Formulae for determining the magnitude of the load overcomed

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

MAVERICK [17]

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

4 0

1 year ago

A block of mass 4 kilograms is initially moving at 5m/s on a horizontal surface. There is friction between the block and the sur

Dmitriy789 [7]

Answer:

d = 2.54 [m]

Explanation:

Through the theorem of work and energy conservation, we can find the work that is done. Considering that the energy in the initial state is only kinetic energy, while the energy in the final state is also kinetic, however, this is zero since the body stops.

$E_{k1}+W=E_{k2}\\$

where:

W = work [J]

Ek1 = kinetic energy at initial state [J]

Ek2 = kinetic energy at the final state = 0.

We must remember that kinetic energy can be calculated by means of the following expression.

$\frac{1}{2} *m*v^{2}-W=0\\W= \frac{1}{2} *4*(5)^{2}\\W= 50 [J]$

We know that work is defined as the product of force by distance.

$W=F*d$

where:

F = force [N]

d = distance [m]

But the friction force is equal to the product of the normal force (body weight) by the coefficient of friction.

$f=m*g*0.5\\f = 4*9.81*0.5\\f = 19.62 [N]$

Now solving the equation for the work.

$d=W/F\\d = 50/19.62\\d = 2.54[m]$

4 0

2 years ago

What is solar made from

Eduardwww [97]

Solar cells are made out of silicon wafers. These are made out of the element silicon, a hard and brittle crystalline solid that is the second most abundant element in the Earth's crust after oxygen. If you're at the beach and see shiny black specks in the sand, that's silicon.

Hope this helps!

Please give brainliest!

5 0

2 years ago

PLEASE HELP! Compare and contrast between reflection and refraction. Be sure to be specific in your explanation and state in you

horsena [70]

Most ultrasound technicians train in an associates degree, but can also get a bachelors degree in the field. There is also certification programs you can go through. Hope this helped!!! :)

7 0

3 years ago

The three forces shown act on a particle. what is the direction of the resultant of these three forces?

melisa1 [442]

Missing figure: http://d2vlcm61l7u1fs.cloudfront.net/media/f5d/f5d9d0bc-e05f-4cd8-9277-da7cdda3aebf/phpJK1JgJ.png

Solution:
We need to find the magnitude of the resultant on both x- and y-axis.

x-axis) The resultant on the x-axis is
$F_x = 65 N\cdot cos 30^{\circ} - 30 N - 20 N\cdot sin 20^{\circ} = 19.45 N$
in the positive direction.

y-axis) The resultant on the y-axis is
$F_y = 65 N \cdot sin 30^{\circ} - 20 N \cdot cos 20^{\circ} = 13.70 N$
in the positive direction.

Both Fx and Fy are positive, so the resultant is in the first quadrant. We can find the angle and so the direction using
$\tan \alpha = \frac{F_y}{F_x} = \frac{13.70 N}{19.45 N}=0.7$
from which we find
$\alpha=35^{\circ}$

7 0

3 years ago

Formulae for determining the magnitude of the load overcomed​

Formulae for determining the magnitude of the load overcomed