The intensity of the electric field is 30,000 N/C
Explanation:
The strength of the electric field produced by a single-point charge is given by the equation
where:
is the Coulomb's constant
q is the magnitude of the charge
r is the distance from the charge
In this problem, we have:
is the magnitude of the charge
r = 3 cm = 0.03 m is the distance at which we are calculating the field intensity
Substituting, we find:
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2.57 joule energy lose in the bounce .
<u>Explanation</u>:
when ball is the height of 1.37 m from the ground it has some gravitational potential energy with respect to hits the ground
Formula for gravitational potential energy given by
Potential Energy = mgh
Where ,
m = mass
g = acceleration due to gravity
h = height
Potential energy when ball hits the ground
m= 0.375 kg
h = 1.37 m
g = 9.8 m/s²
Potential Energy = 5.03 joule
Potential energy when ball bounces up again
h= 0.67 m
Potential Energy = 2.46 joule
Energy loss = 5.03 - 2.46 = 2.57 joule
2.57 joule energy lose in the bounce