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Shalnov [3]
1 year ago
13

A solid steel cylinder is standing (on one of its ends) vertically on the floor. The length of the cylinder is 3.2 m and its rad

ius is 59 cm. When an object is placed on top of the cylinder, the cylinder compresses by an amount of 5.6 10-7 m. What is the weight of the object?
Physics
1 answer:
maksim [4K]1 year ago
4 0

To solve this problem it is necessary to apply the concepts related to Young's Module and its respective mathematical and modular definitions. In other words, Young's Module can be expressed as

\Upsilon = \frac{F/A}{\Delta L/L_0}

Where,

F = Force/Weight

A = Area

\Delta L= Compression

L_0= Original Length

According to the values given we have to

\Upsilon_{steel} = 200*10^9Pa

\Delta L = 5.6*10^{-7}m

L_0 = 3.2m

r= 0.59m \rightarrow A = \pi r^2 = \pi *0.59^2 = 1.0935m^2

Replacing this values at our previous equation we have,

\Upsilon = \frac{F/A}{\Delta L/L_0}

200*10^9 = \frac{F/1.0935}{5.6*10^{-7}/3.2}

F = 38272.5N

Therefore the Weight of the object is 3.82kN

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Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper,
balandron [24]

Answer with Explanation:

We are given that

Radius of  solid core wire=r=2.28 mm=2.28\times 10^{-3} m

1mm=10^{-3} m

Radius of each strand  of thin wire=r'=0.456 mm=0.456\times 10^{-3} m

Current density of each wire=J=3750 A/m^2

a.Area =\pi r^2

Where \pi=3.14

Using the formula

Cross section area of copper wire has solid core =3.14\times (2.28\times 10^{-3})^2=16.3\times 10^{-6} m^2

Current density =J=\frac{I}{A}

Using the formula

3750=\frac{I}{16.3\times 10^{-6}}

I=3750\times 16.3\times 10^{-6}=0.061 A

Total number of strands=19

Area of strand wire=A'=19\times 3.14\times (0.456\times 10^{-3})^2=12.4\times 10^{-6} m^2

J'=\frac{I'}{A'}

3750=\frac{I'}{19\times 3.14(0.456\times 10^{-3})^2}

I'=3750\times 19\times 3.14(0.456\times 10^{-3})^2

I'=0.047 A

b.Resistivity of copper wire=\rho=1.69\times 10^{-8}\Omega-m

Length of each wire =6.25 m

Resistance, R=\frac{\rho l}{A}

Using the formula

Resistance of solid core wire=R=\frac{1.69\times 10^{-8}\times 6.25}{16.3\times 10^{-6}}=6.5\times 10^{-3}\Omega

Resistance of strand wire=R'=\frac{1.69\times 10^{-8}\times 6.25}{12.4\times 10^{-6}}=8.5\times 10^{-3}\Omega

7 0
1 year ago
Consider a vibrating system described by the initial value problem. (A computer algebra system is recommended.) u'' + 1 4 u' + 2
GarryVolchara [31]

Answer:

Therefore the required solution is

U(t)=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega} cos\omega t +\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega} sin \omega t

Explanation:

Given vibrating system is

u''+\frac{1}{4}u'+2u= 2cos \omega t

Consider U(t) = A cosωt + B sinωt

Differentiating with respect to t

U'(t)= - A ω sinωt +B ω cos ωt

Again differentiating with respect to t

U''(t) =  - A ω² cosωt -B ω² sin ωt

Putting this in given equation

-A\omega^2cos\omega t-B\omega^2sin \omega t+ \frac{1}{4}(-A\omega sin \omega t+B\omega cos \omega t)+2Acos\omega t+2Bsin\omega t = 2cos\omega t

\Rightarrow (-A\omega^2+\frac{1}{4}B\omega +2A)cos \omega t+(-B\omega^2-\frac{1}{4}A\omega+2B)sin \omega t= 2cos \omega t

Equating the coefficient of sinωt and cos ωt

\Rightarrow (-A\omega^2+\frac{1}{4}B\omega +2A)= 2

\Rightarrow (2-\omega^2)A+\frac{1}{4}B\omega -2=0.........(1)

and

\Rightarrow -B\omega^2-\frac{1}{4}A\omega+2B= 0

\Rightarrow -\frac{1}{4}A\omega+(2-\omega^2)B= 0........(2)

Solving equation (1) and (2) by cross multiplication method

\frac{A}{\frac{1}{4}\omega.0 -(-2)(2-\omega^2)}=\frac{B}{-\frac{1}{4}\omega.(-2)-0.(2-\omega^2)}=\frac{1}{(2-\omega^2)^2-(-\frac{1}{4}\omega)(\frac{1}{4}\omega)}

\Rightarrow \frac{A}{2(2-\omega^2)}=\frac{B}{\frac{1}{2}\omega}=\frac{1}{(2-\omega^2)^2+\frac{1}{16}\omega}

\therefore A=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega}   and        B=\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega}

Therefore the required solution is

U(t)=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega} cos\omega t +\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega} sin \omega t

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1 year ago
A technician is troubleshooting a problem. The technician tests the theory and determines the theory is confirmed. Which of the
vladimir1956 [14]

Answer:

b)  Document lessons learned.

Explanation:

First he should do documentation

then C

then D

then A

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1 year ago
A force F of strength 20 N acts on an object of mass 3 kg as it moves a distance of 4 m. If F is perpendicular to the 4 m displa
forsale [732]
Work = (displacement) x (force in the direction of the displacement)

Since none of the force is in the direction of the displacement, the work it does
is zero. Something else is making the object move.  It's not the 20N force.
3 0
1 year ago
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A force of 1.35 newtons is required to accelerate a book by 1.5 meters/second2 along a frictionless surface. What is the mass of
malfutka [58]

B. seems like the answer, I'm not sure.

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2 years ago
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