It takes approximately 8 minutes and 20 seconds.
<span>Place a test charge in the middle. It is 2cm away from each charge.
The electric field E= F/Q where F is the force at the point and Q is the charge causing the force in this point.
The test charge will have zero net force on it. The left 30uC charge will push it to the right and the right 30uC charge will push it to the left. The left and right force will equal each other and cancel each other out.
THIS IS A TRICK QUESTION.
THe electric field exactly midway between them = 0/Q = 0.
But if the point moves even slightly you need the following formula
F= (1/4Piε)(Q1Q2/D^2)
Assume your test charge is positive and make sure you remember two positive charges repel, two unlike charges attract. Draw the forces on the test charge out as vectors and find the magnetude of the force, then divide by the total charge to to find the electric field strength:)</span>
Answer:
(a) Angular acceleration is 1.112 rad/s².
(b) Average angular velocity is 2.78 rad/s .
Explanation:
The equation of motion in Rotational kinematics is:
θ = θ₀ + 0.5αt²
Here θ is angular displacement at time t, θ₀ is angular displacement at time t=0, t is time and α is constant angular acceleration.
(a) According to the problem, θ is 13.9 rad, θ₀ is zero as it is at rest and t is 5 s. Put these values in the above equation:
13.9 = 0 + 0.5α(5)²
α = 1.112 rad/s²
(b) The equation of average angular velocity is:
ω = Δθ/Δt
ω = 
ω = 2.78 rad/s
Explanation:
Given that,
Distance, s = 47 m
Time taken, t = 8.6 s
Final speed of the truck, v = 2.3 m/s
Let u is the initial speed of the truck and a is its acceleration such that :
.............(1)
Now, the second equation of motion is :

Put the value of a in above equation as :




u = 8.63 m/s
So, the original speed of the truck is 8.63 m/s. Hence, this is the required solution.
The first step of the scientific method is to MAKE OBSERVATIONS