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Ne4ueva [31]
3 years ago
14

What it means for an object to be negatively charged

Physics
1 answer:
zepelin [54]3 years ago
4 0

Answer: the total charge of the object is negative

Explanation: This means that this object has added many electrons ( negative charge) from other object.

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Helium in the sun is produced by which process?.
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Nuclear Fusion is a process by which the nucleus of the two atoms fuses to form a completely new element. The Helium in the sun is produced by the process of nuclear fusion.

<h2></h2><h2>Nuclear Fusion:</h2>

It is a process by which the nucleus of the two atoms fuses to form a completely new element.

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2 years ago
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6 0
3 years ago
Suppose in the previous example (from Machining 1 lecture) that cutting force and thrust force are measured during an orthogonal
mojhsa [17]

Complete Question

Suppose in the previous example (from Machining 1 lecture) that cutting force and thrust force are measured during an orthogonal cutting operation: Fc = 1559 N and Ft = 1271 N.  The chip thickness before the cut =0.5mm and the width of the orthogonal cutting operation is 3.0mm . Rake angle is equal to 10 and shear plane angle is 25.4. Based on these data, determine the shear strength of the work material.

Answer:

The shear strength of the work material is  \sigma = 245.09 N/mm^2

Explanation:

From the question we are told that the

    The the cutting force is F_c = 1559N

     The thrust force is F_t = 1271\ N

    The width of the octagonal cutting operation is  w = 3.0\ mm

     The thickness is t = 0.5mm

     The shear plane angle is  \theta = 25.4^o

The shear strength of the work materiel is mathematically represented as

                       \sigma = \frac{F_s}{A}

Where  F_s is the shear force which is mathematically evaluated as

                   F_s = F_c cos\ \theta -F_t sin\ \theta

Substituting values we have

                  F_s = 1559 cos (25.4) - 1271sin(25.4)

                        = 857.841N

While A is the area which is mathematically evaluated as

                        A = \frac{t* w}{sin \theta }

Substituting values

                      A = \frac{0.5 *3}{sin (25.4)}

                         = 3.50

Substituting this into the equation for shear strength

               \sigma = \frac{857.8}{3.5}

                  \sigma = 245.09 N/mm^2

8 0
3 years ago
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