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Ede4ka
2 months ago
5

Some of the fastest dragsters (called "top fuel) do not race for more than 300-400m for safety reasons. Consider such a dragster

in this problem, starting a race at rest. After the light turns green, the dragster completes a 400 m race in a time t = 8.6 5. Otheexpertta.com A 50% Part (a) How many times larger is the dragster's acceleration during this period than the acceleration due to gravity? alg Grade Summary Deductions Potential 16096 sino cos fano cotano asin) acos atan acotan) sinh cosho tanho cotanho Deprees Radians Submissions Attempts remaining: 5 ( per attempt) detailed view Sabrit Feedback: deduction per feedback Hinta o deduction per hist. Hints remaining 2 H A 50% Part (b) If the dragster could continue with this average acceleration, what would its speed be, in miles per hour, after it has travelled a total distance of 1.6 km (-1.0 mile)? All LLC
Physics
1 answer:
Sav [3.1K]2 months ago
5 0

Answer:

1.10261 times the acceleration due to gravity

416.17506 mph

Explanation:

t = duration

u = initial speed

v = speed at end

s = distance moved

a = acceleration

g = gravity acceleration = 9.81 m/s²

s=ut+\frac{1}{2}at^2\\\Rightarrow 400=0\times 8.6+\frac{1}{2}\times a\times 8.6^2\\\Rightarrow a=\frac{400\times 2}{8.6^2}\\\Rightarrow a=10.81665\ m/s^2

When divided by g

\dfrac{a}{g}=\dfrac{10.81665}{9.81}\\\Rightarrow \dfrac{a}{g}=1.10261\\\Rightarrow a=1.10261g

The acceleration measures 1.10261 times that of g

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 10.81665\times 1.6\times 10^3+0^2}\\\Rightarrow v=186.04644\ m/s

Expressed in mph

186.04644\times \dfrac{3600}{1609.34}=416.17506\ mph

The dragster's speed reaches 416.17506 mph

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Response:

The resistivity can be expressed as \rho = 2.5 *10^{-8} \ \Omega \cdot m

Clarification:

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When tension is doubled:

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