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vichka [17]
3 years ago
11

When is thermal equilibrium achieved between two objects?

Physics
2 answers:
ki77a [65]3 years ago
4 0
Thermal equilibrium is when a hot and cold object (could also be air) come in contact with each other and over time become the same temperature.
thermal equilibrium is reached when both objects are the same temperature.
anygoal [31]3 years ago
4 0

Answer:

Thermal equilibrium is achieved when two identical objects have the same temperature after contacting and exchanging energy.

Explanation:

By bringing two bodies into contact at a different temperature, the one with the highest temperature yields part of its energy to the one with the lowest temperature until its temperatures equalize. This is what we call "thermal equilibrium". In other words, thermal equilibrium is the state in which the temperatures of two bodies that initially had different temperatures are equalized. When the temperatures equalize, the heat flow is suspended, and this is how the equilibrium situation is reached.

Thermal equilibrium is achieved when two identical objects have the same temperature after contacting and exchanging energy.

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Pressure = Force/ Area = 3000/2 = 1500 pascal.
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A space station has a large ring-like component that rotates to simulategravity for the crew. This ring has a massM= 2.1×105kg a
ivann1987 [24]

Answer:

Each thruster has to applied a force of 294.5N in tangential direction

Explanation:

Mass of the ring ,M =2.1×105kg

Mass of the ship ,m = 3.5×104kg

Radius of the ring R = 86.0 m

distance of ship from center of the ring, r =31.0 m

Let force applied by each thruster be F

Time taken to reach  gravity ,t = 3hrs = 3600× 3 =10800sec

The movement of ring make the object kept at the edge feel a force of centrifuge in outward direction.

Centrifugal force = weight of the object on earth

Assume the ring is moving with angular speed ω

Centripetalforce of the object kept at ring

m₁R ω²=m₁g  (m₁=mas of object)

⇒Rω² = g

⇒ω = √g/R

The ring start from 0 angular speed with constant angular acceleration

Let the constant angular acceleration be ∝

∝ = ω  / t

(ω = √g/R)

∝ = \frac{1}{t } \sqrt{\frac{g}{R} } }

Consider Torque on the ring and ship system

T = FR + FR = 2FR

Moment of inertia of ring ship system

I = I(ring)+I(ship)+I(ship)

= MR² + mr² + mr² = MR² + 2mr²

angular acceleration of the ring ship system

∝ = \frac{2FR}{MR^2 + 2mr^2}

Now we have ,

∝ = \frac{1}{t} \sqrt{\frac{g}{R} }  , ∝ = \frac{2FR}{MR^2+2mr^2}

equating both values

We have,

F = \frac{MR^2+2mr^2}{2Rt} \sqrt{\frac{g}{R} }

where,

m = 2.1 × 10⁵kg, R = 86m, m = 3.5 × 10₄kg,

r = 31m, g = 9.8m/s² , t = 10800sec

F = \frac{(2.1 \times 10^5 \times86^2)+(2\times3.5v10^4\times31^2) }{2\times86\times10800} \times\sqrt{\frac{9.8}{86} } \\\\F = 294.47N\\\approx294.5N

Each thruster has to applied a force of 294.5N in tangential direction

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3 years ago
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Answer:

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7 0
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A diffraction pattern forms when light passes through a single slit. The wavelength of the light is 691 nm. Determine the angle
expeople1 [14]

Explanation:

Given that,

Wavelength of the light, \lambda=691\ nm=691\times 10^{-9}\ m

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The angle that locates the first dark fringe is given by :

sin\theta=\dfrac{\lambda}{a}

sin\theta=\dfrac{691\times 10^{-9}}{3.8\times 10^{-4}}

\theta=0.104^{\circ}

(b) Slit width, a=3.8\times 10^{-6}\ m

The angle that locates the first dark fringe is given by :

sin\theta=\dfrac{\lambda}{a}

sin\theta=\dfrac{691\times 10^{-9}}{3.8\times 10^{-6}}

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Hence, this is the required solution.

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