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Jlenok [28]
3 years ago
5

Why could you hold a stack of books on your head without pain, whereas it would hurt if someone put a small pebble between the b

ook and your head? Why is it so easy to cut yourself with the edge of a piece of paper (a "paper cut")?
Physics
2 answers:
trapecia [35]3 years ago
7 0

Answer:

Both the cases deal with the principle of pressure due to force acting on an area.

Explanation:

When a stone is place between a pile of book and our head when we carry it then the area subjected to the force decreases while the magnitude of the force remains same and therefore the pressure on our head increases.

As we know that pressure is force per unit area is called pressure.

Mathematically:

Pressure=\frac{Force}{Area}

It is easy to cut our skin with an edge of a paper because our skin is relatively softer than the edge of a paper and the edge of a paper is thin which applies more pressure due to  less area of contact with our skin.

tiny-mole [99]3 years ago
5 0

Answer:

Explanation:

When we hold a stack of books on the head it does not cause pain whereas it hurt when a small pebble is placed between books and head.

This happens because the surface area of books is much larger as compared to the small pebble.

The pressure is given by Force per unit area and for pebble, the area is very small causing too much pressure intensity on the head which leads to pain while holding it.        

Similar is the case with paper. As the edge of the paper is sharp so a small amount of force is sufficient to cause a cut.      

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3 years ago
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Answer:

The acceleration of M_2 is  a =  0.7156 m/s^2

Explanation:

From the question we are told that

    The mass of first block is  M_1 =  2.25 \ kg

    The angle of inclination of first block is  \theta _1 =  43.5^o

    The coefficient of kinetic friction of the first block is  \mu_1  = 0.205

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     The angle of inclination of the second block is  \theta _2 =  32.5^o

      The coefficient of kinetic friction of the second block is \mu _2 = 0.105

The acceleration of M_1 \ and\  M_2 are same

The force acting on the mass M_1 is mathematically represented as

     F_1 = T -  M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1

=> M_1 a = T -  M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1

Where T is the tension on the rope

The force acting on the mass M_2 is mathematically represented as    

  F_2 =  M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2

   M_2 a =  M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2

At equilibrium

  F_1 =  F_2

So

 T -  M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1 =M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2

making a the subject of the formula

    a =  \frac{M_2 g sin \theta_2 - M_1 g sin \theta_1 - \mu_1 M_1g cos \theta - \mu_2 M_2 g cos \theta_2 }{M_1 +M_2}

substituting values a =  \frac{(5.45) (9.8) sin (32.5) - (2.25) (9.8) sin (43.5) - (0.205)*(2.25) *9.8cos (43.5) - (0.105)*(5.45) *(9.8) cos(32.5) }{2.25 +5.45}

    => a =  0.7156 m/s^2

     

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