Hope this helps!!!!!!!!!!!!!
Answer:
the electric field strength on the second one is 2.67 N/C.
Explanation:
the electric fiel on the first one is:
E1 = k×q/(r^2)
r^2 = k×q/(E1)
= (9×10^9)×(q)/(24.0)
= 375000000q
then the electric field on the second one is:
E2 = k×q/(R^2)
we know that R = 3r
R^2 = 9×r^2
E2 = k×q/(9×r^2)
= k×q/(9×375000000q)
= k/(9×375000000)
= (9×10^9)/(9×375000000)
= 2.67 N/C
Therefore, the electric field strength on the second one is 2.67 N/C.
Answer:
48.6°
Explanation:
The forward force, F equals the component of the weight along the slope.
So mgsinθ = ma where a = acceleration and θ = angle between the slope and the horizontal.
So a = gsinθ
Since we are given that a = 75%g = 0.75g,
0.75g = gsinθ
sinθ = 0.75
θ = sin⁻¹(0.75)
= 48.6°
Answer:
5880lb-ft of work is done
Explanation:
The length of the heavy rope is given as 60ft and the weight per length is 0.7lb/ft.
Therefore, the total weight of the heavy rope is
60×0.7 =42lb.
The work done in pulling the heavy rope to the top of the building is w = Fd.
Where
F is force is measured in pounds;42lb
d is distance through which the heavy rope is to be pulled measured in feet; 140ft
w= 42lb×140ft= 5880lb-ft
