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m_a_m_a [10]
3 years ago
12

The transfer of thermal energy between two bodies which are at different temperatures. The si unit for this is the joule. Genera

lly considered the raising of temperature due to thermal energy.
Physics
2 answers:
weeeeeb [17]3 years ago
8 0

Answer:

Heat

Explanation:

The given term is nothing but definition of Heat.

Heat is is defined as the energy that is transferred  between two bodies which are at different temperatures.It always flows from the body at higher temperature to the body at lower temperature.  The SI  unit for this is the joule. Generally considered the raising of temperature due to thermal energy.

ANTONII [103]3 years ago
8 0

Answer:

heat

Explanation:

did it on usa test prep and like duh

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What is measured using micrometer?<br><br>​
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6 0
3 years ago
Suppose that the current in the solenoid is i(t). The self-inductance L is related to the self-induced EMF E(t) by the equation
Artemon [7]

Answer:

L =   μ₀ n r / 2I

Explanation:

This exercise we must relate several equations, let's start writing the voltage in a coil

        E_{L} = - L dI / dt

 

Let's use Faraday's law

       E = - d Ф_B / dt

in the case of the coil this voltage is the same, so we can equal the two relationships

        - d Ф_B / dt = - L dI / dt

The magnetic flux is the sum of the flux in each turn, if there are n turns in the coil

        n d Ф_B = L dI

we can remove the differentials

      n Ф_B = L I

magnetic flux is defined by

     Ф_B = B . A

in this case the direction of the magnetic field is along the coil and the normal direction to the area as well, therefore the scalar product is reduced to the algebraic product

      n B A = L I

the loop area is

      A = π R²

     

we substitute

       n B π R² = L I                    (1)

To find the magnetic field in the coil let's use Ampere's law

        ∫ B. ds = μ₀ I

where B is the magnetic field and s is the current circulation, in the coil the current circulates along the length of the coil

           s = 2π R

we solve

              B 2ππ R =  μ₀ I

              B =  μ₀ I / 2πR

we substitute in

       n ( μ₀ I / 2πR) π R² = L I

       n  μ₀ R / 2 = L I

       L =   μ₀ n r / 2I

4 0
3 years ago
How much work (in J) is done against the gravitational force on a 3.9 kg briefcase when it is carried from the ground floor to t
katrin2010 [14]

Answer:

W = 14523.6 J

Explanation:

Given,

Mass = 3.9 Kg

Vertical height , h = 380 m

Work done against gravitational force

W = m g h

g is acceleration due to gravity

W = 3.9 x 9.8 x 380

W = 14523.6 J

Hence, the work done by the gravitational force is equal to W = 14523.6 J

3 0
3 years ago
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