Answer:
e.)At twice the distance, the strength of the field is E/4.
Explanation:
The strength of the electric field at a certain distance from a point charge is given by:
where
k is the Coulomb's constant
Q is the charge
r is the distance from the point charge
In this problem, the distance from the point charge is doubled:
r' = 2r
So the new electric field strength is
so, at twice the distance the strength of the field is E/4.
Answer:
a_total = 2 √ (α² + w⁴) , a_total = 2,236 m
Explanation:
The total acceleration of a body, if we use the Pythagorean theorem is
a_total² = a_T²2 + ²
where
the centripetal acceleration is
a_{c} = v² / r = w r²
tangential acceleration
a_T = dv / dt
angular and linear acceleration are related
a_T = α r
we substitute in the first equation
a_total = √ [(α r)² + (w r² )²]
a_total = 2 √ (α² + w⁴)
Let's find the angular velocity for t = 2 s if we start from rest wo = 0
w = w₀ + α t
w = 0 + 1.0 2
w = 2.0rad / s
we substitute
a_total = r √(1² + 2²) = r √5
a_total = r 2,236
In order to finish the calculation we need the radius to point A, suppose that this point is at a distance of r = 1 m
a_total = 2,236 m