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2 years ago

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A steel ring is 50mm diameter and 2mm thick. it must be fitted onto a shaft 50.04 mm diameter. calculate the temperature to whic

h it must be heated in order to fit the shaft. the initial temperature is 20 deg C and the coefficient of linear expansion is 15x10^-6 per degrees celcius. calculate the stress and forced induced in the ring
okay so the stress and forced induced I can do that but the first part is confusing me because there is no final temperature.

1 answer:

MAVERICK [17]2 years ago

4 0

The temperature to which it must be heated in order to fit the shaft is 73.33 ⁰C.

<h3>Linear expansivity </h3>

The temperature to which it must be heated in order to fit the shaft is calculated as follows;

$\frac{\Delta L}{L} = \alpha \Delta T\\\\\Delta T = \frac{\Delta L}{\alpha L}$

where;

ΔT is change in temperature
ΔL is change in length = 50.04 mm - 50 mm = 0.04 mm
α is coefficient of linear expansion
L is original length

ΔT = (0.04)/(50 x 15 x 10⁻⁶)

ΔT = 53.3 ⁰C

<h3>Final temperature</h3>

T₂ - T₁ = ΔT

T₂ = ΔT + T₁

where;

T₂ is final temperature
T₁ is initial temperature

T₂ = 53.3 + 20

T₂ = 73.33 ⁰C

Learn more about linear expansivity here: brainly.com/question/14325928

#SPJ1

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Answer:

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3 years ago

The total energy of a block—spring system is 0.18 J. The amplitude is 14.0 cm and the maximum speed is 1.25 m/s. Find: (a) the m

algol13

a) The mass is 0.23 kg

b) The spring constant is 1.25 N/m

c) The frequency is 1.42 Hz

d) The speed of the block is 1.08 m/s

Explanation:

a)

We can find the mass of the block by applying the law of conservation of energy: in fact, the total mechanical energy of the system (which is sum of elastic potential energy, PE, and kinetic energy, KE) is constant:

$E=PE+KE=const.$

The potential energy is given by

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Re-arranging eq.(1), we can find the mass:

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b)

The maximum speed in a spring-mass system is also given by

$v_{max} =\sqrt{\frac{k}{m}} A$

where

k is the spring constant

m is the mass

A is the amplitude

Here we have:

$v_{max}=1.25 m/s$ is the maximum speed

m = 0.23 kg is the mass

A = 14.0 cm = 0.14 m is the amplitude

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c)

The frequency in a spring-mass system is given by

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$

where

k is the spring constant

m is the mass

In this problem, we have:

k = 18.3 N/m is the spring constant (found in part b)

m = 0.23 kg is the mass (found in part a)

Substituting and solving for f, we find the frequency of the system:

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d)

We can solve this part by using the law of conservation of energy; in fact, we have

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E = 0.18 J

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k = 18.3 N/m is the spring constant

m = 0.23 kg is the mass

Substituting

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$v=\sqrt{\frac{2E-kx^2}{m}}=\sqrt{\frac{2(0.18)-(18.3)(0.07)^2}{0.23}}=1.08 m/s$

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Answer:

Explanation: I think...

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