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SVEN [57.7K]
3 years ago
14

A particular interaction force does work wint inside a system. the potential energy of the interaction is u. which equation rela

tes u and wint
Physics
1 answer:
larisa86 [58]3 years ago
5 0

ΔU = -Wint

Consdier the work of of interaction is W =m*g*h - equation -1

and the Potential energy U.

Final Potential energy Uf =0 , And the Initial Potential Energy Ui =m*g*h

<span>Now we will write the equation for a Change in Potential energy ΔU,</span>

ΔU = Uf - Ui

= 0-m*g*h

<span>  ΔU = -m*g*h --Equation 2</span>

Now compare the both equation

<span>Wint = -ΔU</span>

we can rewrite the above equation

ΔU = -W.

<span>So our Answer is ΔU = -W. .</span>

<span> </span>

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Read 2 more answers
Water flows through a horizontal pipe of varying cross-section. In the first section, the cross-sectional area is 10 cm2 and flo
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Answer:

(a) the flow speed of the second section is 11 m/s

(b) the pressure of the second section is 6.33 x 10⁴ Pa

Explanation:

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area of the first cross section, A₁ = 10 cm²

pressure in the first cross section, P₁ = 1.2 x 10⁵ Pa

area of the second section, A₂ = 2.5 cm²

(a) the flow speed of the second section (V₂)

Apply continuity equation;

Q₁ = Q₂

Q₁ = A₂V₂

V₂ = Q₁ / A₂

V₂ = (2750) / (2.5)

V₂ = 1100 cm/s = 11 m/s

(b) the pressure of the second section (P₂)

Apply Bernoulli's equation;

P₁ + ¹/₂ρV₁² = P₂ + ¹/₂ρV₂²

where;

ρ is density of water = 1000 kg/m³

V₁ is the speed of water in the first section;

Q₁ = A₁V₁

V₁ = Q₁ / A₁

V₁ = (2750) / (10)

V₁ = 275 cm/s = 2.75 m/s

P₂ = P₁ + ¹/₂ρV₁² - ¹/₂ρV₂²

P₂ = P₁ + ¹/₂ρ(V₁² - V₂²)

P₂ = 1.2 x 10⁵ Pa + ¹/₂ x 1000 (2.75² - 11²)

P₂ = 1.2 x 10⁵ Pa + 500(-113.438)

P₂ = 1.2 x 10⁵ Pa - 0.567  x 10⁵ Pa

P₂ = 0.633 x 10⁵ Pa

P₂ = 6.33 x 10⁴ Pa

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