Normally, the water pressure inside a pump is higher than the vapor pressure: in this case, at the interface between the liquid and the vapor, molecules from the liquid escapes into vapour form. Instead, when the pressure of the water becomes lower than the vapour pressure, molecules of vapour can go inside the water forming bubbles: this phenomenon is called
cavitation.
So, cavitation occurs when the pressure of the water becomes lower than the vapour pressure. In our problem, vapour pressure at

is 1.706 kPa. Therefore, the lowest pressure that can exist in the pump without cavitation, at this temperature, is exactly this value: 1.706 kPa.
Answer:
Net force on the block is 32 N.
Acceleration of the object is 6.4 m/s².
Explanation:
Let the acceleration of the object be
m/s².
Given:
Mass of the block is, 
Force of pull is, 
Frictional force on the block is, 
The free body diagram of the object is shown below.
From the figure, the net force in the forward direction is given as:

Now, from Newton's second law of motion, net force is equal to the product of mass and acceleration. So,

Therefore, the acceleration of the object in the forward direction is 6.4 m/s².
1) Current in each bulb: 0.1 A
The two light bulbs are connected in series, this means that their equivalent resistance is just the sum of the two resistances:

And so, the current through the circuit is (using Ohm's law):

And since the two bulbs are connected in series, the current through each bulb is the same.
2) 4 W and 8 W
The power dissipated by each bulb is given by the formula:

where I is the current and R is the resistance.
For the first bulb:

For the second bulb:

3) 12 W
The total power dissipated in both bulbs is simply the sum of the power dissipated by each bulb, so:
