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2 years ago

By considering action-reacton forces,identify why water rises up a thin capillary tube

1 answer:

6 0

The forces of attraction between water molecules and the glass walls and within the molecules of water themselves are what enable the water to rise in a thin tube immersed in water.

<h3>What is force?</h3>

Force is defined as the push or pulls applied to the body. Sometimes it is used to change the shape, size, and direction of the body.

Force is defined as the product of mass and acceleration. Its unit is Newton.

Surface or interfacial forces lead to capillarity. The forces of attraction between the water molecules and the glass walls and among the water molecules themselves are what causes the water in a thin tube submerged in water to rise.

Hence, the water rises up a thin capillary tube can be explained by Newton's third law.

To learn more about the force refer to the link;

brainly.com/question/26115859#SPJ1

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3 years ago

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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3 years ago

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