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GalinKa [24]
3 years ago
11

What is the best description of temperature?

Physics
1 answer:
Nonamiya [84]3 years ago
5 0

Explanation:

o has Celsius of indirect measurment

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An object is moving along the x axis. At t = 5.6 s, the object is at x = +3.0 m and has a velocity of +5.7 m/s. At t = 8.5 s, it
Debora [2.8K]

The average acceleration between t = 5.6 s and t = 8.5 s is 2.31  m/s²

<h3>What is acceleration?</h3>

Acceleration is defined as the rate change of velocity with time.

acceleration a = (Δv) / (Δt)

An object is moving with initial velocity u =5.7 m/s and its final velocity v= -1.0 m/s.

Time taken for the change in speed, t= 8.5 - 5.6 = 2.9 seconds

The acceleration is given by

a = (-1 - 5.7)/ 2.9

a = -  2.31 m/s²

|a | = 2.31 m/s²

Thus, the object's acceleration is 2.31 m/s²

Learn more about acceleration.

brainly.com/question/12550364

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3 0
2 years ago
Michael has a substance that he puts in Container 1. The substance has a volume of 5 cubic meters. He then puts the substance in
m_a_m_a [10]
Gas because liquids and solids volumes don't change from switching containers.
3 0
3 years ago
To practice Problem-Solving Strategy 23.2 for continuous charge distribution problems. A straight wire of length L has a positiv
Lesechka [4]

Answer:

             E = k Q / [d(d+L)]

Explanation:

As the charge distribution is continuous we must use integrals to solve the problem, using the equation of the elective field

       E = k ∫ dq/ r² r^

"k" is the Coulomb constant 8.9875 10 9 N / m2 C2, "r" is the distance from the load to the calculation point, "dq" is the charge element  and "r^" is a unit ventor from the load element to the point.

Suppose the rod is along the x-axis, let's look for the charge density per unit length, which is constant

         λ = Q / L

If we derive from the length we have

        λ = dq/dx       ⇒    dq = L dx

We have the variation of the cgarge per unit length, now let's calculate the magnitude of the electric field produced by this small segment of charge

        dE = k dq / x²2

        dE = k λ dx / x²

Let us write the integral limits, the lower is the distance from the point to the nearest end of the rod "d" and the upper is this value plus the length of the rod "del" since with these limits we have all the chosen charge consider

        E = k \int\limits^{d+L}_d {\lambda/x^{2}} \, dx

We take out the constant magnitudes and perform the integral

        E = k λ (-1/x){(-1/x)}^{d+L} _{d}

   

Evaluating

        E = k λ [ 1/d  - 1/ (d+L)]

Using   λ = Q/L

        E = k Q/L [ 1/d  - 1/ (d+L)]

 

let's use a bit of arithmetic to simplify the expression

     [ 1/d  - 1/ (d+L)]   = L /[d(d+L)]

The final result is

     E = k Q / [d(d+L)]

3 0
3 years ago
If you put two identical cars on opposite sides of a large magnet, what happens
FrozenT [24]

Answer:

Depends on what pole it is.

Explanation:

If the poles of the cars and magnets are the same they will repel, if different, attracts.

7 0
3 years ago
Read 2 more answers
2.0 mol of monatomic gas A initially has 5000 J of thermal energy. It interacts with 3.0 mol of monatomic gas B, which initially
Naddika [18.5K]

Answer:

 E_particle = 1,129 10⁻²⁰ J / particle

  T= 817.5 K

Explanation:

Energy is a scalar quantity so it is additive, let's look for the total energy of each gas

Gas a

         E_a = 2 5000 = 10000 J

Gas b

         E_b = 3 8000 = 24000 J

When the total system energy is mixed it is

          E_total = E_a + E_b

          E_total = 10000 + 24000 = 34000

The total mass is

           M = m_a + m_b

           M = 2 +3 = 5

The average energy among the entire mass is

           E_averge = E_total / M

            E_averago = 34000/5

            E_average = 6800 J

One mole of matter has Avogadro's number of atoms 6,022 10²³ particles

Therefore, each particle has an energy of

                E_particle = E_averag / 6.022 10²³ = 6800 /6.022 10²³

                E_particle = 1,129 10⁻²⁰ J / particle

For  find the temperature let's use equation

               E = kT

               T = E / k

     

               T = 1,129 10⁻²⁰ / 1,381 10⁻²³

               T = 8.175 102 K

               T= 817.5 K

5 0
3 years ago
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