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Alchen [17]
3 years ago
12

To maintain a constant speed, the force provided by a car's engine must equal the drag force plus the force of friction of the r

oad (the rolling resistance). The density of air is 1.2 kg/m3.
(a) What are the drag forces in newtons at 77 km/h and 106 km/h for a Toyota Camry? (Drag area = 0.70 m2 and drag coefficient = 0.28.) at 77 km/h N at 106 km/h N
(b) What are the drag forces in newtons at 77 km/h and at 106 km/h for a Hummer H2? (Drag area = 2.44 m2 and drag coefficient = 0.57.) at 77 km/h N at 106 km/h N Supporting Materials
Physics
1 answer:
sweet [91]2 years ago
6 0

Answer:

a). 53.75 N and 101.92 N

b). 381.44 N and 723.25 N

Explanation:

V= 77 \frac{km}{h}* \frac{1h}{3600 s} *\frac{1000m}{1 km} = 21.38 \frac{m}{s} \\V=106 \frac{km}{h}* \frac{1h}{3600 s} *\frac{1000m}{1 km} = 29.44 \frac{m}{s}

a).

ρ= 1.2 \frac{kg}{m^{3} }, A_{t}= 0.7 m^{2}, D_{t}= 0.28

F_{t1} = \frac{1}{2} * D_{t} * A_{t}* p_{t}* v_{t}^{2}

F_{t1} = \frac{1}{2} * 0.28 * 0.7m^{2} * 1.2\frac{kg}{m^{3} }* 21.38^{2}= 53.75 N

F_{t1} = \frac{1}{2} * 0.28 * 0.7m^{2} * 1.2\frac{kg}{m^{3} }* 29.44^{2}= 101.92 N

b).

ρ= 1.2 \frac{kg}{m^{3} }, A_{h}= 2.44 m^{2}, D_{h}= 0.57

F_{t1} = \frac{1}{2} * D_{h} * A_{h}* p_{h}* v_{h}^{2}

F_{t1} = \frac{1}{2} * 0.57 * 2.44 m^{2} * 1.2\frac{kg}{m^{3} }* 21.38^{2}= 381.44 N

F_{t1} = \frac{1}{2} * 0.57 * 2.44 m^{2} * 1.2\frac{kg}{m^{3} }* 29.44^{2}= 723.25 N

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Answer:

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Explanation:

From the question we are told that

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The height of the window from the ground is h  =  36.5 \  m

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T =  t +  t_1

Here t_1 is the time duration that elapsed after Henrietta has passed below the window the value is given as 4 s

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T = 2.73  +  4

=> T = 6.73 \  s

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s=  v  *  T

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Generally the speed with which Bruce threw her lunch is mathematically represented as

v  =  \frac{18.17}{2.73}

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Answer:

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