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lukranit [14]
3 years ago
15

Determine the average normal stress developed in rod AB if the load has a mass of 55 kg . The diameter of rod AB is 8 mm.

Physics
1 answer:
Fudgin [204]3 years ago
3 0

Answer:

The normal stress is 10.7[MPa]

Explanation:

The normal stress can be calculated with the following equation:

S_{norm} =\frac{F}{A} \\where:\\F= force [Newtons]\\A=area [m^2]\\S_{norm} = Normal stress [\frac{N}{m^{2} }] or [Pa]

The area of the rod can be calculated using the equation:

A=\frac{\pi }{4}*d^{2}  \\d=8[mm]=0.008[m]\\A=\frac{\pi }{4}*(0.008)^{2}  \\A=5.02*10^{-5} [m^{2} ]

The force is the result of the mass multiplied by the gravity.

F=55[kg]*9.81[m/s^{2} ] = 539.6[N]\\\\S_{norm} = 539.6/5.02*10^{-5} \\S_{norm} = 10.7*10^{6}[Pa] = 10.7[MPa]

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Ans of this question A test charge of 1 couloumb moved from 30cm against the field of intensity 50N/c find the energy store in i
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Answer:

A. Zero

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                                                                                                        = 0.3 m

The electric field intensity, E = 50 N/C

The energy stored in the charge at 0.3 m is given by the formula,

                                V = k q/r

Where,                        

                                     = 9 x 10⁹ Nm²C⁻²

The charge is moved from the potential V₁ to V₂ at 30 cm

Substituting the given values in the above equation

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And,

                            V₂ = 1.5 x 10¹² J

The energy stored in it is,

                             W = V₂ - V₁

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Hence, the energy stored in the charge is, W = 0        

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