The electric field in a certain region of Earth's atmosphere is directed vertically down. At an altitude of 241 m the field has
magnitude 46.8 N/C; at an altitude of 93.9 m, the magnitude is 114 N/C. Find the net amount of charge contained in a cube 147.1 m on edge, with horizontal faces at altitudes of 93.9 m and 241 m.
2 answers:
Answer:
Explanation:
Electric field at height 241 m , E = 46.8 N/C
Electric field at height 93.1 m, E' = 114 N/C
length of edge, a = 147.1 m
The total flux
Фnet = Ф + Ф'
Фnet = E A Cos 180 + E' A Cos 0
Фnet = A ( E' - E)
According to the Gauss theorem
Фnet = enclosed charge / ∈o
∈o x A (E' - E) = q
q = 8.854 x 10^-12 x 147.1 x 147.1 x (114 - 46.8)
q = 1.29 x 10^-5 C
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Answer:
1185 N
Explanation:
From Newton’s second law of motion,
F=ma where m= mass of motorcycle, a is acceleration of the motorcycle and F=Force
Net force acting on motorcycle is given by
Where F is force acting on motorcycle and f is frictional force
Substituting F-f for
hence ma= F- f Substituting a with 3, m with 245Kg and f with 450N as provided
245*3= F- 450
F=245*3 +450= 1185 N