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expeople1 [14]
3 years ago
12

A cylindrical wire made of an unknown alloy hangs from a support in the ceiling. You measure the relaxed length of the wire to b

e 16 m long; and the radius of the wire to be 3.5 m. When hang a 5 kg mass from the wire, you measure that it stretches a distance of 4 x 10 m The average bond length between atoms is 2.3 x 10^0 m for th alloy.
Required:
What is the stiffness of a typical interatomic bond in the alloy
Physics
1 answer:
IceJOKER [234]3 years ago
8 0

Answer: hello  some of your values are wrongly written hence I will resolve your question using the right values

answer:

stiffness =  1.09 * 10^-6 N/m

Explanation:

Given data:

Length ( l ) = 16 m

radius of wire ( r ) = 3.5 m

mass ( m ) = 5kg

<u>Distance stretched (  Δl ) = 4 * 10^-3 m </u> ( right value )

<u>average bond length ( between atoms ) = 2.3 * 10^-10 m </u>( right value)

first step : calculate the area

area ( A ) = πr^2 = π * ( 3.5)^2 = 38.48 m^2

        γ          = MgL / A Δl

                    = [ (5 * 9.81 * 16 ) / ( 38.48 * (4.3*10^-3) ) ]

                    = 784.8 / 0.165 = 4756.36 N/m^2

hence : stiffness =   γ  * bond length

                           =  4756.36 * 2.3 * 10^-10  = 1.09 * 10^-6 N/m

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Since it is inside the conductor

E=0N/C

2. Since the entire charge us inside the surface, then the electric field at a distance r (5.2m) away form the surface is given as

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The potential energy is given by:

U=(M+m)gh       (1)

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m: mass of the bullet = 6g = 6.0*10^-3 kg

g: gravitational constant = 9.8m/s^2

h: distance to the ground

The distance to the ground is calculate d by using the information about the length of the cord and the degrees of the cord respect to the vertical:

h=l-lsin\theta\\\\h=2.2m-2,2m\ sin60\°=0.29m

The potential energy is:

U=(1kg+6*10^{-3}kg)(9.8m/s^2)(0.29m)=2.85J

Next, the potential energy is equal to kinetic energy of the block and the bullet at the beginning of its motion:

U=\frac{1}{2}(M+m)v^2\\\\v=\sqrt{2\frac{U}{M+m}}=\sqrt{2\frac{2.85J}{1kg+6*10^{-3}kg}}=2.38\frac{m}{s}

Next, you use the momentum conservation law, in order to calculate the speed of the bullet before the impact:

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You solve the equation (2) for v2:

M(0)+mv_2=(M+m)v    

v_2=\frac{M+m}{m}v=\frac{1kg+6*10^{-3}kg}{6*10^{-3}kg}(2.38m/s)\\\\v_2=399.04\frac{m}{s}

The speed of the bullet before the impact with the wood block is 399.04 m/s

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The impulse is 2.38 kgm/s

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The force on the cord after the impact is 2.59N

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