Answer:
yes ............................................ks
Explanation: is good
Answer: option 1 : the electric potential will decrease with an increase in y
Explanation: The electric potential (V) is related to distance (in this case y) by the formulae below
V = kq/y
Where k = 1/4πε0
Where V = electric potential,
k = electric constant = 9×10^9,
y = distance of potential relative to a reference point, ε0 = permittivity of free space
q = magnitude of electronic charge = 1.609×10^-19 c
From the formulae, we can see that q and k are constants, only potential (V) and distance (y) are variables.
We have that
V = k/y
We see the potential(V) is inversely proportional to distance (y).
This implies that an increase in distance results to a decreasing potential and a decrease in distance results to an increase in potential.
This fact makes option 1 the correct answer
Bigger is the correct answer. the faster an igneous rock cools the bigger the bigger the crystal size will be
Answer
Given,
Energy absorbed, 
Energy expels,
Temperature of cold reservoir, T = 27°C
a) Efficiency of engine



b) Work done by the engine



c) Power output
t = 0.296 s


