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serg [7]
2 years ago
5

A runner of mass 56.0kg runs around the edge of a horizontal turntable mounted on a vertical, frictionless axis through its cent

er. The runner's velocity relative to the earth has magnitude 3.10m/s . The turntable is rotating in the opposite direction with an angular velocity of magnitude 0.210rad/s relative to the earth. The radius of the turntable is 3.10m , and its moment of inertia about the axis of rotation is 81.0kg?m2
Question: Find the final angular velocity of the system if the runner comes to rest relative to the turntable. (You can treat the runner as a particle.)
Physics
1 answer:
SCORPION-xisa [38]2 years ago
5 0

Answer: -0.84 rad/sec (clockwise)

Explanation:

Assuming no external torques act on the system (man + turntable), total angular momentum must be conserved:

L1 = L2

L1 = It ω + mm. v . r = 81.0 kg . m2 .21 rad/s – 56.0 kg. 3.1m/s . 3.1 m  

L1 = -521.15 kg.m2/sec (1)

(Considering to the man as a particle that is moving opposite to the rotation of  the turntable, so the sign is negative).

Once at rest, the runner is only a point mass with a given rotational inertia respect from the axis of rotation, that can be expressed as follows:

Im = m. r2 = 56.0 kg. (3.1m)2 = 538.16 kg.m2

The total angular momentum, once the runner has come to an stop, can be written as follows:

L2= (It + Im) ωf = -521.15 kg.m2/sec  

L2= (81.0 kg.m2 + 538.16 kg.m2) ωf = -521.15 kg.m2/sec  

Solving for ωf, we get:

ωf = -0.84 rad/sec  (clockwise)

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Two rigid tanks of equal size and shape are filled with different gases. The tank on the left contains oxygen, and the tank on t
fredd [130]

Answer:

The number of oxygen molecules in the left container greater than the number of hydrogen molecules in the right container.

Explanation:

Given:

Molar mass of oxygen, M_O=32

Molar mass of hydrogen, M_H=2

We know ideal gas law as:

PV=nRT

where:

P = pressure of the gas

V = volume of the gas

n= no. of moles of the gas molecules

R = universal gs constant

T = temperature of the gas

∵n=\frac{m}{M}

where:

m = mass of gas in grams

M = molecular mass of the gas

∴Eq. (1) can be written as:

PV=\frac{m}{M}.RT

P=\frac{m}{V}.\frac{RT}{M}

        as: \frac{m}{V}=\rho\ (\rm density)

So,

P=\rho.\frac{RT}{M}

Now, according to given we have T,P,R same for both the gases.

P_O=P_H

\rho_O.\frac{RT}{M_O}=\rho_H.\frac{RT}{M_H}

\Rightarrow \frac{\rho_O}{32}=\frac{\rho_H}{2}

\rho_O=16\rho_H

∴The molecules of oxygen are more densely packed than the molecules of hydrogen in the same volume at the same temperature and pressure. So, <em>the number of oxygen molecules in the left container greater than the number of hydrogen molecules in the right container.</em>

5 0
3 years ago
What are the only elements that exist in nature as isolated atoms
agasfer [191]

Answer:

The only such elements are the Noble Gases (He, Ne, Ar, Kr, Xe, Rn)

(that is helium, neon, argon, krypton, xenon and radon)

Term: Monoatomic

Explanation:

6 0
3 years ago
The following equation is an example of decay?<br><br> 232/90 TH---4/2 HE +228/88 RA?
DaniilM [7]

Answer:

Alpha decay

Explanation:

  • Alpha decay is one of the three major types of decays, others being, beta decay and gamma decay.
  • <em><u>When a radioactive isotope undergoes alpha decay it emits alpha particles. An alpha particle is equivalent to the nucleus of Helium atom.</u></em>
  • <em><u>Therefore, an atom undergoing decay, its atomic mass is decreased by 4 and its atomic number is decreased by 2. </u></em>
  • Thus, since 232/90 Th, has undergone alpha decay its mass number is reduced by 4 to 228 and its atomic number by 2 to 88, and becomes 228/88 Ra.
5 0
2 years ago
Which statement correctly explains molecular motion in different states of matter using the kinetic theory?
Gelneren [198K]

the answer is the second one

8 0
3 years ago
Two astronauts, each with a mass of 50 kg, are connected by a 7 m massless rope. Initially they are rotating around their center
kiruha [24]

Answer:

The angular  velocity is w_f =  1.531 \ rad/ s

Explanation:

From the question we are told that

     The mass of each astronauts is  m =  50 \ kg

      The initial  distance between the two  astronauts  d_i  =  7 \  m

Generally the radius is mathematically represented as r_i  =  \frac{d_i}{2} = \frac{7}{2}  =  3.5 \  m

      The initial  angular velocity is  w_1 = 0.5 \  rad /s

       The  distance between the two astronauts after the rope is pulled is d_f =  4 \  m

Generally the radius is mathematically represented as r_f  =  \frac{d_f}{2} = \frac{4}{2}  =  2\  m

Generally from the law of angular momentum conservation we have that

           I_{k_1} w_{k_1}+ I_{p_1} w_{p_1} = I_{k_2} w_{k_2}+ I_{p_2} w_{p_2}

Here I_{k_1 } is the initial moment of inertia of the first astronauts which is equal to I_{p_1} the initial moment of inertia of the second astronauts  So

      I_{k_1} = I_{p_1 } =  m *  r_i^2

Also   w_{k_1 } is the initial angular velocity of the first astronauts which is equal to w_{p_1} the initial angular velocity of the second astronauts  So

      w_{k_1} =w_{p_1 } = w_1

Here I_{k_2 } is the final moment of inertia of the first astronauts which is equal to I_{p_2} the final moment of inertia of the second astronauts  So

      I_{k_2} = I_{p_2} =  m *  r_f^2

Also   w_{k_2 } is the final angular velocity of the first astronauts which is equal to w_{p_2} the  final angular velocity of the second astronauts  So

      w_{k_2} =w_{p_2 } = w_2

So

      mr_i^2 w_1 + mr_i^2 w_1 = mr_f^2 w_2 + mr_f^2 w_2

=>   2 mr_i^2 w_1 = 2 mr_f^2 w_2

=>   w_f =  \frac{2 * m * r_i^2 w_1}{2 * m *  r_f^2 }

=>    w_f =  \frac{3.5^2 *  0.5}{  2^2 }

=>   w_f =  1.531 \ rad/ s

       

3 0
2 years ago
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