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Tasya [4]
3 years ago
5

A steel rectangular tube has outside dimensions of 150 mm x 50 mm and a wall thickness of 4 mm. State the inside dimensions, the

area of its cross section, and the weight of a piece 1.22 m long.
Engineering
1 answer:
lara [203]3 years ago
6 0

Answer:

 inside dimension  = 142 mm \times 42 mm

cross section area = 7.5\times 10^{-3} m^2

mass of 1.2 meter log steel  = 1.843\times 10^{-3} \rho

Explanation:

given data:

Outside dimension of steel rectangular = 150 mm\times 50mm

Thickness = 4 mm

Long = 1.22 m

inside dimension will be = (150- 8)mm \times ( 50-8)mm

                                         = 142 mm \times 42 mm

cross section area = 150\times 50 mm^2

                               = 7500\times 10^{-6} m^2

                               = 7.5\times 10^{-3} m^2

let the density be assumed as \rho

mass of 1.2 meter log steel will be = 1.2  \times (7.5\times 10^{-3} -  0.142\times 0.048)\times \rho

                                                       = 1.843\times 10^{-3} \rho

                                                       

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g Let the charges start infinitely far away and infinitely far apart. They are placed at (6 cm, 0) and (0, 3 cm), respectively,
irina1246 [14]

Answer:

a) V =10¹¹*(1.5q₁ + 3q₂)

b) U = 1.34*10¹¹q₁q₂

Explanation:

Given

x₁ = 6 cm

y₁ = 0 cm

x₂ = 0 cm

y₂ = 3 cm

q₁ = unknown value in Coulomb

q₂ = unknown value in Coulomb

A) V₁ = Kq₁/r₁

where   r₁ = √((6-0)²+(0-0)²)cm = 6 cm = 0.06 m

V₁ = 9*10⁹q₁/(0.06) = 1.5*10¹¹q₁

V₂ = Kq₂/r₂

where   r₂ = √((0-0)²+(3-0)²)cm = 3 cm = 0.03 m

V₂ = 9*10⁹q₂/(0.03) = 3*10¹¹q₂

The electric potential due to the two charges at the origin is

V = ∑Vi = V₁ + V₂ = 1.5*10¹¹q₁ + 3*10¹¹q₂ = 10¹¹*(1.5q₁ + 3q₂)

B) The electric potential energy associated with the system, relative to their infinite initial positions, can be obtained as follows

U = Kq₁q₂/r₁₂

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U = 9*10⁹q₁q₂/(3√5/100)

⇒ U = 1.34*10¹¹q₁q₂

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