Question

A brick of gold is 0.1 m wide, 0.1 m high, and 0.2 m long. The density of gold is 19,300 kg/m3. What pressure does the brick exert on the table if the brick is resting on its side?

Answer 1

`P=18,933.3Pa=18.9kPa`

Step-by-step explanation:

Hello!

In this case, since we can compute the volume of the brick as shown below:

`V=0.1m*0.1m*0.2m=0.002m^3`

Next, we can compute the mass of the brick given its density:

`\rho =m/V\\\\m=V*\rho\\\\m=0.002m^3*19,300kg/m^3\\\\m=38.6kg`

Now, since the force exerted on the table corresponds to the weight of the brick, we use the gravity to obtain:

`W=38.6kg*9.81m/s^2=378.7N`

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:

`A=0.1m*0.2m=0.02m^2`

Therefore, the pressure is:

`P=(F)/(A)=(W)/(A)=(378.7N)/(0.02m^2)\\\\P=18,933.3Pa=18.9kPa`

Best regards!