A student grabs a piece of paper and tears it from the middle. Which statement best explains the forces involved?

Physics

2 answers:

PolarNik [594]2 years ago

3 0

Answer:

b. The normal force between the molecules of the paper is overcome by the contact force of the hands.

Explanation:

The paper molecules are held together by a weak bond. When the student holds the paper on both sides with the center of the paper in between, the student applies two equal forces in the opposite direction of the paper making the paper molecules weaken and separate.

stich3 [128]2 years ago

3 0

Answer:

The electrostatic forces from the contact of the hands with the paper causes the paper molecules to separate.

Here's the rest of the answers

1.) Which statement is correct about the strength of forces?

Electrostatic forces are trillions of times stronger than gravitational forces.

2.) A student tries to push a book across a table. Which forces acting on the book are a result of electrostatic forces?

normal force and friction

3.) What type of force can explain how a gecko climbs up a wall?

contact forces

5.) A heavy hammer is sitting on top of a board. Which statement correctly describes why the hammer does not go through the board?

The electrostatic forces between the molecules of the board prevent the force of gravity from breaking the board apart.

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If the moment acting on the cross section is M=630N⋅m, determine the maximum bending stress in the beam. Express your answer to

-BARSIC- [3]

Answer:

2.17 Mpa

Explanation:

The location of neutral axis from the top will be

$\bar y=\frac {(240\times 25)\times \frac {25}{2}+2\times (20\times 150)\times (25+(\frac {150}{2}))}{(240\times 25)+2\times (20\times 150)}=56.25 mm$

Moment of inertia from neutral axis will be given by $\frac {bd^{3}}{12}+ ay^{2}$

Therefore, moment of inertia will be

$\frac {240\times 25^{3}}{12}+(240\times 25)\times (56.25-25/2)^{2}+2\times [\frac {20\times 150^{3}}{12}+(20\times 150)\times ((25+150/2)-56.25)^{2}]=34.5313\times 10^{6} mm^{4}}$

Bending stress at top= $\frac {630\times 10^{3}\times (175-56.25)}{34.5313\times 10^{6}}=2.1665127\approx 2.17 Mpa$

Bending stress at bottom= $\frac {630\times 10^{3}\times 56.25}{34.5313\times 10^{6}}=1.026242858\approx 1.03$ Mpa

Comparing the two stresses, the maximum stress occurs at the bottom and is 2.17 Mpa

8 0

3 years ago

A 95kg clock initially at rest on a horizontal floor requires a 560N horizontal force to set it in motion. After the clock is in

miv72 [106K]

Complete Question

A 95 kg clock initially at rest on a horizontal floor requires a 650 N horizontal force to set it in motion. After the clock is in motion, a horizontal force of 560 N keeps it moving with a constant velocity. Find the coefficient of static friction and the coefficient of kinetic friction.

Answer:

The value for static friction is $\mu_s = 0.60$