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Answer:
The magnitude of the electric field is 8.47 N/C
Explanation:
Given;
uniform charge density, λ = 100 nC/m³
inner radii of the cylinder, r = 1.0 mm and 3.0 mm
distance from the symmetry axis, R = 2.0 mm
Area = 2πrl
Area =2π(2 x 10⁻)l
Volume = A x d
d = Volume / Area
the magnitude of the electric field
Therefore, the magnitude of the electric field is 8.47 N/C
Answer:
Ek= 35.392 kJ or 35.392 10³ J
Explanation:
Formula for Kinetic energy:
Ek= 
× m × 
since m= 65kg and v= 33m/s
Ek= × 65×
Ek= 35.392 kJ or 35.392 10³ J
Answer:
3540.5N
Explanation:
Step one:
given data
mass m= 0.196kg
speed v= 31m/s
distance r= 5.32cm = 0.0532m
Step two
The expression relating force, mass, velocity and distance is
F= mv^2/r
substitute we have
F=0.196*31^2/0.0532
F=0.196*961/0.0532
F=188.356/0.0532
F=3540.5N
