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Ber [7]
2 years ago
9

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.
Physics
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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taurus [48]

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The possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.

Explanation:

Given that,

The notes produced by a tuba range in frequency from approximately 45 Hz to 375 Hz.

The speed of sound in air is 343 m/s.

To find,

The wavelength range for the corresponding frequency.

Solution,

The speed of sound is given by the following relation as :

v=f_1\lambda_1

Wavelength for f = 45 Hz is,

\lambda_1=\dfrac{v}{f_1}

\lambda_1=\dfrac{343}{45}=7.62\ m

Wavelength for f = 375 Hz is,

\lambda_2=\dfrac{v}{f_2}

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6 0
4 years ago
A wad of clay of mass m1 = 0.49 kg with an initial horizontal velocity v1 = 1.89 m/s hits and adheres to the massless rigid bar
notka56 [123]

Answer:

<h2>The angular velocity just after collision is given as</h2><h2>\omega = 0.23 rad/s</h2><h2>At the time of collision the hinge point will exert net external force on it so linear momentum is not conserved</h2>

Explanation:

As per given figure we know that there is no external torque about hinge point on the system of given mass

So here we will have

L_i = L_f

now we can say

m_1v_1\frac{L}{2} = (m_2L^2 + m_1(\frac{L}{2})^2)\omega

so we will have

0.49(1.89)(0.45) = (2.13(0.90)^2 + 0.49(0.45)^2)\omega

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3 years ago
A 0.68 kg squirrel is resting on a branch 8 meters above the ground. What is the gravitational potential energy of a squirrel? A
ANEK [815]

Answer:

The gravitational potential energy of a squirrel is 53.312 J.

Explanation:

We have,

Mass of a squirrel is 0.68 kg

It is placed at a height of 8 m above the ground.

It is required to find the gravitational potential energy of a squirrel. It is possessed by an object due to its position. Its formula is given by :

E=mgh\\\\E=0.68\times 9.8\times 8\\\\E=53.312\ J

So, the gravitational potential energy of a squirrel is 53.312 J.

7 0
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