the potential at the center of curvature of the arc = v = Q ∕ (4πε∘a) or 
ATQ,
We have density of charge,
λ = 
Where L is the rod's length, in this case the semicircle's length L = πr
Q is the charge on the rod
The potential created at the center by an differential element of charge is:
where k is the coulomb's constant
r is the distance from dQ to center of the circle
v = ∫ 
, Where a = radius, k = 1 / 4πε∘
v = or Q ∕ (4πε∘a)
To learn more about potential from the given link
brainly.com/question/25923373
#SPJ4
B. move away from the mercury and notify the teacher :)
you should never use a broom to sweep mercury, it breaks the mercury in smaller droplets and spread them.
Pouring the mercury in the sink would likely be logged in the plumbing. and pickingthings up with the napkin may cause contamination and cuts from the glass.
Answer:
1.5T
Explanation:
Since, the amplitude of SHM is A. So, in one time period T the object will travel travel a distance of 4A.
6A-4A= 2A.
Now, this 2A distance must be traveled in T/2 time period.
So, the total time taken to travel a distance of 6A is T+T/2 = 3T/2 = 1.5T
Answer:
(a) 20 m
(b) 6 m/s²
(c) Between t=0 and t=2, the body moves to the left.
Between t=2 and t=4, the body moves to the right.
Explanation:
v = 3t² − 6t
x(0) = 4
(a) Position is the integral of velocity.
x = ∫ v dt
x = ∫ (3t² − 6t) dt
x = t³ − 3t² + C
Use initial condition to find value of C.
4 = 0³ − 3(0)² + C
4 = C
x = t³ − 3t² + 4
Find position at t = 4.
x = 4³ − 3(4)² + 4
x = 20
(b) Acceleration is the derivative of velocity.
a = dv/dt
a = 6t − 6
Find acceleration at t = 2.
a = 6(2) − 6
a = 6
(c) v = 3t² − 6t
v = 3t (t − 2)
The velocity is 0 at t = 0 and t = 2. Evaluate the intervals.
When 0 < t < 2, v < 0.
When t > 2, v > 0.
