Answer:
a)
, b) 
Explanation:
a) The potential energy is:



b) Maximum final speed:

The final speed is:


For simplicity, let's call vector B-A vector C Then C is
Cx = (-6.1 - 2.2)
Cy = (-2.2 - (-6.9)) Or,
Cx = -8.3 Cy = 4.7
The magnitude is found with the Pythagorean theorem
||C|| = √(-8.3² + 4.7²) = 9.538
Answer:
4.2 J
Explanation:
Specific heat capacity: This is defined as the amount of a heat required to rise a unit mass of a substance through a temperature of 1 K
From specific heat capacity,
Q = cmΔt.............................. Equation 1
Where Q = amount of energy absorbed or lost, c = specific heat capacity of water, m = mass of water, Δt = Temperature rise.
Given: m = 1 g = 0.001 kg, Δt = 1 °C
Constant : c = 4200 J/kg.°C
Substitute into equation 1
Q = 0.001×4200(1)
Q = 4.2 J.
Hence the energy absorbed or lost = 4.2 J