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SVEN [57.7K]
2 years ago
10

Consider two ideal gases, A and B, at the same temperature. The rms speed of the molecules of gas A is twice that ofgas B. How d

oes the molecular mass of A compare to that of B?a. The molecular mass of A is one half that of B.b. The molecular mass of A is one fourth that of B.c. The molecular mass of A is 1.4 times that of B.d.The molecular mass of A is four times that of B.e. The molecular mass of A is twice that of B.
Physics
1 answer:
katen-ka-za [31]2 years ago
8 0

The rms speed of the molecules of gas A is twice that of gas B. The molecular mass of A is one fourth to that of B.

Answer: Option B

<u>Explanation:</u>

Measuring the speed of particles at a given point in time results in a large distribution of values. Some molecules can move very slowly, others very fast, and because they are still moving in different directions, the speeds may be zero. (Velocity, vector quantity that corresponds to the speed and direction of the molecule.)

To correctly estimate the average velocity, you must take the squares of the mean velocity and take the square root of this value. This is known as the root mean square (rms) velocity and is shown as follows:

                 V_{r m s}=\sqrt{\frac{3 R T}{M}}

Where,

M – Gas’s molar mass

R – Molar mass constant

T – Temperature (in Kelvin)

Given data is rms speed for gas molecule A is twice that of gas molecule B. So,

                 \left(V_{r m s}\right)_{A}=2\left(V_{r m s}\right)_{B}

Therefore, equating the molecule’s rms speed formula for both A and B,

                  \sqrt{\frac{3 R T}{M_{A}}}=2(\sqrt{\frac{3 R T}{M_{B}}})

On squaring both sides, we get,

                 \frac{3 R T}{M_{A}}=4\left(\frac{3 R T}{M_{B}}\right)

By solving the above equations, we get,

                 M_{A}=\frac{M_{B}}{4}

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An airplane accelerates from a velocity of 22 m/s to 40 m/s with an acceleration of 2 m/s2. How long does it
andrey2020 [161]

The time it takes the plane to change its velocity is 9s.

<h3>What is time?</h3>

Time can be defined the measured or measurable period during which an action, process, or condition exists or continues.

To calculate the time it takes the airplane to change its velocity, we use the formula below.

Formula:

  • t = (v-u)/a.......... Equation 1

Where:

  • a = Acceleration
  • v = Final velocity
  • u = Initial velocity
  • t = time

From the question,

  • v = 40 m/s
  • u = 22 m/s
  • a = 2 m/s²

Substitute these values into equation  1

  • t = (40-22)/2
  • t = 18/2
  • t = 9s

Hence, the time it takes the plane to change its velocity is 9s.

Learn more about time here: brainly.com/question/2854969

4 0
1 year ago
A baseball is hit that just goes over a wall that is 45.4m high. If the baseball is traveling at 46.2 m/s at an angle of 32.7° b
mario62 [17]

Answer:

54.9 m/s at 44.9 degrees

Explanation:

If the ball has a total velocity of 46.2 m/s, at an angle of -32.7 degrees, we can decompose its speed into its horizontal and vertical components.

Vx = V * cos(a) = 46.2 * cos(-32.7) = 38.9 m/s

Vy = V * sin(a) = 46.2 * sin(-32.7) = -25 m/s

SInce there is no force on the horizontal direction (omitting air drag), we can assume constant horizontal speed.

Since a ball thrown is at free fall, only affected by gravity (omitting air drag), we can say it is affected by constant acceleration, therefore we can use

Y(t) = Y0 + Vy0 *t + 1/2 * a * t^2

We consider t=0 as the moment when the ball was hit, so in this case Y0 = 1 m

If we take the first derivative of the equation of position, we get the equation for speed

V(t) = Vy0 + a * t

We know that being t2 the moment the ball goes over the wall

V(t2) = -25 m/s

Y(t2) = 45.4 m

So:

45.4 = 1 + Vy0 * t2 + 1/2 * a * t2^2

-25 = Vy0 + a * t2

Then:

Vy0 = -25 - a * t2

So:

45.4 = 1 + (-25 - a * t2) * t2 + 1/2 * a * t2^2

0 = -44.4 - 25 * t2 - 1/2 * a * t2^2

a = -9.81 m/s^2

0 = -44.4 - 25 * t2 + 4.9 * t2^2

Solving this quadratic equation we get:

t1 = -1.39 s

t2 = 6.5 s

Since we are looking for a positive value we disregard t1.

Now we can obtain Vy0:

Vy0 = -25 + 9.81 * 6.5 = 38.76 m/s

Since horizontal speed is constant Vx0 = 38.9 m/s

By Pythagoras theorem we obtain the value of the initial speed:

V0 = \sqrt{Vx0^2 + Vy0^2} = \sqrt{38.9^2 + 38.76^2} = 54.9 m/s

The angle is in the the first quadrant because both comonents ate positive, so: 0 < a < 90

a = atan(Vy0/Vx0) = 44.9 degrees

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3 years ago
What is the principle of potentiometer?​
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Answer:

The principle of a potentiometer is that the potential dropped across a segment of a wire of uniform cross-section carrying a constant current is directly proportional to its length. The potentiometer is a simple device used to measure the electrical potentials (or compare the e.m.f of a cell).

Explanation:

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True or False? Materials that are good conductors of heat are usually poor conductors of electricity.
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False

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lanet R47A is a spherical planet where the gravitational acceleration on the surface is 3.45 m/s2. A satellite orbitsPlanet R47A
qaws [65]

2.6×10^6\:\text{m}

Explanation:

The acceleration due to gravity g is defined as

g = G\dfrac{M}{R^2}

and solving for R, we find that

R = \sqrt{\dfrac{GM}{g}}\:\:\:\:\:\:\:(1)

We need the mass M of the planet first and we can do that by noting that the centripetal acceleration F_c experienced by the satellite is equal to the gravitational force F_G or

F_c = F_G \Rightarrow m\dfrac{v^2}{r} = G\dfrac{mM}{r^2}\:\:\:\:\:(2)

The orbital velocity <em>v</em> is the velocity of the satellite around the planet defined as

v = \dfrac{2\pi r}{T}

where <em>r</em><em> </em>is the radius of the satellite's orbit in meters and <em>T</em> is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

\dfrac{4\pi^2 r}{T^2} = G\dfrac{M}{r^2}

Solving for <em>M</em>, we get

M = \dfrac{4\pi^2 r^3}{GT^2}

Putting this expression back into Eqn(1), we get

R = \sqrt{\dfrac{G}{g}\left(\dfrac{4\pi^2 r^3}{GT^2}\right)}

\:\:\:\:=\dfrac{2\pi}{T}\sqrt{\dfrac{r^3}{g}}

\:\:\:\:=\dfrac{2\pi}{(1.44×10^4\:\text{s})}\sqrt{\dfrac{(5×10^6\:\text{m})^3}{(3.45\:\text{m/s}^2)}}

\:\:\:\:= 2.6×10^6\:\text{m}

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3 years ago
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