Answer:
a) the metal removal rate is 14.4 in³/min
b) the cutting time is 0.98 min
Explanation:
Given the data from the question
first we find the rpm for the spindle of the drilling tool, using the equation
Ns = 12V/πD
V is the cutting speed(120 fpm) and D is the diameter of the hole( 2 in)
so we substitute
Ns = 12 × 120 / π2
Ns = 1440 / 6.2831
Ns = 229.18 rmp
Now we find the metal removal rate using the equation
MRR = (πD²/4) Fr × Ns
Fr is the feed rate( 0.02 ipr ),
so we substitute
MRR = ((π × 2²)/4) × 0.02 × 229.18
MRR = 14.3998 ≈ 14.4 in³/min
Therefore the metal removal rate is 14.4 in³/min
Next we find the allowance for approach of the tip of the drill
A = D/2
A = 2/2
= 1 in
now find the time required to drill the hole
Tm = (L + A) / (Fr × Ns)
Lis the the depth of the hole( 3.5 in)
so we substitute our values
Tm = (3.5 + 1) / (0.02 × 229.18 )
Tm = 4.5 / 4.5836
Tm = 0.98 min
Therefore the cutting time is 0.98 min
What test should be performed on abrasive wheels
Answer:
before wheel is put on it should be looked at for damage and a sound or ring test should be done to check for cracks, to test the wheel it should be tapped with a non metallic instrument (I looked it up)
The test that should be performed on abrasive wheels is the ring test.
What is the purpose of the ring test on the abrasive wheels?The ring test can be regarded as one of the mechanical test that is used to know whether the wheel is cracked or damaged.
To carry out this test , the wheel will be arranged to be in the 45 degrees each side and it is then aligned to be at a specific diameter, this can be done by the expert in this field to know the state of that wheel.
Learn more about ring test on:
https://brainly.com/question/4621112
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You are playing guitar on a stool that is 22" tall. How tall is the stool if it is expressed as a combination of feet and inches?
Answer:
1 foot 10 inches
Explanation:
1 foot = 12 inches + 10 inches = 22 inches
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A spherical Gaussian surface of radius R is situated in space along with both conducting and insulating charged objects. The net electric flux through the Gaussian surface is:______
Answer:
Ф = [tex]\frac{Q}{e_{0} } [ \frac{\frac{4\pi }{3 }(R)^3 }{\frac{4}{3}\pi (R)^3 } ][/tex]
Explanation:
Radius of Gaussian surface = R
Charge in the Sphere ( Gaussian surface ) = Q
lets take the radius of the sphere to be equal to radius of the Gaussian surface i.e. R
To determine the net electric flux through the Gaussian surface
we have to apply Gauci law
Ф = 4[tex]\pi r^2 E[/tex]
Ф = [tex]\frac{Q_{enc} |}{e_{0} }[/tex]
= [tex]\frac{Q}{e_{0} } [ \frac{\frac{4\pi }{3 }(R)^3 }{\frac{4}{3}\pi (R)^3 } ][/tex]
the pressure rise, across a pump can be expressed as where D is the impeller diameter, p, is the fluid density, w is the rotational speed, adn q is the flowrate. determine a suitable set of dimensionless parameters
Answer:
hello your question is incomplete below is the complete question
The pressure rise Δp across a pump can be expressed as Δp = f(D, p, w, Q) where D is the impeller diameter, p is the fluid density, w is the rotational speed, and Q is the flowrate. determine a suitable set of dimensionless parameters
answer : Δp / D^2pw^2 = Ф (Q / D^3w )
Explanation:
k ( number of variables ) = 5
r ( number of reference dimensions ) = 3
applying the pi theorem
hence the number of pi terms = k - r = 5 - 3 = 2
To provide some perspective on the dimensions of atomic defects, consider a metal specimen that has a dislocation density of 105 mm^-2 . Suppose that all the dislocations in 1000 mm^3 (1 cm^3) were somehow removed and linked end to end.
Required:
a. How far (in miles) would this chain extend?
b. Now suppose that the density is increased to 1010 mm^-2 by cold working. What would be the chain length of dislocations in 1000 mm^3 of material?
Answer:
[tex]62.14\ \text{miles}[/tex]
[tex]6213727.37\ \text{miles}[/tex]
Explanation:
The distance of the chain would be the product of the dislocation density and the volume of the metal.
Dislocation density = [tex]10^5\ \text{mm}^{-2}[/tex]
Volume of the metal = [tex]1000\ \text{mm}^3[/tex]
[tex]10^5\times 1000=10^8\ \text{mm}\\ =10^5\ \text{m}[/tex]
[tex]1\ \text{mile}=1609.34\ \text{m}[/tex]
[tex]\dfrac{10^5}{1609.34}=62.14\ \text{miles}[/tex]
The chain would extend [tex]62.14\ \text{miles}[/tex]
Dislocation density = [tex]10^{10}\ \text{mm}^{-2}[/tex]
Volume of the metal = [tex]1000\ \text{mm}^3[/tex]
[tex]10^{10}\times 1000=10^{13}\ \text{mm}\\ =10^{10}\ \text{m}[/tex]
[tex]\dfrac{10^{10}}{1609.34}=6213727.37\ \text{miles}[/tex]
The chain would extend [tex]6213727.37\ \text{miles}[/tex]