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irina1246 [14]
4 years ago
15

Interactive Solution 11.13 presents a model for solving this problem. A solid concrete block weighs 100 N and is resting on the

ground. Its dimensions are 0.400 m x 0.250 m x 0.130 m. A number of identical blocks are stacked on top of this one. What is the smallest number of whole bricks (including the one on the ground) that can be stacked so that their weight creates a pressure of at least two atmospheres on the ground beneath the first block
Physics
1 answer:
mash [69]4 years ago
3 0

Answer:

The value is }  N  =  66 \  blocks

Explanation:

From the question we are told that

The weight of the block is W_b  = 100 \  N

The dimension of the block is d =  0.400 m  \ X  \ 0.250 \  m  \  X  \ 0.130 \ m

Generally two atmosphere is equivalent to

P_{2atm} =  2 *  1.013 *10^{5} =  202600 \  N/m^2

Generally 1 atm = 1.013 *10^{5} N/m^2

The area of the block would be evaluated using width and height because we need for the smaller surface to be in contact with the ground in order to maximize the pressure and minimize number of blocks

So

A =  0.250 *  0.130

=> A =  0.0325 \  m^2

Generally the force due to this blocks is mathematically represented as

F =  N  *  W_b

Here N is the number of blocks

So

}  202600 =  \frac{N  *  100 }{ 0.0325}

=>   }  N  =  66 \  blocks

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At the instant the traffic light turns green, an automobile starts with a constant acceleration a of 2.70 m/s2. At the same inst
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Answer:

66.85 m

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3 years ago
give the mathematical expression for coulombs force if q1,q2 are the magnitude of charges and r is the distance between them.
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Two water jets are emerging from a vessel at a height of 50 centimeters and 100 centimeters. If their horizontal velocities at t
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t2 = √2h2/g = √2*1.0/9,8 = 0.451 sec 

In which t = times for the vertical movement
h = height
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d1 = 1*0.319 = 0.319 m 
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in which d = Horizontal distance

ratio
= di : d2
= 0.319 : 0.225    

 = 3.19 : 2.25
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