8: One gallon of paint (volume = 3.78 × 103 cm3) coversan area of 25.0 m2. What is the thickness of the fresh painton the wall? [volume=Area×tickness].(a) 0.51 × 103 m(b) 1.51 × 10−4 m(c) 0.51 × 103 m(d) 0.51 × 10−4 m(e) 9.45 × 103 mQ9: The Andam auditorium in Dungu Campus of UDSmeasures 40.0 m × 20.0 m × 12.0 m. The density of air is1.20 kg/m3. What is the weight of air in the room in pounds?[1 lb = 4.448 N].(a) 1.13 × 104 lb(b) 2.54 × 104 lb(c) 1.13 × 105 lb(d) 2.54 × 10−4 lb(e) 1.13 × 10−5 lbQ10: Assume the equation x = A t3 + B t describes themotion of a particular object, with x having the dimensionof length and t having the dimension of time. Determine thedimensions of the constants A and B.5

Sure, let's solve these problems one by one.

Q8: We are given the volume of the paint and the area it covers. We can find the thickness by rearranging the formula volume = area × thickness to thickness = volume / area.

Volume = 3.78 × 10^3 cm^3 = 3.78 × 10^-3 m^3 (since 1 m^3 = 10^6 cm^3) Area = 25.0 m^2

So, thickness = (3.78 × 10^-3 m^3) / 25.0 m^2 = 1.51 × 10^-4 m

So, the answer is (b) 1.51 × 10^-4 m.

Q9: The weight of the air in the room can be found by first finding the volume of the room, then multiplying by the density of air to find the mass, and finally converting to weight.

Volume = length × width × height = 40.0 m × 20.0 m × 12.0 m = 9600 m^3 Mass = volume × density = 9600 m^3 × 1.20 kg/m^3 = 11520 kg Weight = mass × gravity = 11520 kg × 9.8 m/s^2 = 112896 N Weight in pounds = 112896 N / 4.448 N/lb = 2.54 × 10^4 lb

So, the answer is (b) 2.54 × 10^4 lb.

Q10: The dimensions of A and B can be found by matching the dimensions on both sides of the equation.

For the term At^3, the dimensions of A must be [L]/[T]^3, where [L] is the dimension of length and [T] is the dimension of time.

For the term Bt, the dimensions of B must be [L]/[T].

So, the dimensions of A are [L]/[T]^3 and the dimensions of B are [L]/[T].