To determine the best answer for the question, let's follow these steps:
1. Identify the parameters:
- The mass of the load ([tex]\( m \)[/tex]) is 350 kg.
- The area ([tex]\( A \)[/tex]) subjected to the load is 1 [tex]\( m^2 \)[/tex].
- The acceleration due to gravity ([tex]\( g \)[/tex]) is 9.8 [tex]\( m/s^2 \)[/tex].
2. Calculate the force exerted by the load:
The force ([tex]\( F \)[/tex]) due to the load is given by the product of the mass and the acceleration due to gravity.
[tex]\[
F = m \times g
\][/tex]
Substituting the known values:
[tex]\[
F = 350 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 3430 \, \text{N}
\][/tex]
Therefore, the force exerted by the load is 3430 Newtons (N).
3. Calculate the pressure exerted on the liquid:
Pressure ([tex]\( P \)[/tex]) is defined as the force per unit area. It can be calculated using:
[tex]\[
P = \frac{F}{A}
\][/tex]
Using the given area:
[tex]\[
P = \frac{3430 \, \text{N}}{1 \, \text{m}^2} = 3430 \, \text{Pa}
\][/tex]
Thus, the pressure on the liquid is 3430 Pascals (Pa).
4. Convert the pressure from Pascals to kilopascals:
Since 1 kilopascal (kPa) is equal to 1000 Pascals (Pa), we convert the pressure to kilopascals by dividing by 1000:
[tex]\[
\text{Pressure in kPa} = \frac{3430 \, \text{Pa}}{1000} = 3.43 \, \text{kPa}
\][/tex]
5. Conclusion:
The external pressure on the upper surface of the liquid is 3.43 kPa.
Therefore, the best answer for the question is:
D. 3.43 kPa
