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Ghella [55]
3 years ago
13

The sports car has a weight of 4900 lblb and center of gravity at GG. If it starts from rest it causes the rear wheels to slip a

s it accelerates. The coefficients of static and kinetic friction at the road are μsμs = 0.5 and μkμkmu_k = 0.35, respectively. Neglect the mass of the wheelDetermine how long it takes for it to reach a speed of 10 ft/s.What are the normal reactions at each of the rear wheels on the road?What are the normal reactions at each of the front wheels on the road?

Engineering
1 answer:
Zinaida [17]3 years ago
3 0

Answer:

2.27secs

Explanation:

The car will take 2.27secs to reach a speed of 10 ft/s.

Kindly go through the attached files for a step by step solution that will show you how the answer was gotten and the normal reaction of both the front and rear wheels.

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The hot and cold inlet temperatures to a concentric tube heat exchanger are Th,i = 200°C, Tc,i = 100°C, respectively. The outlet
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Answer:Counter,

0.799,

1.921

Explanation:

Given data

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T_{h_o}=120^{\circ}C

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T_{c_o}=125^{\circ}C

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m_hc_{ph}\left [ T_{h_i}-T_{h_o}\right ]=m_cc_{pc}\left [ T_{c_o}-T_{c_i}\right ]

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we can see that heat capacity of hot fluid is minimum

Also from energy balance

Q=UA\Delta T_m=\left ( mc_p\right )_{h}\left ( T_{h_i}-T_{h_o}\right )

NTU=\frac{UA}{\left ( mc_p\right )_{h}}=\frac{\left ( T_{h_i}-T_{h_o}\right )}{T_m}

T_m=\frac{\left ( 200-125\right )-\left ( 120-100\right )}{\ln \frac{75}{20}}

T_m=41.63^{\circ}C

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And\ effectiveness \epsilon =\frac{1-exp\left ( -NTU\left ( 1-c\right )\right )}{1-c\left ( -NTU\left ( 1-c\right )\right )}

\epsilon =\frac{1-exp\left ( -1.921\left ( 1-0.3125\right )\right )}{1-0.3125exp\left ( -1.921\left ( 1-0.3125\right )\right )}

\epsilon =\frac{1-exp\left ( -1.32068\right )}{1-0.3125exp\left ( -1.32068\right )}

\epsilon =\frac{1-0.2669}{1-0.0834}

\epsilon =0.799

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