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Rufina [12.5K]
3 years ago
11

A tube with sealed ends has some sand at one end. When the tube is turned upside down the sand falls 0.6m then settles down agai

n. The energy stays in the sand as thermal energy. Use g=9.81m/s2. 1) If the mass of the sand is 0.5kg, how much thermal energy is added on each turn? Increase of U=
Physics
2 answers:
barxatty [35]3 years ago
4 0

Answer:

2.943 J

Explanation:

Data:

acceleration due to gravity, g = 9.81 m/s²

mass of sand  = 0.5 kg

distance  = 0.6

Thermal energy  = the energy possessed by the falling sand

The force of sand = ma

                              = 0.5 * 9.81\\= 4.905 N

However, the energy is given as the force multiplied by the distance, so

E =fd

   4.905 * 0.6\\= 2.943 J

Therefore, the thermal energy is 2.943 joules

 

jok3333 [9.3K]3 years ago
3 0

Answer:

<em>The thermal energy added would be 2.94 J</em>

Explanation:

Part 1

The increase in thermal energy is equal to the new potential energy of the body. As the sand falls and stop at 0.6 m it gains thermal energy in form of the potential energy;

Given the mass = 5 kg

g acceleration due to gravity = 9.8 m/s^{2}

increase in U would be  

+U = mgh

       = 0.5 x 0.6 x 9.8

      = 2.94 J

Therefore 2.94 J is added to the thermal energy of the body.

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Answer:

i. The radius 'r' of the electron's path is 4.23 × 10^{-5} m.

ii. The frequency 'f' of the motion is 455.44 KHz.

Explanation:

The radius 'r' of the electron's path is called a gyroradius. Gyroradius is the radius of the circular motion of a charged particle in the presence of a uniform magnetic field.

                 r = \frac{mv}{qB}

Where: B is the strength magnetic field, q is the charge, v is its velocity and m is the mass of the particle.

From the question, B = 1.63 × 10^{-5}T, v = 121 m/s, Θ = 90^{0} (since it enters perpendicularly to the field), q = e  = 1.6 × 10^{-19}C and m = 9.11 × 10^{-31}Kg.

Thus,

         r = \frac{mv}{qB} ÷ sinΘ

But,  sinΘ =  sin 90^{0} = 1.

So that;

          r = \frac{mv}{qB}

            = (9.11 × 10^{-31} × 121) ÷ (1.6 × 10^{-19}  × 1.63 × 10^{-5})

            = 1.10231 × 10^{-28}   ÷ 2.608 × 10^{-24}

            = 4.2266 × 10^{-5}

            = 4.23 × 10^{-5} m

The radius 'r' of the electron's path is 4.23 × 10^{-5} m.

B. The frequency 'f' of the motion is called cyclotron frequency;

           f = \frac{qB}{2\pi m}

             =  (1.6 × 10^{-19}  × 1.63 × 10^{-5}) ÷ (2 ×\frac{22}{7} × 9.11 × 10^{-31})

             =  2.608 × 10^{-24} ÷  5.7263 × 10^{-30}

             = 455442.4323

          f  = 455.44 KHz

The frequency 'f' of the motion is 455.44 KHz.

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