Answer:
θ = θ₀ + ½ w₀ (t -t_1) + α (t -t_1)²
Explanation:
This is an angular kinematic exercise the equation for the angular position
the particle A
θ = θ₀ + ω₀ t + ½ α t²
They say for the particle B
w₀B = ½ w₀
αB = 2 α
In addition, the particle begins at a time t_1 after particle A, in order to use the same timer, we must subtract this time from the initial
t´ = t - t_1
l
et's write the equation of particle B
θ = θ₀ + w₀B t´ + ½ αB t´2
replace
θ = θ₀ + ½ w₀ (t -t_1) + ½ 2α (t -t_1)²
θ = θ₀ + ½ w₀ (t -t_1) + α (t -t_1)²
Explanation:
Formula depicting relation between total flux and total charge Q is as follows.
(Gauss's Law)
Putting the given values into the above formula as follows.
Q = 
= 
= 
= -8.4 nC
Therefore, when the unknown charge is q then,
-14.0 nC + 33.0 nC + q = -8.4 nC
q = -27.4 nC
Thus, we can conclude that charge on the third object is -27.4 nC.
That would be an asteroid
Answer:
I = Δq / t
Explanation:
The quantity of electricity i.e charge is related to current and time according to the equation equation:
Q = It
Δq = It
Where:
Q => is the quantity of electricity i.e charge
I => is the current.
t => is the time.
Thus, we can rearrange the above expression to make 'I' the subject. This is illustrated below:
Δq = It
Divide both side by t
I = Δq / t
