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ikadub [295]
3 years ago
5

A rod is made up of copper and wood joined together.

Physics
1 answer:
IgorLugansk [536]3 years ago
4 0

my views

Explanation:

Figure (8) shows a rod made up of copper and wood joined together. The rod is heated at the joint in the centre for about a minute. At which point – A, B, C or D, would it show the lowest temperature? PLEASE GIVE REASONS ALSO(MINIMUM 2)

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Rod AB has a diameter of 200mm and rod BC has a diameter of 150mm. Find the required temperature increase to close the gap at C.
Leni [432]

Answer:

T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}

Explanation:

The given data :-

  • Diameter of rod AB ( d₁ ) = 200 mm.
  • Diameter of rod BC ( d₂ ) = 150 mm.
  • The linear co-efficient of thermal expansion of copper ( ∝ ) = 1.6 × 10⁻⁶ /°C
  • The young's modulus of elasticity of copper ( E ) = 120 GPa = 120 × 10³ MPa.
  • Consider the required temperature increase to close the gap at C = T °C
  • Consider the change in length of the rod = бL

Solution :-

\begin{aligned}\sum H& =0\\-R_A+R_C&=0\\R_A&=R_C\\R_A&=R\\R_C&=R\\R_{A}&=\text{reaction\:force\:at\:A}\\R_{C}&=\text{reaction\:force\:at\:C}\\\sigma_{AB}&=\text{axial\:stress\:at\:A}\\\sigma_{BC}&=\text{axial\:stress\:at\:B}\\\sigma_{AB}&=\frac{R}{A_{A}}\\&=\frac{R_{A}}{A_{A}}\\\sigma_{BC}&=\frac{R_{B}}{A_{B}}\\&=\frac{R}{A_{B}}\\\frac{\sigma_{AB}}{\sigma_{BC}}&=\frac{A_{B}}{A_{B}}\\&=\frac{\frac{\pi}{4}\cdot 150^{2}}{\frac{\pi}{4}\cdot 200^{2}}\\&=\frac{9}{16}\end{aligned}

\begin{aligned}\delta L&= (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\0& = (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\&=\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{AB}+\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{BC}\\&=2\:\alpha\:T\:L-\frac{L}{E}(\sigma_{AB}+\sigma_{BC})\\T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}\end{aligned}

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4 years ago
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Get adequate amounts of liquid
3 0
3 years ago
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Which type of wave has a frequency just higher than violet visible light and is used by the skin to produce Vitamin D?
ANTONII [103]
Ultraviolet rays can help us get vitamin D, but they can also damage or skin if w are exposed for too long.

Your answer is D.
4 0
3 years ago
Una pelota de béisbol de 142g de masa, luego de ser arrojada por el pitcher lleva una velocidad de 90mph. Luego de ser bateada s
V125BC [204]

Answer:

a)    I = 13.38 kg m / s, b)    F = 1,373 10³ N

Explanation:

The impulse is given by the relation

          I = ∫ F dt = Δp

          I = p_f -p₀

          I = m (v_f - v₀)

take the ball's exit direction as positive, whereby the ball velocities

v₀ = -90mph, the final velocity v_f = + 54 m / s

Let's reduce the units to

         I = 0.142 [54- (-40.23) ]

      the SI system

        v₀ = - 90 mph (1609.34 m / 1 mile) (1h / 3600 s = -40.23 m / s

        m = 142 g (1kg / 1000) = 0.142 kg

we calculate  

          I = 0.142 [54- (-40) ]

          I = 13.38 kg m / s

b) let's use the definition of momentum

         I = ∫ F .dt

         I = F ∫ dt

         F = I / t

         F = 13.38 / 0.008

         F = 1,373 10³ N

5 0
3 years ago
An electric charge q of mass m in an oscillating electric field Eosinot experiences force q Eosinot. Suppose it starts from rest
Masja [62]

Answer:

speed of the charge electric is  v = - (Eo q/m) cos t

Explanation:

The electric charge has a very small mass so it follows the oscillations of the electric field. We force ourselves on the load,

          F = q Eo sint

a) To find the velocity of the particle, let's use Newton's second law to find the acceleration and of this by integration the velocity

        F = ma

        q Eo sint = ma

        a = Eo q / m sint

        a = dv / dt

        dv = adt

        ∫ dv = ∫ a dt

        v-vo = I (Eoq / m) sin  t dt

        v- vo = Eo q / m (-cos t)

We evaluate the integral from the initial point, as the particle starts from rest Vo = 0, for t = 0

        v = - (Eo q / m) cos t

b) Kinetic energy

       

         K = ½ m v2

          K = ½ m (Eoq / m)²2 (sint)²

         K = ¹/₂  Eo² q² / m sin² t

c) The average kinetic energy over a period

          K = ½ m v2

         <v2> = (Eoq / m) 2 <cos2 t>

The average of cos2 t = ½, substitute and calculate

          K = ½ m (Eoq / m)²  ½

          K = ¼ Eo² q² / m

7 0
4 years ago
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