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
So we can know what is in space maybe weird or interesting stuff
Answer:
The magnitude of the electric force on a protein with this charge is 
Explanation:
Given that,
Electric field = 1500 N/C
Charge = 30 e
We need to calculate the magnitude of the electric force on a protein with this charge
Using formula of electrostatic force

Where, F = force
E = electric field
q = charge
Put the value into the formula


Hence, The magnitude of the electric force on a protein with this charge is 
Answer:
α = 1930.2 rad/s²
Explanation:
The angular acceleration can be found by using the third equation of motion:

where,
α = angular acceleration = ?
θ = angular displacement = (1500 rev)(2π rad/1 rev) = 9424.78 rad
ωf = final angular speed = 0 rad/s
ωi = initial angular speed = (960 rev/s)(2π rad/1 rev) = 6031.87 rad/s
Therefore,

<u>α = - 1930.2 rad/s²</u>
<u>negative sign shows deceleration</u>