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ArbitrLikvidat [17]
4 years ago
8

Gaseous helium is in thermal equilibrium with liquid helium at 6.4 K. The mass of a helium atom is 6.65 × 10−27 kg and Boltzmann

’s constant is 1.38066 × 10−23 J/K. Determine the most probable speed of a helium atom. Answer in units of m/s.
Physics
1 answer:
chubhunter [2.5K]4 years ago
4 0

Answer:

162.78 m/s is the most probable speed of a helium atom.

Explanation:

The most probable speed:

v_{mp}=\sqrt{\frac{2K_bT}{m}}

K_b= Boltzmann’s constant =1.38066\times 10^{-23} J/K

T = temperature of the gas

m = mass of the gas particle.

Given, m = 6.65\times 10^{-27} kg

T = 6.4 K

Substituting all the given values :

v_{mp}=\sqrt{\frac{2\times 1.38066\times 10^{-23} J/K\times 6.4 K}{6.65\times 10^{-27} kg}}

v_{mp}=162.78 m/s

162.78 m/s is the most probable speed of a helium atom.

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A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0
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The given question is incomplete. The complete question is as follows.

A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 m/s each second. The push force has a horizontal component of 20 N and a vertical component of 25 N downward. Calculate the coefficient of kinetic friction between the box and the floor.

Explanation:

The given data is as follows.

    F_{1} = 20 N, F_{2} = 25 N, a = -0.9 m/s^{2}

             W = 83 N

         m = \frac{83}{9.81}

             = 8.46

Now, we will balance the forces along the y-component as follows.

       N = W + F_{2}

           = 83 + 25 = 108 N

Now, balancing the forces along the x component as follows.

       F_{1} - F_{r} = ma

        20 - F_{r} = 8.46 \times (-0.9)

             F_{r} = 7.614 N

Also, we know that relation between force and coefficient of friction is as follows.

             F_{r} = \mu \times N

          \mu = \frac{F_{r}}{N}

                    = \frac{7.614}{108}

                    = 0.0705

Thus, we can conclude that the coefficient of kinetic friction between the box and the floor is 0.0705.

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Epsilon Eridani is a star that is 10.5 light years away from Earth. One light year is about 9.46 trillion kilometers The distanc
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Given the relation that 1 light year = 9.46 trillion kilometers, and the distance from Epsilon Eridani to Earth which is 10.5 light years, we can solve for the distance between the star and Earth in light years by multiplying both values together. This is shown below

Dsitance (in trillion km) = 9.46 trillion km * (10.5) = 99.33 trillion kilometers

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A huge cannon is assembled on an airless planet (ignore any effects due to the planet's rotation). The planet has a radius of 5.
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Answer:

The projectile's speed as it passes the satellite is 1497.8 m/s.

Explanation:

Given that,

Radius of planet r=5.00\times10^{6}\ m

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Speed = 2000 m/s

Height = 1000 km

We need to calculate the projectile's speed as it passes the satellite

Using conservation of energy

E_{1}=E_{2}

\dfrac{1}{2}mv_{1}^2+\dfrac{GmM}{r_{1}}=\dfrac{1}{2}mv_{2}^2+\dfrac{GmM}{R+h}

\dfrac{v_{1}^2}{2}+\dfrac{GM}{r_{1}}=\dfrac{v_{2}^2}{2}+\dfrac{GM}{R+h}

-\dfrac{v_{2}^2}{2}=-(\dfrac{GM}{R}-\dfrac{GM}{R+h}-\dfrac{v_{1}^2}{2})

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