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3 years ago

8

The rigid link is supported by a pin at A and two A-36 steel wires, each having an unstretched length of 12 in. and cross-sectio

nal area of 0.0125 in2. Determine the force developed in the wires when the link supports the vertical load of 350 lb.

1 answer:

Leokris [45]3 years ago

4 0

Answer:

Extension of the wire is not indicated

Explanation:

Young modulus (YM)= (stress/Strain); Tensile Stress/tensile Strain

=> YM *Strain =Stress=F/A; A is cross sectional area and F is the force

(YM * Strain * A)/36 =F (Force developed in each wire)

F = (350 *e*0.0125)/12

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In a Young's double-slit experiment, a set of parallel slits with a separation of 0.142 mm is illuminated by light having a wave

Alexandra [31]

Answer:

$1.152\ \mu m$

$1.44\ \mu m$

Explanation:

d = Gap between slits = 0.142 mm

$\lambda$ = Wavelength of light = 576 nm

L = Distance between light and screen = 3.5 m

m = Order = 2

Difference in path length is given by

$\delta=dsin\theta=m\lambda\\\Rightarrow \delta=2\times 576\times 10^{-9}\\\Rightarrow \delta=0.000001152\ m\\\Rightarrow \delta=1.152\times 10^{-6}\ m=1.152\ \mu m$

The difference in path lengths is $1.152\ \mu m$

For dark fringe the difference in path length is given by

$\delta=(m+\dfrac{1}{2})\lambda\\\Rightarrow \delta=(2+\dfrac{1}{2})\times 576\times 10^{-9}\\\Rightarrow \delta=0.00000144\ m=1.44\ \mu m$

The difference in path length is $1.44\ \mu m$

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3 years ago

1. What is evaporation and how does it affect weather?

alina1380 [7]

Answer:

Explanation:don’t know

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3 years ago

The origin of an x axis is placed at the center of a nonconducting solid sphere of radius R that carries a charge +qsphere distr

MA_775_DIABLO [31]

Answer:

$q=49Q/64$

and

$x =16R/15$

Explanation:

See attached figure.

$E_{Q}=$ E due to sphere

$E_{q}=$ E due to particule

$E_{total}=E_{Q}-E_{q}=0$ (1)

according to the law of gauss and superposition Law:

$E_{Q}=E_{1}+E_{2}=E_{2}$ ; electric field due to the small sphere with r1=R/4

$E_{Q}=kq_{2}/(r_{1}^{2})=$

$q_{2}=density*4/3*pi*r_{1}^{3}=Q/(4/3*pi*R^{3})*4/3*pi*r_{1}^{3}=Q*r_{1}^{3}/R^{3}$

then: $E_{Q}=kq_{2}/(r_{1}^{2})=k*Q*r_{1}^{3}/(R^{3}*r_{1}^{2}) = kQ/(4*R^{2})$ (2)

on the other hand, for the particule:

$E_{q}=kq/(r_{p}^{2})$

$r_{p}=2R-R/4=7R/4$ ⇒ $E_{q}=16kq/(49R^{2})$ (3)

We replace (2) y (3) in (1):

$E_{total}=E_{Q}-E_{q}=0=kQ/(4*R^{2}) - 49kq/(16R^{2})$

$q=49Q/64$

--------------------

if R<x<2R AND $E_{total}=E_{Q}-E_{q}=0$

$E_{total}=E_{Q}-E_{q}=0=kQ/(x^{2}) - kq/(2R-x^{2})$

remember that $q=49Q/64$

then:

$Q(2R-x^{2})=49/64*x^{2}$

solving:

$x_{1} =16R/15$

$x_{2} =16R$

but: R<x<2R

so : $x =16R/15$

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Answer:

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Explanation:

This is answers

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2 years ago

Read 2 more answers