Water, of density 1000 kg/m3, is flowing in a drainage channel of rectangular cross-section. The width of the channel is 15 m, t

#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.

Answer:

Consequently, the temperature difference across the material will be

Explanation:

In this case, we apply the Fourier Law of heat conduction expressed by the following equation:

  (1)

Where k = thermal conductivity = 0.2 W/ mK

A= 1m^2 denotes the cross-sectional area

Q= 3KW signifies the heat transfer rate

is the temperature difference we need to determine

represents the thickness of the material

Initially, we convert 3KW to W, resulting in:

With all variables accounted for, we can substitute and calculate:

Thus, the temperature difference across the material will be

Answer:

        h = 12.8 cm

Explanation:

The initial parameters are as follows:

distance = 6.4 cm

• when the object descends, its weight matches the spring's force

        weight = spring force

         mg = ky... equation 1

• potential energy stored in a stretched spring = work done by the spring

        mgh = 0.5 x k x h^{2}....equation 2

• Substituting from equation 1 into equation 2

                kyh =  0.5 x k x h^{2}

                y =  0.5 x h

                2y = h

• where y is 6.4, yielding the maximum elongation as

          h = 2 x 6.4 = 12.8 cm