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4 years ago

7

the sample of an ideal gas is in a tank of constant volume the sample absorbs heat energy so that its temperature changes from 3

00 to 600 K if v is the average speed of the gas molecules before the absorption of heat and v is their average speed after the absorption of heat what is the ratio of v2/v1

1 answer:

valkas [14]4 years ago

4 0

Answer:

Explanation

Given that

Temperature changes from

Initial temperature T1 = 300K

Final temperature T2 = 600K

v1 = average speed of gas molecule before absorption of heat

v2 = average speed of gas molecule after absorption of heat

Ratio v2/v1 =?

Related to average speed,

Vrms = √(3RT/M)

We will take 3, T and M as constant,

Then,

Vrms ∝ √T

Then,

Vrms / √T = k

So,

V1 / √T1 = V2 / √T2

Cross multiply

V1 • √T2 = V2 • √T1

Make V2 / V1 subject of formula

V2 / V1 = √T2 / √ T1

V2 / V1 = √600 / √300

We know from indices that

√a / √b = √(a/b)

Then,

V2 / V1 = √(600 / 300)

V2 / V1 = √2

Then, The ratio v2 / v1 is √2

So, if T is double, then, V increases by a factor of √2

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Ciara is swinging a 0.015 kg ball tied to a string around her head in a flat, horizontal circle. The radius of the circle is 0.5

Marat540 [252]

Answer:

Option C

Explanation:

From the question we are told that:

Mass $m=0.015kg$

Radius $r=0.5m$

Time $t=0.70$

Generally the equation for Tension is mathematically given by

$T=\frac{4mr}{t^2} \pi$

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Therefore

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Option C

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3 years ago

The idea is to get as much EMF produced from the sprinter running through it. If you were the Olympic coach on a year when there

never [62]

Answer:

the greater the speed, the greater the electromotive force

* The metal pole must be parallel to the field

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Explanation:

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* the greater the speed, the greater the electromotive force

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3 years ago

A playground merry-go-round has radius 2.10m and moment of inertia 2500kg*m^2 about a vertical axle through its center, and it t

vladimir2022 [97]

Answer

Radius of the wheel r = 2.1 m

Moment of inertia I = 2500 Kg m²

Tangential force applied F = 18 N

Time interval t = 16 s

Initial angular speed ω1 = 0

Final angular speed ω2 = ?

Let α be the angular acceleration.

Torque applied τ = Iα

F r = Iα

Angular acceleration α = F r/I

= $\dfrac{18\times 2.1}{2500}$

= 0.015 rad/s²

(a)From rotational kinematic relation

Final angular speed ω₂ = ω₁ + αt

= 0 + (0.015 rad/s^2 * 16 s)

= 0.24 rad/s

(b) Work done W = 0.5 Iω₂² - (1/2)Iω₁²

= 0.5*( 2500 Kg m²)(0.24 rad/s)^2 - 0

= 72 J

(c) Average power supplied by the child P = W/t = $\dfrac{72}{16}$

= 4.5 watt

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3 years ago

Five things that were invented by tesla?<br><br> please answer!!!!

algol13

Answer:

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2 years ago

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The 21 kg mass is attached by a cord to a mass hanging over the edge of the table. The frictional force between the mass and the

sasho [114]

Answer:

<em>147.3 N</em>

Explanation:

Two-Mass Systems

To solve a system where two masses are interacting with each other, we must set up the formulas by applying Newton's second law for each mass. Then, we find the required magnitudes by solving a system of equations.

Our system consists of a hanging object with mass $m_2$ attached to an object of mass $m_1$ lying in a table which applies a known friction force of $F_R=121 N$ . We also know the system accelerates at $0.73\ m/sec^2$ . The situation is pictured in the image below.

Analyzing the forces acting upon mass $m_1$ we have, in the horizontal axis, where movement is taking place:

$\displaystyle T-F_R=m_1\ a$

Where T is the rope's tension force. Now taking the vertical axis of the second mass, we have

$\displaystyle T-W_2=-m_2\ a$

The acceleration is negative since it's directed downwards, contrary to the positive default direction (right and up). Subtracting both equations:

$\displaystyle W_2-F_R=m_1\ a+m_2\ a$

Solving for $W_2$

$\displaystyle w_2=F_R+m_1\ a+m_2\ a$

We know that

$\displaystyle W_2=m_2\ g$

so, the above formula becomes

$\displaystyle m_2\ g=F_R+m_1\ a+m_2\ a$

Rearranging and factoring

$\displaystyle m_2(g-a)=F_R+m_1\ a$

Solving for $m_2$

$\displaystyle m_2=\frac{F_R+m_1\ a}{g-a}$

Let's use our known data

$\displaystyle m_2=\frac{121+21(0,73)}{9,8-0,73}=\frac{136.33}{9.07}$

$\displaystyle m_2=15.03\ kg$

Finally, we compute the object's weight

$\displaystyle W_2=m_2.g=15.03(9.8)=147.3\ N$

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3 years ago