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Sergeu [11.5K]
3 years ago
14

What does the area between the line and the x axis represent on a velocity vs time graph

Physics
1 answer:
pychu [463]3 years ago
5 0

the area bounded by the line and the axes of a velocity-time graph is equal to the displacement of an object during that particular time period

Thank you

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The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50∘C is 2000 m/s. Note that 1.0 mol of diatomic hydrogen at
denis-greek [22]

Answer:

A) d. (1/4)(2000m/s) = 500 m/s

B) c. 4000 J

C) f. None of the above (2149.24 m/s)

Explanation:

A)

The translational kinetic energy of a gas molecule is given as:

K.E = (3/2)KT

where,

K = Boltzman's Constant = 1.38 x 1^-23 J/K

T = Absolute Temperature

but,

K.E = (1/2) mv²

where,

v = root mean square velocity

m = mass of one mole of a gas

Comparing both equations:

(3/2)KT = (1/2) mv²

v = √(3KT)/m  _____ eqn (1)

<u>FOR HYDROGEN:</u>

v = √(3KT)/m = 2000 m/s  _____ eqn (2)

<u>FOR OXYGEN:</u>

velocity of oxygen = √(3KT)/(mass of oxygen)  

Here,

mass of 1 mole of oxygen = 16 m

velocity of oxygen = √(3KT)/(16 m)

velocity of oxygen = (1/4) √(3KT)/m

using eqn (2)

<u>velocity of oxygen = (1/4)(2000 m/s) = 500 m/s</u>

B)

K.E = (3/2)KT

Since, the temperature is constant for both gases and K is also a constant. Therefore, the K.E of both the gases will remain same.

K.E of Oxygen = K.E of Hydrogen

<u>K.E of Oxygen = 4000 J</u>

C)

using eqn (2)

At, T = 50°C = 323 k

v = √(3KT)/m = 2000 m/s

m = 3(1.38^-23 J/k)(323 k)/(2000 m/s)²

m = 3.343 x 10^-27 kg

So, now for this value of m and T = 100°C = 373 k

v = √(3)(1.38^-23 J/k)(373 k)/(3.343 x 10^-27 kg)

<u>v = 2149.24 m/s</u>

<u></u>

8 0
3 years ago
Which statement accurately represents the arrangement of electrons in Bohr’s atomic model?
noname [10]

Answer: Electrons move around the nucleus in fixed orbits of equal levels of energy

Explanation:

The statement that accurately represents the arrangement of electrons in Bohr’s atomic model is that the electrons move around the nucleus in fixed orbits of equal levels of energy.

It should be noted that the electrons have a fixed energy level when they travel around the nucleus in with energies which varies for different levels.

Higher energy levels are depicted by the orbits that are far from the nucleus. There's emission of light when the electrons then return back to a lower energy level.

8 0
3 years ago
Consider a cyclotron in which a beam of particles of positive charge q and mass m is moving along a circular path restricted by
Ulleksa [173]

A) v=\sqrt{\frac{2qV}{m}}

B) r=\frac{mv}{qB}

C) T=\frac{2\pi m}{qB}

D) \omega=\frac{qB}{m}

E) r=\frac{\sqrt{2mK}}{qB}

Explanation:

A)

When the particle is accelerated by a potential difference V, the change (decrease) in electric potential energy of the particle is given by:

\Delta U = qV

where

q is the charge of the particle (positive)

On the other hand, the change (increase) in the kinetic energy of the particle is (assuming it starts from rest):

\Delta K=\frac{1}{2}mv^2

where

m is the mass of the particle

v is its final speed

According to the law of conservation of energy, the change (decrease) in electric potential energy is equal to the increase in kinetic energy, so:

qV=\frac{1}{2}mv^2

And solving for v, we find the speed v at which the particle enters the cyclotron:

v=\sqrt{\frac{2qV}{m}}

B)

When the particle enters the region of magnetic field in the cyclotron, the magnetic force acting on the particle (acting perpendicular to the motion of the particle) is

F=qvB

where B is the strength of the magnetic field.

This force acts as centripetal force, so we can write:

F=m\frac{v^2}{r}

where r is the radius of the orbit.

Since the two forces are equal, we can equate them:

qvB=m\frac{v^2}{r}

And solving for r, we find the radius of the orbit:

r=\frac{mv}{qB} (1)

C)

The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.

It can be calculated as the ratio between the length of the circumference (2\pi r) and the velocity of the particle (v):

T=\frac{2\pi r}{v} (2)

From eq.(1), we can rewrite the velocity of the particle as

v=\frac{qBr}{m}

Substituting into(2), we can rewrite the period of revolution of the particle as:

T=\frac{2\pi r}{(\frac{qBr}{m})}=\frac{2\pi m}{qB}

And we see that this period is indepedent on the velocity.

D)

The angular frequency of a particle in circular motion is related to the period by the formula

\omega=\frac{2\pi}{T} (3)

where T is the period.

The period has been found in part C:

T=\frac{2\pi m}{qB}

Therefore, substituting into (3), we find an expression for the angular frequency of motion:

\omega=\frac{2\pi}{(\frac{2\pi m}{qB})}=\frac{qB}{m}

And we see that also the angular frequency does not depend on the velocity.

E)

For this part, we use again the relationship found in part B:

v=\frac{qBr}{m}

which can be rewritten as

r=\frac{mv}{qB} (4)

The kinetic energy of the particle is written as

K=\frac{1}{2}mv^2

So, from this we can find another expression for the velocity:

v=\sqrt{\frac{2K}{m}}

And substitutin into (4), we find:

r=\frac{\sqrt{2mK}}{qB}

So, this is the radius of the cyclotron that we must have in order to accelerate the particles at a kinetic energy of K.

Note that for a cyclotron, the acceleration of the particles is achevied in the gap between the dees, where an electric field is applied (in fact, the magnetic field does zero work on the particle, so it does not provide acceleration).

6 0
3 years ago
What force is required to accelerate a body with a mass of 15 kilograms at a rate of 8 m/s²?
mihalych1998 [28]

Newton's Second law of motion: 

                  Force = (mass) x (acceleration)

                  Force = (15kg) x (8m/s²) = 120 kg-m/s² = 120 newtons
6 0
3 years ago
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What are symptoms of Gomorrah
marishachu [46]
Consumed by Fire and Brimstone would be 1 symptom of Gomorrah lol
5 0
3 years ago
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