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3 years ago

For a monatomic ideal gas, temperature is proportional to : the square of the average atomic velocity.

Physics

1 answer:

Alex777 [14]3 years ago

3 0

Answer:

the square of the average atomic velocity.

Explanation:

From the formulas for kinetic energy and temperature for a monoatomic gas, which has three translational degrees of freedom, the relationship between root mean square velocity and temperature is as follows:

$v_{rms}=\sqrt{\frac{3RT}{M}}$ (1)

Where $v_{rms}$ is the root mean square velocity, M is the molar mass of the gas, R is the universal constant of the ideal gases and T is the temperature.

The root mean square velocity is a measure of the velocity of the particles in a gas. It is defined as the square root of the mean square velocity of the gas molecules:

$v_{rms}=\sqrt{}$ (2)

substituting 2 in 1, we find the relationship between mean square speed and temperature:

$\sqrt{}=\sqrt{\frac{3RT}{M}}\\T=\frac{M}{3R}\\\\T\sim$

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A 80kg astronaut is training in human centrifuge to prepare for a launch. The astronaut uses the centrifuge to practice having a

inessss [21]

The answers on the model of the human centrifuge ready for the launch to each question of the statement are listed below:

a) A force of 2479.210 newtons is acting on the astronaut's back.

b) A <em>net centripetal</em> force of 2479.210 newtons is acting on the astronaut.

c) The <em>centripetal</em> acceleration of the astronaut is 30.990 meters per square second.

d) The astronaut has a <em>linear</em> speed of approximately 19.284 meters per second.

e) The <em>angular</em> speed of the astronaut is 1.607 radians per second (15.346 revolutions per minute).

<h3>How to apply Newton's laws to analyze a process in a human centrifuge training</h3>

The human centrifuge experiments a <em>centripetal</em> acceleration when it reaches a <em>peak</em> angular speed. In this question we must apply Newton's laws of motion and concepts of <em>centripete</em> and <em>centrifugal</em> forces to answer the questions. Now we proceed to answer the questions:

<h3>How much force is acting on the astronaut's back?</h3>

By the third Newton's law the astronaut experiments a <em>rection</em> force (<em>F</em>), in newtons, which has the same magnitude to <em>centrifugal</em> force but opposed to that force. The magnitude of the force acting on the back of the astronaut is equal to:

$F = 3.16\cdot (80\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$F = 2479.210\,N$

A force of 2479.210 newtons is acting on the astronaut's back. $\blacksquare$

<h3>What is the net centripetal force on the astronaut?</h3>

By the second and third Newton's laws we know that the <em>net centripetal</em> force on the astronaut is equal to the magnitude of the force found in the previous question. Thus, a <em>net centripetal</em> force of 2479.210 newtons is acting on the astronaut. $\blacksquare$

<h3>What is the astronaut's centripetal acceleration?</h3>

The centripetal acceleration of the astronaut (<em>a</em>), in meters per square second, is found by dividing the result of the previous question by the mass of the astronaut (<em>m</em>), in kilograms:

$a = \frac{F}{m}$ (1)

If we know that F = 2479.210 newtons and m = 80 kilograms, then the centripetal acceleration of the astronaut is:

$a = \frac{2479.210\,N}{80\,kg}$

$a = 30.990\,\frac{m}{s^{2}}$

The <em>centripetal</em> acceleration of the astronaut is 30.990 meters per square second. $\blacksquare$

<h3>What is the astronaut's linear speed?</h3>

By definition of <em>uniform circular</em> motion, we have the following formula for the <em>linear</em> velocity of the astronaut (<em>v</em>):

$v = \sqrt{a\cdot r}$ (1)

Where <em>r</em> is the radius of the human centrifuge, in meters.

If we know that $a = 30.990\,\frac{m}{s^{2}}$ and $r = 12\,m$ , then linear velocity of the astronaut is:

$v = \sqrt{\left(30.990\,\frac{m}{s^{2}} \right)\cdot (12\,m)}$

The astronaut has a <em>linear</em> speed of approximately 19.284 meters per second. $\blacksquare$

<h3>What is the astronaut's angular speed? </h3>

The <em>angular</em> speed of the astronaut (ω), in radians per second, is found by the following <em>kinematic</em> relationship:

$\omega = \frac{v}{R}$ (1)

If we know that <em>v ≈ 19.284 m/s</em> and <em>R = 12 m</em>, then the angular speed is:

$\omega = \frac{19.284\,\frac{m}{s} }{12\,m}$

The <em>angular</em> speed of the astronaut is 1.607 radians per second (15.346 revolutions per minute). $\blacksquare$

To learn more on centripetal forces, we kindly invite to check this verified question: brainly.com/question/11324711

6 0

2 years ago

Now in "real life," this automobile is cruising at 20.5 m/s (equal to 73.8 km/hr) when it is about to hit a pedestrian stuck in

algol13

Answer:

He needs 1.53 seconds to stop the car.

Explanation:

Let the mass of the car is 1500 kg

Speed of the car, v = 20.5 m/s

He will not push the car with a force greater than, $F=2\times 10^4\ N$

The impulse delivered to the object is given by the change in momentum as :

$F\times t=mv\\\\t=\dfrac{mv}{F}\\\\t=\dfrac{1500\times 20.5}{2\times 10^4}\\\\t=1.53\ s$

So, he needs 1.53 seconds to stop the car. Hence, this is the required solution.

5 0

4 years ago

At a given instant in time, two cars are traveling at different velocities, one twice as large as the other. Based on this infor

Sati [7]

Answer: No, is not possible.

Explanation:

We have two cars traveling at different velocities, one with velocity v and other with velocity 2*v

We want to know of it is possible to say which of these two cars has the larger acceleration at this instant of time.

Acceleration is the rate of change of the velocity, for example, if acceleration = 0, the velocity does not change.

Notice that knowing only the velocity at a given instant of time, we do not have any way of knowing the acceleration of each car.

Such that acceleration is given by the second Newton's equation:

F = m*a

force equals mass times acceleration.

We do not know any of these quantities for the cars, so we can not say anything about the accelerations.

We also could know the average acceleration if we know the change in velocity over an interval of time, such that:

Av acceleration = (change in velocity)/(interval of time)

But here we do not have a change in velocity, so again, we can't say anything about the acceleration.

3 0

3 years ago

What is the moment of inertia of a cube with mass M=0.500kg and side lengths s=0.030m about an axis which is both normal (perpen

prisoha [69]

Answer:

The moment of inertia I is

I = 2.205x10^-4 kg/m^2

Explanation:

Given mass m = 0.5 kg

And side lenght = 0.03 m

Moment of inertia I = mass x radius of rotation squared

I = mr^2

In this case, the radius of rotation is about an axis which is both normal (perpendicular) to and through the center of a face of the cube.

Calculating from the dimensions of the the box as shown in the image below, the radius of rotation r = 0.021 m

Therefore,

I = 0.5 x 0.021^2 = 2.205x10^-4 kg/m^2

8 0

4 years ago

A block (mass m1) lying on a frictionless inclined plane is connected to a mass (mass m2) by a massless cord passing over a pull

azamat

Answer:

a) System aceleration:

$a=\frac{g(m_2-m_1sin(\theta))}{m_1+m_2}$

b) Direction of movement:

The block $m_1$ down the plane when the acceleration is negative. This occur when:

$m_2-m_1sin(\theta)$

The block $m_2$ up the plane when the acceleration is positive. This occur when:

$m_2-m_1sin(\theta)>0$

Explanation:

For the block $m_1$ the move direction is parallel (||) to the plane

$\sum F_{||}=m_1a=T-sin(\theta)mg$ (1)

For the block $m_2$ the move direction is vertical (y)

$\sum F_y=m_2a=m_2g-T$ (2)

Both blocks are connected with the same cable, therefore, the tension on the cable and the acceleration is the same for both.

Solving the equation 2 for T:

$T=m_2g-m_2a$ (3)

replacing (3) in the equation (1)

$m_2g-m_2a- m_1gsin(\theta)=m_1a$ (4)

solving (4) for a:

$a=\frac{g(m_2-m_1sin(\theta))}{m_1+m_2}$

8 0

3 years ago