Answer:
[tex]P=18,933.3Pa=18.9kPa[/tex]
Explanation:
Hello!
In this case, since we can compute the volume of the brick as shown below:
[tex]V=0.1m*0.1m*0.2m=0.002m^3[/tex]
Next, we can compute the mass of the brick given its density:
[tex]\rho =m/V\\\\m=V*\rho\\\\m=0.002m^3*19,300kg/m^3\\\\m=38.6kg[/tex]
Now, since the force exerted on the table corresponds to the weight of the brick, we use the gravity to obtain:
[tex]W=38.6kg*9.81m/s^2=378.7N[/tex]
Finally, since the surface of the brick in contact with the table corresponds to the 0.1x0.2 area (length and width), the area on which the weight force is exerted is:
[tex]A=0.1m*0.2m=0.02m^2[/tex]
Therefore, the pressure is:
[tex]P=\frac{F}{A}=\frac{W}{A}=\frac{378.7N}{0.02m^2}\\\\P=18,933.3Pa=18.9kPa[/tex]
Best regards!
