Peter left Town A at 13:30 and travelled towards Town B at an

I’ve provided the missing image. We can analyze this scenario by applying the principle of energy conservation. At point A, the car possesses both potential and kinetic energy. As it moves down the track, some initial energy is lost due to friction. Thus, as it approaches point B, we have a specific amount of energy remaining. According to the conservation of energy laws, this remaining energy at point B will equal the sum of its kinetic and potential energy.

Refer to the diagram shown below.

m₁ = 1100 kg represents the mass of the car.
m₂ = 700 kg indicates the combined mass of the trailer and boat.
F = 1900 N is the driving force acting on the vehicle.
N₁ denotes m₁g, the normal force on the car.
N₂ corresponds to m₂g, the normal force on the trailer and boat.
Frictional forces are represented by μN₁ and μN₂, where μ is the coefficient of kinetic friction.
T signifies the force in the connection between the car and the trailer.

Part (a)
Let R₁ signify the total resistance acting against the motion of the car, boat, and trailer.
With the acceleration at 0.550 m/s², it follows that
(m₁ + m₂ kg)*(0.55 m/s²) = F
(1100 + 700 kg)*(0.55 m/s²) = (1900 - R₁) N.
This leads to the equation 990 = 1900 - R.
Therefore, R₁ = 910 N.

Answer: The total resistive force amounted to 910 N.

Part (b)
The trailer and boat experience 80% of the resisting forces.
Let R₂ denote this resistive force.
Thus,
R₂ = 0.8*R₁ = 728 N.
Assuming T is the tension in the hitch connecting the car and trailer, it follows:
T - R₂ = m₂(0.55 m/s²)
(T - 728 N) = (700 kg)*(0.55 m/s²).
This leads to T - 728 = 385.
Thus, T equals 1113 N.

Answer: The tension in the hitch is 1113 N.

The tension exerted in the cable amounts to T = 16653.32 N. Parameters: Cross section area A = 1.3 m² Drag coefficient CD = 1.2 Velocity V = 4.3 m/s The angle formed by the cable with the horizontal is 30 degrees. Density defined as follows: The drag force FD is determined by the equation: FD = (1/2) * ρ * V² * A * CD Calculating the drag force yields 14422.2 N acting opposite to motion. Given the cable's angle of 30 degrees with horizontal, the horizontal component contributes to the drag force calculation: T * cos(30) = F_D Thus, T = 16653.32 N.

The new charge of the ball will amount to 8x10^8C after removing 5x10^27 electrons.

Explanation:

Initially, if the sphere is electrically neutral, its charge stands at 0C.

When an electron with a charge of (-1.6*10^-19 C) is taken away, we effectively add a positive charge, leading to:

1.6*10^-19 C as the sphere's new charge.

For a total of N electrons removed, the sphere's overall charge now becomes:

N*1.6*10^-19 C.

To calculate N when:

N*1.6*10^-19 C = 8.0x 10^8 C.

We find that N is: (8.0/1.6)x10^(8 + 19) = 5x10^27 electrons.

C. The rate of water flow diminishes.