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2 years ago

6

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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Reika [66]

Answer:

$tan \theta = \mu_s$

Explanation:

An object is at rest along a slope if the net force acting on it is zero. The equation of the forces along the direction parallel to the slope is:

$mg sin \theta - \mu_s R =0$ (1)

where

$mg sin \theta$ is the component of the weight parallel to the slope, with m being the mass of the object, g the acceleration of gravity, $\theta$ the angle of the slope

$\mu_s R$ is the frictional force, with $\mu_s$ being the coefficient of friction and R the normal reaction of the incline

The equation of the forces along the direction perpendicular to the slope is

$R-mg cos \theta = 0$

where

R is the normal reaction

$mg cos \theta$ is the component of the weight perpendicular to the slope

Solving for R,

$R=mg cos \theta$

And substituting into (1)

$mg sin \theta - \mu_s mg cos \theta = 0$

Re-arranging the equation,

$sin \theta = \mu_s cos \theta\\\rightarrow tan \theta = \mu_s$

This the condition at which the equilibrium holds: when the tangent of the angle becomes larger than the value of $\mu_s$ , the force of friction is no longer able to balance the component of the weight parallel to the slope, and so the object starts sliding down.

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2 years ago

What is the equation describing the motion of a mass on the end of a spring which is stretched 8.8 cm from equilibrium and then

Darya [45]

Answer:

$y = -8.37 cm$

Explanation:

As we know that the equation of SHM is given as

$y = A cos(\omega t)$

here we know that

$\omega = \frac{2\pi}{T}$

here we have

$T = 0.66 s$

now we have

$\omega = \frac{2\pi}{0.66}$

$\omega = 3\pi$

now we have

$y = (8.8 cm) cos(3\pi t)$

now at t = 2.3 s we have

$y = (8.8 cm) cos(3\pi \times 2.3)$

$y = -8.37 cm$

6 0

1 year ago

Read 2 more answers

A solid sphere of brass (bulk modulus of 14.0 ✕ 1010 N/m2) with a diameter of 2.20 m is thrown into the ocean. By how much does

astra-53 [7]

Answer:

Diameter decreases by the diameter of 0.0312 m.

Explanation:

Given that,

Bulk modulus = 14.0 × 10¹⁰ N/m²

Diameter d = 2.20 m

Depth = 2.40 km

Pressure = ρ g h = 1030 × 9.81 × 2.4 × 1000

= 24.25 × 10⁶ N/m²

Volume = $\dfrac{4}{3} \pi r^3$

$\dfrac{\Delta V}{V}=\dfrac{(\Delta r)^3}{r^3}$

Bulk modulus is equal to

$B = -\dfrac{\Delta P}{\dfrac{\Delta V}{V} }$

now

$B = -\dfrac{24.25 \times 10^6}{\dfrac{(\Delta r)^3}{r^3} }$

$B = -\dfrac{24.25 \times 10^6}{\dfrac{(\Delta r)^3}{1.1^3} }$

$(\Delta r)^3 = \dfrac{24.25 \times 10^6 \times 1.1^3}{14.0 \times 10^{10}}$

Δ r = -0.0156 m

change in diameter

Δ d = -2 × 0.0156

Δ d = -0.0312 m

Diameter decreases by the diameter of 0.0312 m.

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2 years ago

A confused dragonfly flies forward and backward in a straight line. Its motion is shown on the following graph of horizontal pos

sertanlavr [38]

Answer:

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Explanation:

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average speed = displacement/time = 0/(8 s)

average speed = 0

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2 years ago

What is the binding energy (in J/mol or kJ/mol) of an electron in a metal whose threshold frequency for photoelectrons is 2.50 u

Aleksandr-060686 [28]

Answer:

binding energy is 99771 J/mol

Exlanation:

given data

threshold frequency = 2.50 × $10^{14}$ Hz

solution

we get here binding energy using threshold frequency of the metal that is express as

$E=h\nu_{o}$ ..................1

here E is the energy of electron per atom $E=\frac{x}{N}$ and h is plank constant i.e. $6.626\times10^{-34} Js$ and x is binding energy

and here N is the Avogadro constant = $6.023\times10^{23}$

so E will $\frac{x}{6.023\times10^{23}}$

E = $3.19\times10^{-19} J$

so put value in equation 1 we get

$\frac{x}{6.023\times10^{23}}$ = 2.50 × $10^{14}$ × $6.626\times10^{-34} Js$

solve it we get

x = 99770.99

so binding energy is 99771 J/mol

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2 years ago