1 Amp = 1 Coulomb/sec
1 Coulomb/sec = 6.25*10^18 electrons/sec
Therefore,
5.0 A = 5 C/s = 5*6.25*10^18 = 3.125*10^19 e/s
In 10 second, number of electrons are calculated as;
Number of electrons through the device = 3.125*10^19*10 = 3.125*10^20 electrons
F = net force acting on the elevator in upward direction = 3000 N
m = mass of the elevator = 1200 kg
a = acceleration of the elevator = ?
Acceleration of the elevator is given as
a = F/m
a = 3000/1200
a = 2.5 m/s²
v₀ = initial velocity of the elevator = 0 m/s
Y = displacement of the elevator = 15 m
t = time taken
Using the kinematics equation
Y = v₀ t + (0.5) a t²
15 = (0) t + (0.5) (2.5) t²
t = 3.5 sec
Answer:
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Explanation:
the answer is A homogeneous electric field has the same magnitude and direction at any place. A good example of a homogeneous field is the field between two charged metal plates. The field strength depends on the voltage U and the plate distance d.
Answer:
a) E = 8628.23 N/C
b) E = 7489.785 N/C
Explanation:
a) Given
R = 5.00 cm = 0.05 m
Q = 3.00 nC = 3*10⁻⁹ C
ε₀ = 8.854*10⁻¹² C²/(N*m²)
r = 4.00 cm = 0.04 m
We can apply the equation
E = Qenc/(ε₀*A) (i)
where
Qenc = (Vr/V)*Q
If Vr = (4/3)*π*r³ and V = (4/3)*π*R³
Vr/V = ((4/3)*π*r³)/((4/3)*π*R³) = r³/R³
then
Qenc = (r³/R³)*Q = ((0.04 m)³/(0.05 m)³)*3*10⁻⁹ C = 1.536*10⁻⁹ C
We get A as follows
A = 4*π*r² = 4*π*(0.04 m)² = 0.02 m²
Using the equation (i)
E = (1.536*10⁻⁹ C)/(8.854*10⁻¹² C²/(N*m²)*0.02 m²)
E = 8628.23 N/C
b) We apply the equation
E = Q/(ε₀*A) (ii)
where
r = 0.06 m
A = 4*π*r² = 4*π*(0.06 m)² = 0.045 m²
Using the equation (ii)
E = (3*10⁻⁹ C)/(8.854*10⁻¹² C²/(N*m²)*0.045 m²)
E = 7489.785 N/C
I think the answer would be A