What is the magnitude of the total displacement of the school bus.
1 answer:
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Answer:
a)
346.67 N/C, downward
b)
1.3 m
Explanation:
(a)
q = magnitude of charge on the particle = 1.95 x 10⁻⁹ C
r = distance of location from the charged particle = 0.225 m
E = magnitude of electric field at the location
Magnitude of electric field at the location is given as
Inserting the values
E = 346.67 N/C
a negative charge produce electric field towards itself.
Direction : downward
(b)
E = magnitude of electric field at the location = 10.5 N/C
r = distance of location from the charged particle = ?
q = magnitude of charge on the particle = 1.95 x 10⁻⁹ C
Magnitude of electric field at the location is given as
Inserting the values
r = 1.3 m
Let us assume that rocket only runs in initial energy and not using its own to flying.
Also , let upward direction is +ve and downward direction is -ve .
Initial velocity , u = 58.8 m/s .
Acceleration due to gravity , .
Final velocity , v - = 0 m/s .
We know , by equation of motion .
Hence, this is the required solution .