Arrange these source charges in order from least to greatest magnitude.
2 answers:
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Answer:
q = 4.5 nC
Explanation:
given,
electric field of small charged object, E = 180000 N/C
distance between them, r = 1.5 cm = 0.015 m
using equation of electric field
k = 9 x 10⁹ N.m²/C²
q is the charge of the object
now,
q = 4.5 x 10⁻⁹ C
q = 4.5 nC
the charge on the object is equal to 4.5 nC
Answer:
218m
Explanation:
Given parameters:
Initial velocity = 3.3m/s
acceleration = 3.7m/s²
time = 10s
Unknown:
How far will it travel during the time of acceleration = ?
Solution:
We use of the kinematics equations to solve this problem;
S = ut + at²
S is the distance
u is the initial velocity
t is the time
a is the acceleration
So;
S = (3.3x10) + ( x 3.7 x 10²) = 218m