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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Answer:
4.4345× 10^-7V
Explanation:
The computation of the half voltage for a 1.2T magnetic field applied is shown below
The volume of one mole of copper is
v = m ÷p
= 63.5 ÷ 8.92
= 7.12cm
Now the density of free electrons in copper is
n = Na ÷ V
= 6.02 × 10^23 ÷ 7.12
= 8.456× 10^28/m^3
Now the half voltage is
= IB ÷ nqt
= (5 × 1.20) ÷ (8.456× 10^28 × 1.6 × 10^-19 × 0.1× 10^-2)
= 4.4345× 10^-7V
Answer:
2.4 m
Explanation:
Consider the motion along the vertical direction
= initial position of ball above the ground = 4.5 m
= time taken by the ball to hit the smokestack = 0.65 s
= initial velocity of the ball along vertical direction
= acceleration due to gravity = - 9.8 m/s²
= position of ball at the time of hitting the smokestack
Using the kinematics equation
inserting the above values