Mastering Physics Solutions: Question 4.12

Mastering Physics Solutions: Question 4.12

On January 14, 2012, in Chapter 04: Force and Motion, by Mastering Physics Solutions

Part A = Only with constant forces.

The kinematic equation x = x0 + V0t + 1/2at2 can be used:

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Mastering Physics Solutions: Question 4.10

Mastering Physics Solutions: Question 4.10

On January 14, 2012, in Chapter 04: Force and Motion, by Mastering Physics Solutions

Part A = The magnitude of the force of the brick on the glass is equal to the magnitude of the force of the glass on the brick.

A brick hits a glass window. The brick breaks the glass, so:

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Mastering Physics Solutions: Question 4.4

Mastering Physics Solutions: Question 4.4

On January 14, 2012, in Chapter 04: Force and Motion, by Mastering Physics Solutions

Part A = All of the preceding

If the net force on an object is zero, the object could:

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Mastering Physics Solutions: Board Pulled Out from under a Box

Mastering Physics Solutions: Board Pulled Out from under a Box

On January 14, 2012, in Chapter 04: Force and Motion, by Mastering Physics Solutions

Part A = Fmin = μsg(m1 + m2)

Find Fmin, the constant force with the least magnitude that must be applied to the board in order to pull the board out from under the the box (which will then fall off of the opposite end of the board).

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Mastering Physics Solutions: Two Forces Acting at a Point

Mastering Physics Solutions: Two Forces Acting at a Point

On January 14, 2012, in Chapter 04: Force and Motion, by Mastering Physics Solutions

Part A = -7.96N Click to use the calculator/solver for this part of the problem
Part B = 2.34N Click to use the calculator/solver for this part of the problem
Part C = 8.29N Click to use the calculator/solver for this part of the problem
Part D = 16.4 Degrees Click to use the calculator/solver for this part of the problem

Two forces, F1 and F2, act at a point, as shown in the picture.

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Mastering Physics Solutions: Two Hanging Masses

Mastering Physics Solutions: Two Hanging Masses

On January 14, 2012, in Chapter 04: Force and Motion, by Mastering Physics Solutions

Part A = M2g
Part B = (M1 + M2)g
Part C = M2a + M2)g
Part D = (M1 + M2)a + (M1 + M2)g

Two blocks with masses M1 and M2 hang one under the other. For this problem, take the positive direction to be upward, and use g for the magnitude of the acceleration due to gravity.

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