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anastassius [24]
2 years ago
4

It's possible for a determined group of people to pull an aircraft. Drag is negligible at low speeds, and the only force impedin

g motion is the rolling friction of the rubber tires on the concrete runway. In 2000, a team of 60 British police officers set a world record by pulling a Boeing 747, with a mass of 200,000 kg, a distance of 100 m in 53 s. The plane started at rest. Suppose that μr=0.02.
Estimate the force with which each officer pulled on the plane, assuming constant pulling force and constant acceleration.
Physics
1 answer:
harina [27]2 years ago
7 0

Answer:

867 N

Explanation:

Given:

Mass of the aircraft (m) = 200,000 kg

Displacement of the aircraft (d) = 100 m

Time taken (t) = 53 s

Coefficient of friction between the tires and concrete (μ) = 0.02

Number of police officers (n) = 60

Initial velocity of aircraft (u) = 0 m/s

Now, using equation of motion, we can find the acceleration of the aircraft.

The equation of motion is given as:

d=ut+\frac{1}{2}at^2

Here, 'a' is the acceleration of the aircraft.

Plug in the given values and solve for 'a'. This gives,

100=0+\frac{1}{2}a(53)^2\\\\a=\frac{200}{2809}= 0.07\ m/s^2

Now, the frictional force is given as:

f=\mu N=\mu mg=0.02\times 200000\times 9.8=39200\ N

Now, let the force of pull of one police be 'F'.

So, the total force of pull of 60 officers = 60F

Now, the net force acting on the aircraft is given as:

F_{net}=60F-f\\\\F_{net}=60F-39200

From Newton's second law, the net force acting on the aircraft is equal to the product of mass and acceleration. So,

F_{net}=ma\\\\60F-39200=200000\times 0.07\\\\60F=14000+39200\\\\60F=53200\\\\F=\frac{53200}{60}=866.67\approx 867\ N

Therefore, the force with which each officer pulled on the plane is 867 N.

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The question is missing some parts. Here is the complete question.

An ideal gas is contained in a vessel at 300K. The temperature of the gas is then increased to 900K.

(i) By what factor does the average kinetic energy of the molecules change, (a) a factor of 9, (b) a factor of 3, (c) a factor of \sqrt{3}, (d) a factor of 1, or (e) a factor of \frac{1}{3}?

Using the same choices in part (i), by what factor does each of the following change: (ii) the rms molecular speed of the molecules, (iii) the average momentum change that one molecule undergoes in a colision with one particular wall, (iv) the rate of collisions of molecules with walls, and (v) the pressure of the gas.

Answer: (i) (b) a factor of 3;

              (ii) (c) a factor of \sqrt{3};

              (iii) (c) a factor of \sqrt{3};

             (iv) (c) a factor of \sqrt{3};

              (v) (e) a factor of 3;

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n is mols

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As you can see, kinetic energy and temperature are directly proportional: when tem perature increases, so does energy.

So, as temperature of an ideal gas increased 3 times, kinetic energy will increase 3 times.

For temperature and energy, the factor of change is 3.

(ii) Rms is root mean square velocity and is defined as

V_{rms}=\sqrt{\frac{3k_{B}T}{m} }

Calculating velocity for each temperature:

For 300K:

V_{rms1}=\sqrt{\frac{3k_{B}300}{m} }

V_{rms1}=30\sqrt{\frac{k_{B}}{m} }

For 900K:

V_{rms2}=\sqrt{\frac{3k_{B}900}{m} }

V_{rms2}=30\sqrt{3}\sqrt{\frac{k_{B}}{m} }

Comparing both veolcities:

\frac{V_{rms2}}{V_{rms1}}= (30\sqrt{3}\sqrt{\frac{k_{B}}{m} }) .\frac{1}{30} \sqrt{\frac{m}{k_{B}} }

\frac{V_{rms2}}{V_{rms1}}=\sqrt{3}

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Since velocity has a factor of \sqrt{3} and velocity and momentum are proportional, average momentum change increase by a factor of

(iv) Collisions increase with increase in velocity, which increases with increase of temperature. So, rate of collisions also increase by a factor of \sqrt{3}.

(v) According to the Pressure-Temperature Law, also known as Gay-Lussac's Law, when the volume of an ideal gas is kept constant, pressure and temperature are directly proportional. So, when temperature increases by a factor of 3, Pressure also increases by a factor of 3.

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