Pressure Practice

Sample questions

5 of 20 — sign in to practice the rest with adaptive difficulty and mastery tracking.

1. Sample 1difficulty 1/5

   A $200~\text{N}$ force is distributed uniformly over a $0.04~\text{m}^2$ area. What is the pressure?

   • A

     $50~\text{Pa}$

   • B

     $5000~\text{Pa}$

     check_circle
   • C

     $800~\text{Pa}$

   • D

     $8~\text{Pa}$

   Why

   $P = F/A = 200/0.04 = 5000~\text{Pa}$.

   View full solution & statsarrow_forward

2. Sample 2difficulty 1/5

   A pressure of $50~\text{kPa}$ equals

   • A

     $5 \times 10^6~\text{Pa}$

   • B

     $5000~\text{Pa}$

   • C

     $50{,}000~\text{Pa}$

     check_circle
   • D

     $50~\text{Pa}$

   Why

   $1~\text{kPa} = 10^3~\text{Pa}$. $50~\text{kPa} = 50{,}000~\text{Pa}$.

   View full solution & statsarrow_forward

3. Sample 3difficulty 1/5

   Doubling the depth in a uniform-density fluid changes the gauge pressure by a factor of

   • A

     $1$

   • B

     $1/2$

   • C

     $4$

   • D

     $2$

     check_circle

   Why

   $P = \rho g h$, linear in $h$. Doubling $h$ doubles $P$.

   View full solution & statsarrow_forward

4. Sample 4difficulty 2/5

   A flat plate of area $0.10~\text{m}^2$ is submerged horizontally $2.0~\text{m}$ deep in water ($\rho = 1000~\text{kg/m}^3$, $g \approx 10~\text{m/s}^2$). The force from water on its top side (gauge) is

   • A

     $20{,}000~\text{N}$

   • B

     $200~\text{N}$

   • C

     $1000~\text{N}$

   • D

     $2000~\text{N}$

     check_circle

   Why

   $P = \rho g h = 20{,}000~\text{Pa}$. $F = P A = 20{,}000 \cdot 0.10 = 2000~\text{N}$.

   View full solution & statsarrow_forward

5. Sample 5difficulty 2/5

   What is the gauge pressure at a depth of $5.0~\text{m}$ below the surface of a fresh-water lake? ($\rho_\text{water} = 1000~\text{kg/m}^3$, $g \approx 10~\text{m/s}^2$.)

   • A

     $5000~\text{Pa}$

   • B

     $100{,}000~\text{Pa}$

   • C

     $50{,}000~\text{Pa}$

     check_circle
   • D

     $10{,}000~\text{Pa}$

   Why

   $P = \rho g h = (1000)(10)(5.0) = 50{,}000~\text{Pa} = 50~\text{kPa}$.

   View full solution & statsarrow_forward