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melomori [17]
1 year ago
8

imagine you have to drill an accurate bolt pattern into a part using a manual machine. what type of machine would you use and wh

at feature(s) of this machine allow it to locate holes accurately?
Physics
1 answer:
max2010maxim [7]1 year ago
6 0

The drill press's primary function is drilling, but it can also be used for reaming, countersinking, counterboring, and tapping.

<h3>Before drilling a hole, what tool do machinists use to precisely locate the hole?</h3>

Making a tiny divot with a Spotting Drill will help you find the drill's center when you start a plunge. Some machinists use these tools to chamfer the top of drilled holes, which is a different function for them. When inserted, screw heads sit flush with the component thanks to the chamfer.

<h3>What device do we employ to drill a hole in a part?</h3>

Most often, hole-making tools are related with drilling and drill bits. They offer a quick, simple, and affordable technique to make machined holes. Drill bits operate by axially entering the workpiece and creating a hole with a diameter.

To know more about drill visit:-

brainly.com/question/4281675

#SPJ4

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You connect a 250-2 resistor, a 1.20-mH inductor, and a 1.80-uF capacitor in series across a 60.0-Hz, 120-V (peak) source. When
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Answer:

Explanation:

 reactance of inductor = wL = 2 X 3.14 X 60 X 1.2 X 10⁻³ = .45 ohm.

reactance of capacitor = 1/wC = 1/( 2 X 3.14 X 60 X 1.8 X 10⁻⁶ ) = 1474.4

Impedence of the circuit =[ R² + ( I/wC - wL)  ]¹/²  = [250² + ( 1474.4-.45 )]¹/²

Impedence = 1495 ohm.

RMS Voltage = 120/ 1.414 = 84.86 V

current = 84.86 / 1495 = 0.0576

Potential over resistance = 0.0576 x 250 = 14.2 V.

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A particle of mass 2.37 kg is subject to a force that is always pointed towards East or West but whose magnitude changes sinusoi
Pavlova-9 [17]

Answer:

v(1.5)=0.7648\ m/s

Explanation:

<u>Dynamics</u>

When a particle of mass m is subject to a net force F, it moves at an acceleration given by

\displaystyle a=\frac{F}{m}

The particle has a mass of m=2.37 Kg and the force is horizontal with a variable magnitude given by

F=2cos1.1t

The variable acceleration is calculated by:

\displaystyle a=\frac{F}{m}=\frac{2cos1.1t}{2.37}

a=0.8439cos1.1t

The instant velocity is the integral of the acceleration:

\displaystyle v(t)=\int_{t_o}^{t_1}a.dt

\displaystyle v(t)=\int_{0}^{1.5}0.8439cos1.1t.dt

Integrating

\displaystyle v(1.5)=0.7672sin1.1t \left |_0^{1.5}

\displaystyle v(1.5)=0.7672(sin1.1\cdot 1.5-sin1.1\cdot 0)

\boxed{v(1.5)=0.7648\ m/s}

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