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almond37 [142]
3 years ago
6

HELP ASAP!!! So stressed! :(

Physics
1 answer:
nasty-shy [4]3 years ago
7 0
Question #16:. last choice.
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3 0
3 years ago
Problem 12: A 20.0-m tall hollow aluminum flagpole is equivalent in strength to a solid cylinder 4.00 cm in diameter. A strong w
avanturin [10]

Answer:

The deformation in the pole due to force is 0.70 mm.

Explanation:

Given that,

Height = 20.0 m

Diameter = 4.00 cm

Force = 1100 N

We need to calculate the  area

Using formula of area

A=\pi\times r^2

A=\pi\times(2.00\times10^{-2})^2

A=0.00125\ m^2

A=1.25\times10^{-3}\ m^2

We need to calculate the deformation

Using formula of deformation

\Delta x=\dfrac{1}{s}(\dfrac{F}{A}\times L)

Where, s = shear modulus

F = force

l = length

A = area

Put the value into the formula

\Delta x=\dfrac{1}{2.5\times10^{10}}\times(\dfrac{1100}{1.25\times10^{-3}}\times 20.0)

\Delta x=0.000704\ m

\Delta x=7.04\times10^{-4}\ m

\DElta x=0.70\ mm

Hence, The deformation in the pole due to force is 0.70 mm.

3 0
3 years ago
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