All categories

3 years ago

What is a milling machine

Engineering

1 answer:

Olenka [21]3 years ago

3 0

Answer:

Milling machines are a type of machinery for removing material from a workpiece using rotary cutters. These machines can drill, bore, and cut an array of materials.

Explanation:

If this helps you please mark brainliest!

Have a nice day!

You might be interested in

A fluid has a dynamic viscosity of 0.048 Pa.s and a specific gravity of 0.913. For the flow of such a fluid over a flat solid su

sattari [20]

Answer:

Explanation:

First we should recall how Newton's laws relates shear stress to a fluid's velocity profile:

$\tau = \mu \cfrac{\partial v}{\partial y}$

where tau is the shear stress, mu is viscosity, v is the fluid's velocity and y is the direction perpendicular to flow.

Now, in this case we have a parabolic velocity profile, and also we know that the fluid's velocity is zero at the boundary (no-slip condition) and that the vertex (maximum) is at $y=75 \, mm$ and the velocity at that point is $1.125 \, m/s$

We can put that in mathematical terms as:

$v(y)= A+ By +Cy^2 \\v(0) = 0\\v(75 \, mm) = 1.125 \, m/s\\v'(75 \, mm) = 0\\$

From the no-slip condition, we can deduce that $A=0$ and so we are left with just two terms:

$v(y) = By + C y ^2 \\$

We know that the vertex is at $y= 75 \, mm$ and so we can rewrite the last equation as:

$v(y) = k(y-75 \, mm) ^2+h$

where k and h are constants to be determined. First we check that $v( 75 \, mm) = 1.125 \, m/s$ :

$v( 75 \, mm) = k(75 \, mm -75 \, mm) ^2+h = h = 1.125 \, m/s\\\\h= v_{max} = 1.125 \, m/s$

So we found that h was the maximum velocity for the fluid, now we have to determine k, for that we need to make use of the no-slip condition.

$v( 0) = k( -75 \, mm) ^2+ 1.125 \, m/s= 0 \quad (no \, \textendash slip) \\\\k= - \cfrac{ 1.125 \, m/s }{(75 \, mm ) ^2} = - \cfrac{ 1125 \, mm/s }{(75 \, mm ) ^2}\\\\k= - \cfrac{0.2}{mm \times s}$

And thus we find that the final expression for the fluid's velocity is:

$v( y) = 1125- 0.2 ( y -75 ) ^2$

where v is in mm/s and y is in mm.

In SI units it would be:

$v( y) = 1.125- 200 ( y -0.075 ) ^2$

To calculate the shear stress, we need to take the derivative of this expression and multiply by the fluid's viscosity:

$\tau = \mu \cfrac{\partial v}{\partial y}$

$\tau =0.048\, \cdot (-400) ( y-0.075 )$

for $y= 0.050 \, m$ we have:

$\tau =0.048\, \cdot (-400) ( 0.050 -0.075 ) = 0.48\, Pa$

Which is our final result

5 0

4 years ago

Gray cast iron, with an ultimate tensile strength of 31 ksi and an ultimate compressive strength of 109 ksi, has the following s

suter [353]

Using an appropriate failure theory, find the factor of safety in each case. State the name of the theory that you are using the theory is max stress theory.

<h3>Wat is the max stress theory?</h3>

The most shear strain concept states that the failure or yielding of a ductile fabric will arise whilst the most shear strain of the fabric equals or exceeds the shear strain fee at yield factor withinside the uniaxial tensile test.”

Stress states at various critical locations are f= 2.662.

Read more about strain:

brainly.com/question/6390757

#SPJ1

3 0

2 years ago

To make an even better electrical junction, what should you do?

docker41 [41]

Answer:

A:Solder it.

Explanation:

Hopefully this helps!

4 0

3 years ago

A car accelerates from rest with an acceleration of 5 m/s^2. The acceleration decreases linearly with time to zero in 15 s, afte

Tpy6a [65]

Answer: At time 18.33 seconds it will have moved 500 meters.

Explanation:

Since the acceleration of the car is a linear function of time it can be written as a function of time as

$a(t)=5(1-\frac{t}{15})$

$a=\frac{d^{2}x}{dt^{2}}\\\\\therefore \frac{d^{2}x}{dt^{2}}=5(1-\frac{t}{15})$

Integrating both sides we get

$\int \frac{d^{2}x}{dt^{2}}dt=\int 5(1-\frac{t}{15})dt\\\\\frac{dx}{dt}=v=5t-\frac{5t^{2}}{30}+c$

Now since car starts from rest thus at time t = 0 ; v=0 thus c=0

again integrating with respect to time we get

$\int \frac{dx}{dt}dt=\int (5t-\frac{5t^{2}}{30})dt\\\\x(t)=\frac{5t^{2}}{2}-\frac{5t^{3}}{90}+D$

Now let us assume that car starts from origin thus D=0

thus in the first 15 seconds it covers a distance of

$x(15)=2.5\times 15^{2}-\farc{15^{3}}{18}=375m$

Thus the remaining 125 meters will be covered with a constant speed of

$v(15)=5\times 15-\frac{15^{2}}{6}=37.5m/s$

in time equalling $t_{2}=\frac{125}{37.5}=3.33seconds$

Thus the total time it requires equals 15+3.33 seconds

t=18.33 seconds

3 0

3 years ago

There are 22 gloves in a drawer: 5 pairs of red gloves, 4 pairs of yellow, and 2 pairs of green. You select the gloves in the da

antiseptic1488 [7]

Answer:

Best case = 2 gloves

Given Information:

Red gloves = 5 pairs

Yellow gloves = 4 pairs

Green gloves = 2 pairs

smallest number of gloves you need to select to have at least one matching pair in the best case = ?

Explanation:

How many gloves do you need to make one matching pair?

Yes, you are right. 2 gloves makes a matching pair and it is the smallest number of gloves you need to select to have at least one matching pair.

But what about worst case?

lets say

you tried all 5 red gloves either all of them were left or right

then you tried 4 yellow gloves either all of them were left or right

then you tried 2 green gloves either all of them were left or right

Now all the left or right gloves are tried (5 + 4 + 2 = 11) and the 12th one will definitely be either matching red, yellow or green.

Therefore, in the worst case scenario, the smallest number of gloves you need to select to have at least one matching pair is 12.

3 0

3 years ago