{v_2} \times \frac{v_2}{v_3} = 18 \times \frac{1}{1.5} = 12$

$\frac{T_3}{T_4} =(\frac{v_4}{v_3} )^{k-1} = 12^{0.4} = 2.702$

Therefore;

$T_4 = \frac{1383.002}{2.702} =511.859 \ k$

$Q_1 = c_p(T_3-T_2)$

Q₁ = 1.005(1383.002 - 922.00139) = 463.306 kJ/jg

Heat rejected per kilogram is given by the following relation;

$c_v(T_4-T_1)$ = 0.718×(511.859 - 290.15) = 159.187 kJ/kg

The efficiency is given by the following relation;

$\eta = 1-\frac{\beta ^{k}-1}{\left (\beta -1 \right )r_{v}^{k-1}}$

Where:

β = Cut off ratio

Plugging in the values, we get;

$\eta = 1-\frac{1.5 ^{1.4}-1}{\left (1.5 -1 \right )18^{1.4-1}}= 0.5191$

Therefore;

$\eta = \frac{\sum Q}{Q_1}$

$\therefore 0.5191 = \frac{150}{Q_1}$

Heat supplied = $\frac{150}{0.5191} = 288.978 \ hp$

Therefore, heat supplied = 215491.064 W

Heat supplied ≈ 215.5 kW

The rate of heat addition = 215.5 kW.

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Answer: 15.8

Explanation:

You are given that the

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Substitute F and V into the formula

1/30.1 = 1/32 + 1/V

1/V = 1/30.1 - 1/32

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Reciprocate both sides

V = 506.94 cm

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M = V/U

substitute the values of V and U into the formula

M = 506.94/32

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