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2 years ago

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A 50-kg load is suspended from a steel wire of diameter 1.0 mm and length 11.2 m. By what distance will the wire stretch? Young'

s modulus for steel is 2.0 x 1011 Pa.
Please show work.

1 answer:

lbvjy [14]2 years ago

5 0

Answer:

3.5 cm

Explanation:

mass, m = 50 kg

diameter = 1 mm

radius, r = half of diameter = 0.5 mm = 0.5 x 10^-3 m

L = 11.2 m

Y = 2 x 10^11 Pa

Area of crossection of wire = π r² = 3.14 x 0.5 x 10^-3 x 0.5 x 10^-3

= 7.85 x 10^-7 m^2

Let the wire is stretch by ΔL.

The formula for Young's modulus is given by

$Y =\frac{mgL}{A\Delta L}$

$\Delta L =\frac{mgL}{A\times Y}$

ΔL = 0.035 m = 3.5 cm

Thus, the length of the wire stretch by 3.5 cm.

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A beam of electrons moves at right angles to a magnetic field of 4.5 × 10-2 tesla. If the electrons have a velocity of 6.5 × 106

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Answer:

$4.7\cdot 10^{-14}N$

Explanation:

For a charge moving perpendicularly to a magnetic field, the force experienced by the charge is given by:

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a rod of some material 0.20 m long elongates 0.20 mm on heating from 21 to 120°c. determine the value of the linear coefficient

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Answer:

The value of the linear coefficient of thermal expansion is : α=1.01 *10⁻⁵ (ºC)⁻¹

Explanation:

Li = 0.2m

ΔL = 0.2 mm = 0.0002m

T1 = 21ºC

T2 = 120ºC

ΔT =99ºC

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α =0.0002m /(0.2m * 99ºC)

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At 1 atm pressure, the heat of sublimation of gallium is 277 kj/mol and the heat of vaporization is 271 kj/mol. to the correct n

Andreyy89

Below are the choices that can be found elsewhere:

a. 268 kJ
<span>b. 271 kJ </span>
<span>c. 9 kJ </span>
<span>d. 6 kJ
</span>
So the key thing to realize here is what the information given to you actually means. Sublimation is going from a sold to a gas. Vaporization is going from a liquid to a gas. Hence you can create two equations from the information that you have:

<span>Ga (s) --> Ga (g) delta H = 277 kJ/mol </span>

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<span>From these two equations, you can then infer how to get the melting equation be simply finding the difference between the sublimation (two steps) and vaporization (one step). </span>

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<span>*ANSWER* </span>
<span>9 kJ/mol (C)</span>

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2 years ago

A 1.5 m cylinder of radius 1.1 cm is made of a complicated mixture materials. Its resistivity depends on the distance x from the

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Answer:

a) $R = 171$ μΩ

b) $E = 1.7 *10^{-4} V/m$

c) $R_{2} = 1.16 *10^{-4}$ Ω

here * stand for multiplication

Explanation:

length of cylinder = 1.5 m

radius of cylinder = 1.1 cm

resistivity depends on the distance x from the left

$p(x)=a+bx^2$ ............(i)

using equation

$R = \frac{pl}{a}$

let dR is the resistance of thickness dx

$dR =\frac{p(x)dx}{a}$

where p(x) is resistivity l is length

a is area

$\int\limits^R_0 {dR} =\frac{1}{\pi r^2} \int\limits^L_0 {(a+bx^2)} \, dx \\$ .........................(2)

after integration

$R = \frac{[aL+\frac{bL^3}{3}] }{\pi r^2}$ ...............(3)

it is given $p(0) = a = 2.25 * 10 ^{-8}$ Ωm

$p(L) = a + b(L)^2$ = $8.5 * 10 ^{-8}$ Ωm

$8.5 * 10 ^{-8} = 2.25 * 10^{-8}+b(1.5)^2\\$

(here * stand for multiplication )

on solving we get

$b = 2.78* 10^{-8}$ Ωm

put each value of a and b and r value in equation 3rd we get

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E = p(x)L

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for given current I = 1.75 A

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$E = \frac{[a+b(L/2)^2]I }{\pi r^2}$

by substitute the values

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$E = 1.7 *10^{-4} V/m$

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c ).

75 cm means length will be half

that is x = L/2

integrate the second equation with upper limit L/2

Let resistance is $R_{1}$

so after integration we get

$R_{1} = \frac{[a(L/2) +(b/3)(L^3/8)]}{\pi r^2}$

substitute the value of a , b and L we get

$R_{1} = 5.47 * 10 ^{-5}$ Ω

for second half resistance

$R_{2} = R- R_{1}$

$R_{2} = 1.7 *10^{-4} -5.47 *10^{-5}$

$R_{2} = 1.16 *10^{-4}$ Ω

(here * stand for multiplication )

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