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2 years ago

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An airplane flying at 115 m/s due east makes a gradual turn following a circular path to fly south. The turn takes 15 seconds to

complete. What is the magnitude of the centripetal acceleration during the turn?
a. 12 m/s2
b. 9.8 m/s2
c. 0 m/s2
d. 6.9 m/s
e. 8.1 m/s2

1 answer:

ankoles [38]2 years ago

5 0

Answer:

The magnitude of the centripetal acceleration during the turn is $a=12.04\ m/s^2.$

Explanation:

Given :

Speed to the airplane in circular path , v = 115 m/s.

Time taken by plane to turn , t= 15 s.

Also , the plane turns from east to south i.e. quarter of a circle .

Therefore, time taken to complete whole circle is , $T=t\times 4=60\ s.$

Now , Velocity ,

$v=\dfrac{2\pi r}{T}\\\\115=\dfrac{2\times 3.14\times r}{60}\\\\r=1098.73\ m.$

Also , we know :

Centripetal acceleration ,

$a=\dfrac{v^2}{r}$

Putting all values we get :

$a=12.04\ m/s^2.$

Hence , this is the required solution .

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An ideal gas is contained in a vessel at 300 K. The temperature of the gas is then increased to 900 K. (i) By what factor does t

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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;

Explanation: (i) Kinetic energy for ideal gas is calculated as:

$KE=\frac{3}{2}nRT$

where

n is mols

R is constant of gas

T is temperature in Kelvin

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:

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

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

For rms, factor of change is $\sqrt{3}$

(iii) Average momentum change of molecule depends upon velocity:

q = m.v

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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Answer:

<em>a) 6738.27 J</em>

<em>b) 61.908 J</em>

<em>c) </em> $\frac{4492.18}{v_{car} ^{2} }$

<em></em>

Explanation:

The complete question is

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.

Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?

Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?

Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar. Enter an expression for the mass of the car, in terms of the quantities defined here.

moment of inertia is given as

$I$ = $\frac{1}{2}$ $mr^{2}$

where m is the mass of the flywheel,

and r is the radius of the flywheel

for the flywheel with radius 1.1 m

and mass 11 kg

moment of inertia will be

$I$ = $\frac{1}{2}$ $*11*1.1^{2}$ = 6.655 kg-m^2

The maximum speed of the flywheel = 35 m/s

we know that v = ωr

where v is the linear speed = 35 m/s

ω = angular speed

r = radius

therefore,

ω = v/r = 35/1.1 = 31.82 rad/s

maximum rotational energy of the flywheel will be

E = $Iw^{2}$ = 6.655 x $31.82^{2}$ = <em>6738.27 J</em>

<em></em>

b) second flywheel has

radius = 2.8 m

mass = 16 kg

moment of inertia is

$I$ = $\frac{1}{2}$ $mr^{2}$ = $\frac{1}{2}$ $*16*2.8^{2}$ = 62.72 kg-m^2

According to conservation of angular momentum, the total initial angular momentum of the first flywheel, must be equal to the total final angular momentum of the combination two flywheels

for the first flywheel, rotational momentum = $Iw$ = 6.655 x 31.82 = 211.76 kg-m^2-rad/s

for their combination, the rotational momentum is

$(I_{1} +I_{2} )w$

where the subscripts 1 and 2 indicates the values first and second flywheels

$(I_{1} +I_{2} )w$ = (6.655 + 62.72)ω

where ω here is their final angular momentum together

==> 69.375ω

Equating the two rotational momenta, we have

211.76 = 69.375ω

ω = 211.76/69.375 = 3.05 rad/s

Therefore, the energy stored in the first flywheel in this situation is

E = $Iw^{2}$ = 6.655 x $3.05^{2}$ = <em>61.908 J</em>

<em></em>

<em></em>

c) one third of the initial energy of the flywheel is

6738.27/3 = 2246.09 J

For the car, the kinetic energy = $\frac{1}{2}mv_{car} ^{2}$

where m is the mass of the car

$v_{car}$ is the velocity of the car

Equating the energy

2246.09 = $\frac{1}{2}mv_{car} ^{2}$

making m the subject of the formula

mass of the car m = $\frac{4492.18}{v_{car} ^{2} }$

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