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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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A boy is whirling a stone around his head by means of a string. The string makes one complete revolution every second; and the m
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Answer:

(A) The tension's magnitude grows to four times the initial value, 4F.

Explanation:

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                               v is the velocity or speed of the object

                               r signifies the radius of the circular path

Importantly, the tension corresponds to the centripetal force.

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Applying our formula:F =\frac { m{ v }^{ 2 } }{ r }

                               where F indicates the tension in the string

assuming the starting speed is v, after doubling it becomes 2v

Maintaining the circle's radius, we arrive at:

F=\frac { m{ (2v) }^{ 2 } }{ r } =\frac { 4m{ v }^{ 2 } }{ r }

From this equation, it's clear that the initial tension has quadrupled.

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What mass needs to be attached to a spring with a force constant of 7N/m in order to make a simple harmonic oscillator oscillate
Sav [3153]

Answer:

The mass will be 4.437 kg

Explanation:

The force constant k is given as 7 N/m

The time period of oscillation T is 5 sec

Thus, angular frequency \omega =\frac{2\pi }{T}=\frac{2\times 3.14}{5}=1.256rad/sec

It is known that angular frequency is computed via

\omega =\sqrt{\frac{k}{m}}

1.256 =\sqrt{\frac{7}{m}}

Squaring both sides gives us

1.577 =\frac{7}{m}

The mass equals 4.437 kg

6 0
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