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Crazy boy [7]
4 years ago
13

1. The workpart in a turning operation is 88 min in diameter and 400 mm long. A feed of 0.25 mm/rev is used in this operation. I

f cutting speed is 3.5 m/s, the too should be changed in every 3 workparts, but if the cutting speed is 2.5 m/sec, the tool can be used to produce 20 pieces between the tool changes. Determine the cutting speed that will allow the tool to be used for 50 parts between tool changes.

Physics
1 answer:
Vedmedyk [2.9K]4 years ago
4 0

Find the given attachments

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Increasing the mass of an object increases its momentum.

Explanation:

  • Momentum of an object is measured as the quantity of motion done by the object.
  • It is calculated using the formula, p = m × v where m is mass of the object and v is the velocity of the object.
  • As momentum and mass vary proportionally, as seen in the formula, increasing the mass of an object will also increase its momentum.
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How charged objects can affect other objects without touching them. ?
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A car with a mass of 500 kg, gets off road and goes straight into a cliff of 30 m with an
mars1129 [50]

Answer:

The velocity of the car when it hits the ground is approximately 55.574 meters per second.

Explanation:

According to this expression, the car goes straight into the cliff and goes down due to gravity, whose situation is described by the Principle of Energy Conservation and supposing that all non-conservative forces are negligible:

U_{g,1}+K_{1} = U_{g,2}+K_{2} (1)

Where:

U_{g,1}, U_{g,2} - Gravitational potential energies of the car at the top and the bottom, measured in joules.

K_{1}, K_{2} - Translational kinetic energies of the car at the top and the bottom, measured in joules.

By definitions of gravitational potential and translational kinetic energies, we expand and simplify the equation above:

\frac{1}{2}\cdot m\cdot v_{2}^{2} = \frac{1}{2}\cdot m \cdot v_{1}^{2}+m\cdot g \cdot (z_{1}-z_{2}) (2)

v_{2}^{2} = v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2})

v_{2} = \sqrt{v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2})}

Where:

m - Mass, measured in kilograms.

v_{1}, v_{2} - Velocity of the car at the top and the bottom, measured in meters per second.

g - Gravitaitional acceleration, measured in meters per square second.

z_{1}, z_{2} - Height of the car at the top and at the bottom, measured in meters.

If we know that v_{1} = 50\,\frac{m}{s}, g = 9.807\,\frac{m}{s^{2}}, z_{1} = 30\,m and z_{2} = 0\,m, then the velocity of the car when it hits the ground is:

v_{2} = \sqrt{\left(50\,\frac{m}{s} \right)^{2}+2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (30\,m-0\,m)}

v_{2}\approx 55.574\,\frac{m}{s}

The velocity of the car when it hits the ground is approximately 55.574 meters per second.

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Answer:

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Explanatio

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