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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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▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪

Let's solve ~

Given terms :

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The formula to find kinetic Energy is ~

\boxed{ \boxed{ \sf{ \frac{1}{2}  m{v}^{2} }}}

Now, apply the formula according to given situation

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \: \dfrac{1}{2}  \times 7 \times ( {4)}^{2}

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \: \dfrac{1}{2}  \times 7 \times 16

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:7 \times 8

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:56 \:  \: joules

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