A brick weight 17.8N and is resting on the ground. Its dimensions are 0.203mx0.089mx0.0570m. A number of the bricks are then stacked on top of this one.
What is the smallest number of whole bricks (including the one on the ground) that could be used, so that their weight creates a pressure of a least one atmosphere on the ground beneath the first brick? Take P =1.013x105N/m2
P=F/A=NW/A
N=PA/W =
=1.013•10⁵•0.203•0.089/17.8=102.8 ≈103
To find the smallest number of whole bricks required to create a pressure of at least one atmosphere on the ground beneath the first brick, we need to calculate the total weight of the bricks.
First, let's calculate the weight of one brick:
Weight of one brick = mass * gravitational acceleration
Given that the weight of the brick is 17.8 N, and the acceleration due to gravity is approximately 9.8 m/s^2, we can rearrange the formula to find the mass:
Mass of one brick = Weight of one brick / gravitational acceleration
Mass of one brick = 17.8 N / 9.8 m/s^2 = 1.81 kg (rounded to two decimal places)
Next, let's calculate the pressure exerted by one brick on the ground:
Pressure = Force / Area
The force exerted by one brick is its weight (17.8 N), and the area is the base area of the brick.
Base Area = length * width
Base Area = 0.203 m * 0.089 m = 0.01817 m^2 (rounded to five decimal places)
Pressure = 17.8 N / 0.01817 m^2 = 979.32 N/m^2 (rounded to two decimal places)
Now, let's calculate the pressure of one atmosphere in N/m^2.
Pressure of one atmosphere = 1.013 x 10^5 N/m^2
To find the smallest number of whole bricks needed, we can divide the pressure of one atmosphere by the pressure exerted by one brick:
Number of bricks = Pressure of one atmosphere / Pressure exerted by one brick
Number of bricks = 1.013 x 10^5 N/m^2 / 979.32 N/m^2 ≈ 103.6
Since we can't have a fraction of a brick, the smallest number of whole bricks that would create a pressure of at least one atmosphere on the ground beneath the first brick is 104.
Therefore, at least 104 whole bricks (including the one on the ground) would be required.