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elena-s [515]
4 years ago
15

Air rushing over the wings of high-performance race cars generates unwanted horizontal air resistance but also causes a vertical

downforce, which helps cars hug the track more securely. The coefficient of static friction between the track and the tires of a 678-kg car is 0.843. What is the magnitude of the maximum acceleration at which the car can speed up without its tires slipping when a 3620-N downforce and an 1270-N horizontal air resistance force act on it?
Physics
1 answer:
Nookie1986 [14]4 years ago
5 0

Answer:

10.897 m/s²

Explanation:

f_s = Slipping force

F_d = Downforce = 3620 N

\mu = Coefficient of static friction = 0.843

m = Mass of car = 678 kg

f_h = Horizontal force = 1270 N

g = Acceleration due to gravity = 9.81 m/s²

a=\frac{f_s-f_h}{m}\\\Rightarrow a=\frac{\mu(F_D+mg)-f_h}{m}\\\Rightarrow a=\frac{0.843(3620+678\times 9.81)-1270}{678}\\\Rightarrow a=10.897\ m/s^2

Hence, magnitude of the maximum acceleration is 10.897 m/s²

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NikAS [45]

Answer:

a) velocity v = 322.5m/s

b) time t = 19.27s

Explanation:

Note that;

ads = vdv

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a is acceleration

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Given;

a = 6 + 0.02s

so,

\int\limits^s_0 {a} \, ds  = \int\limits^v_0 {v} \, dv\\ \int\limits^s_0 {6+0.02s} \, ds  = \int\limits^v_0 {v} \, dv\\ 6s + \frac{0.02s^{2} }{2} = \frac{1}{2} v^{2} \\v = \sqrt{12s + 0.02s^{2} } .....................1 \\\\\\

Remember that

v = \frac{ds}{dt} \\\frac{ds}{v} = dt\\\int\limits^s_0 {\frac{ds}{\sqrt{12s+0.02s^{2} } } } \, ds = \int\limits^t_0 {} \, dt \\t=  (5\sqrt{2} ) ln  \frac{| [s + 300 + \sqrt{(s^{2}  + 600s)} ] |}{300} .......2

substituting s = 2km =2000m, into equation 1

v = 322.5m/s

substituting s = 2000m into equation 2

t = 19.27s

8 0
4 years ago
Which statement describes the relationship of resistance and current?
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Choices 'B'; and 'D' both begin with the correct words.
But they should end with the equation

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8 0
3 years ago
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A 15 kg dog jumps out a stationary sled which has a mass of 40 kg. If
zzz [600]
Velocity of the sled is 3.2 m/s
5 0
2 years ago
There. That is better.
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6 0
4 years ago
A total resistance of 3.03 Ω is to be produced by connecting an unknown resistance to a 12.18 Ω resistance. (a) What must be the
insens350 [35]

Answer:

(a) 4.0334Ω

(b)parallel

Explanation:

for resistors connected in parallel;

\frac{1}{R_{eq} } =\frac{1}{R1}+\frac{1}{R2}

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\frac{1}{3.03 } =\frac{1}{12.18}+\frac{1}{R2}

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3 0
4 years ago
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