A very long straight wire has charge per unit length 1.56×10−10 C/m .
1 answer:
Answer:
The magnitude of the electric field equal to 2.40 N/C at 1.1537m from the wire.
Explanation:
using Guass law,
(guessing that a cylinder of radius r and length L around wire such that wire is at centre )
E. A = qin / e0
E ( 2πr L ) = (1.56 x x L) / (8.85 x
)
E = (1.56 x ) / (2πr x 8.85 x
)
so 2.40 = (1.54 x ) / (2πr x 8.85 x
)
2.40 (6.284r) = 0.174 x 10²
15.0816r = 17.4
r = 1.1537m
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Answer:
the work done by gravity on the boy is 604.62 J
Explanation:
Given;
distance the boy slides, d = 3 m
angle of inclination of the playground, θ = 40⁰
mass of the boy, m = 32 kg
The vertical height, h, above the ground through which the boy falls represents the height of the triangle which is the opposite side.
The distance through which the boy slides, d, represents the hypotenuse side of the right triangle.
The work done by gravity on the boy is calculated as;
W = P.E = mgh
= 32kg x 9.8m/s² x 1.928m
= 604.62 J
Therefore, the work done by gravity on the boy is 604.62 J
Answer:
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,
Explanation:
The cube root of the complex number can determined by the following De Moivre's Formula:
Where angles are measured in radians and k represents an integer between and
.
The magnitude of the complex number is and the equivalent angular value is
. The set of cubic roots are, respectively:
k = 0
k = 1
k = 2