Isabella drops a pen off her balcony by accident while celebrating the successful completion of a physics problem. assuming air

The city evaluates the continuous increase of carbon monoxide from different origins each year. According to calculations, in the year "C: 2019"<span> (rounded to the closest whole number), the concentration of CO will surpass the allowed threshold.

If this is not correct, feel free to inform me and I will find out the right answer. However, I am confident this is accurate.:) </span>

#1

The volume of lead measures 100 cm^3

with a density of lead at 11.34 g/cm^3

. Thus, the mass of the lead block equals density multiplied by volume

Therefore, its weight in air is noted as

Next, the buoyant force acting on the lead is defined as

We know that

After solving, we find

V = 11.22 cm^3

(ii) This corresponding volume of water exerts the same weight as the buoyant force, resulting in 0.11 N

(iii) The buoyant force measures 0.11 N

(iv) The lead block sinks in water due to its density being greater than that of water.

#2

The buoyant force acting on the lead block counterbalances its weight

(ii) This volume of mercury corresponds to the buoyant force weight, confirming that the block floats within mercury, resulting in 11.11 N as its weight.

(iii) The buoyant force is recorded as 11.11 N

(iv) Given that lead's density is less than mercury's, the lead will float in the mercury medium.

#3

Indeed, an object that has lesser density than a liquid will float; otherwise, it will sink in the liquid.

The new force F3 is added in the same direction as F2. To analyze the forces acting on an object in this scenario, we observe that they operate along the vertical axis, with F1 acting upward and F2 downward. To determine the necessary vector F3 to counteract the net force, it's important to calculate the length difference between F1 and F2. The direction of F3 will match that of the smaller force. If F2 is less than F1, F3 will align with F2.

Response:

a) , b)

Clarification:

a) The absolute pressure at a depth of 27.5 meters is:

b) The force applied by the water is: