A circuit contains a 6.0-v battery, a 4.0-w resistor, a 0.60-µf capacitor, an ammeter, and a switch all in series. what will be

The soccer ball's initial speed stands at 16.38 m/s. Given that the vertical distance is y = 2.44 m, the horizontal span x = 10.0 m, and the angle of launch θ = 25.0°. The initial velocity comprises two components, Vₓ and V. The calculations are as follows: Vₓ = V cosθ and V = V sinθ. The formula for horizontal distance becomes x = Vₓt. Since g is deemed 0, we can state that: x = Vₓt or 10 = V cos 25 * t. Solving for V gives us 10 = 0.906V * t, thus V * t = 10 / 0.906 = 11.038 m. Regarding the vertical distance (with g being negative due to the upward movement opposing gravity), we use y = V sinθ * t - 1/2 * g * t². Following through with the calculations leads us to determine that the soccer ball's initial speed is indeed 16.38 m/s.

#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.

Let us denote as the highest acceleration of the truck by itself. The force generated by the engine to propel the truck in this scenario is

where m is the mass of the truck alone.

Upon attaching a bus that has double the mass of the truck, the total mass of the entire system (truck+bus) becomes (m+2m)=3m. In this situation, the force generated by the engine is

where a2 represents the new acceleration. 
Since the engine remains unchanged, the force generated stays the same, allowing us to set the force equations equal to each other:

and solving for a2 gives us: