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docker41 [41]
3 years ago
15

Router bits with cutting lips on the ends are for

Engineering
1 answer:
Vlad1618 [11]3 years ago
3 0

Answer: plunging

Explanation: from modern cabinet making fifth edition: with plunge bits, the end of the flute has cutting edges

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mote1985 [20]
I just deleted among us :/
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3 years ago
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*13–52. A girl, having a mass of 15 kg, sits motionless relative to the surface of a horizontal platform at a distance of r = 5
Dennis_Churaev [7]

Answer:

  w = 0.626 rad / s, v = 3.13 m/s

Explanation:

For this exercise let's use Newton's second law

        F = m a

Where the force is a friction force and the acceleration is centripetal,

      a = v² / r = w² r

The formula for friction force

     fr = μ N

  In a free body diagram

       N- W = 0

      W = N

The frictiμon outside goes from zero to the maximum value, let's calculate the speed for the maximum value of the friction force, replace

       μ m g = m w² r

       w = √ μ g / r

Let's calculate

     w = √(0.2 9.8 / 5)

     w = 0.626 rad / s

angular and linear velocity are related

      v = w r

      v = 0.626 5

      v = 3.13 m/s

6 0
3 years ago
. True or False: It is possible with feedback control to obtain a stable closed-loop system even when the underlying open-loop s
zheka24 [161]

Answer:

True.

Explanation:

8 0
2 years ago
In Millikan's oil drop experiment, if the electricfield between the plates was of just the right magnitude, it wouldexactly bala
Vilka [71]

Answer:

The electric field to balance the weight is approximately equals to 3.49x10^5 Newton/Coulumb

Explanation:

In order to be stationary position, magnitude of the total force due to electric field should be equal to the gravitational force that is, |F_{electrostatic}| = |F_{gravitational}|

where

F_{electrostatic}=q.E

where <em>q</em> is the charge of the droplet and E is the electric field. On the other hand

F_{gravitational}=m.g=V.d.g=\frac{4}{3}.\pi.r^3.d.g

where <em>m</em>,<em>V ,d</em> and<em> r</em>  are the mass, volume, density and radius of the oil droplet respectively and <em>g </em>is the gravitational acceleration (g=9,80665 m/sn^2). By using the first equation and solving it for the electric field we can write,

q.E=\frac{4}{3}.\pi.r^3.d.g\\E=\frac{4}{3}.\pi.\frac{r^3.d.g}{q}\\E=\frac{4}{3}.\pi\frac{(1,6x10^{-4})^3.(0,85x10^{-3}).(9,81)}{1,6x10^{-19}}\\E\approx 3,49x10^{5}  (N/C)

(Note: d=0.85 x 10^-3 kg/cm^3 and the unit of electric field is Newton per coulumb (N/C))

8 0
4 years ago
Why is a Screw Pump a quiet operating pump?
Julli [10]

Answer:

Screw pumps like those that are used in viscous fluids transportation have lubricating properties. The fluid flows in a line axially without any disturbence. Since, the motion of fluid is free from any sort of rotation even at high speeds there is no turbulence and motion of the pump is quite. Thus screw Pumps provides smooth operation with extremely low pulsation, lower noise levels and higher efficiency.

6 0
4 years ago
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