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dalvyx [7]
3 years ago
11

How do two objects at different temperatures attain thermal equilibrium?

Physics
1 answer:
Lady bird [3.3K]3 years ago
3 0

Answer:

C.As the two objects touch, thermal energy flows as heat from the warmer block to the colder block until particles in both blocks move at the same rate and reach the same temperature.

Explanation:

Heat is the transfer of thermal energy from an object at higher temperature to an object at colder temperature.

The temperature of an object is a measure of how fast the particles in the object move: the higher its temperature, the faster the particles move, the higher the average kinetic energy of the particles in the object. As a result, the particles of the object at higher temperature tend to transfer more energy (called thermal energy) to the particles of the object at colder temperature by colliding with them: this process continues until the particles of the colder object reach the same average kinetic energy as the particles of the warmer object, and this means that the two objects have reached the same temperature.

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An electric fan is running on HIGH. After fan has been running for 1.3 minutes, the LOW button is pushed. The fan slows down to
mafiozo [28]

Answer:

    wo = 18.75 rev / s

Explanation:

This is an exercise in endowment kinematics, it indicates that the final angular velocity is w_f = 109 rad / s, the time to reach this velocity is t = 1.87 s and the deceleration a = 4.7 rad / s²

         w_f = w₀ - a t

         w₀ = w_f + a t

         w₀ = 109 + 4.7 1.87

         w₀ = 117.8 rad / s

let's reduce to revolutions / s

         w₀ = 117.8 rad / s (1 rev / 2pi rad)

         w₀ = 18.75 rev / s

8 0
3 years ago
A 2 kg rubber ball is thrown at a wall horizontally at 3 m/s, and bounces back the way it came at an equal speed. A 2 kg clay ba
Lyrx [107]

Answer:

THE RUBBER BALL

Explanation:

From the question we are told that

      The mass of the rubber ball is m_r   =  2 \ kg

      The  initial  speed of the rubber ball is  u =  3 \ m/s

      The final speed at which it bounces bank v  - 3 \ m/s

      The mass of the clay ball  is  m_c =  2  \ kg

       The  initial  speed of the clay  ball is u = 3 \ m/s

       The final speed of the clay ball is  v = 0 \  m/s

Generally Impulse is mathematically represented as

       I  =  \Delta p

where \Delta  p is the change in the linear momentum so  

       I  =  m(v-u)

For the rubber  is  

        I_r  =  2(-3 -3)

       I_r  = -12\ kg \cdot  m/s

=>     |I_r|  = 12\ kg \cdot  m/s

For the clay ball

       I_c  =  2(0-3)

        I_c =  -6 \ kg\cdot \ m/s

=>    | I_c| =  6 \ kg\cdot \ m/s

So from the above calculation the ball with the a higher magnitude of impulse is the rubber ball

       

8 0
3 years ago
Jonathan has six strings that are the same thickness and are all the same material. He cuts the strings to different lengths and
harkovskaia [24]
The shortest string will have the highest pitch.

7 0
3 years ago
Read 2 more answers
I need help me with my question
lisov135 [29]

The tilt of the moon's axis does not allow for monthly alignment, so the lunar and solar eclipse do not happen every month.

<h3>How do the lunar and solar eclipse occur?</h3>
  • For the occurrence of lunar and solar eclipse, the sun, moon and the earth must remain in a plan and along a straight line.
  • When the earth appears in between the sun and the moon, lunar eclipse occurs.
  • When the moon appears in between the sun and the earth, solar eclipse occurs.
  • The moon and earth are rotating not only around the sun, but also around the black hole of Milky way galaxy.
  • So they are not present in a plan as well as in a straight line in every full moon and new moon time.

Thus, we can conclude that the option D is correct.

Learn more about the lunar eclipse and solar eclipse here:

brainly.com/question/8643

#SPJ1

4 0
2 years ago
A)If your torsion balance deflects to an angle of 10° when your two spheres are 40cm apart, what angle will it deflect to when t
svp [43]

Answer:

Explanation:

Given

for \theta=10^{\circ}

Sphere are d=40\ cm

when sphere d_2=10\ cm apart suppose deflection is \theta _2

We know

F=k_t\cdot \theta

Where F=force between charged particle

\theta =Deflection

F=\frac{kQ_1Q_2}{r^2}=k_t\cdot \theta

\theta =\frac{k}{k_t}\times \frac{Q_1Q_2}{r^2}----1

thus \theta \propto \frac{1}{r^2}

for \theta _2

\frac{\theta _1}{\theta _2}=(\frac{r_2}{r_1})^2

\theta _2=16\times \theta _1

\theta _2=160^{\circ}

(b)for 10^{\circ} deflection Potential v_1=8\ kV

Electric Potential is V=\frac{kQ}{r}

Q=\frac{V\cdot r}{k}

where V=voltage

k=constant

r=distance between charges

Put value of Q in equation 1

\theta =\frac{k}{k_t}\times \frac{V^2r^2}{k^2}

\theta =\frac{V^2r^2}{k\cdot k_t}

thus \theta \propto V^2

therefore

\frac{\theta _1}{\theta _2}=(\frac{V_1}{V_2})^2

\frac{10}{\theta _2}=(\frac{8}{4})^2

\theta _2=\frac{10}{4}=2.5^{\circ}

5 0
3 years ago
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