Answer:
(a)There are no magnetic monopoles. true
(c) There must be a vector potential.
true
(d) Charges create electric fields.2. true
Explanation:
a) there are no magnetic monopoles because magnetic field is created by charges (electrons) and these electrons have dipole field so it is not possible to have magnetic dipoles, more ever Gauss's law always explained that there are not magnetic dipoles. furthermore magnetic monopoles aree caused by magnetic charges and we have electric charges.
c)vector potential is a vector field which serves as a potential for magnetic field so, the magnetic field B by Faraday and Gauss's law is also known as vector potential.
d) electric field is solely generated by charges be it static charges or moving charges if there are no charges it is not possible to have an electric field.
Answer:
<em>600N.</em>
Explanation:
From the question, we are to calculate the net force acting on the car.
According to Newton's second law of motion:
F = ma
m is the mass of the car
a is the acceleration = change in velocity/Time
a = v-u/t
F = m(v-u)/t
v is the final velocity = 30m/s
u is the initial velocity = 20m/s
t is the time = 5secs
m = 300kg
Get the net force:
Recall that: F = m(v-u)/t
F = 300(30-20)/5
F = 60(30-20)
F = 60(10)
<em>F = 600N</em>
<em>Hence the net force acting on the car is 600N.</em>
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Answer:
Explanation:
The power of each of the speakers is 0.535 W. At a distance d intensity of sound can be found by the following formula
Intensity of sound = Power / 4π d²
= .535 / 4 x 3.14 x (27.3/2)²
= 2.286 x 10⁻⁴ J m⁻² s⁻¹
Intensity of sound due to other source = 5.715 x 10⁻⁵J m⁻² s⁻¹
Total intensity = 2 x 2.286 x 10⁻⁴J m⁻² s⁻¹
= 4.57 x 10⁻⁴J m⁻² s⁻¹
b ) In this case, man is standing at distances 18.15 m and 9.15 m from the sources .
The total intensity of sound reaching him is as follows
0.535 / (4 π x18.15² ) + 0.535 / (4 π x9.15² )
= 1.293 x 10⁻⁴ + 5.087 x 10⁻⁴
= 6.38 x 10⁻⁴J m⁻² s⁻¹
Answer:
Explanation:
i )
When it is disconnected with the battery , the charge stored in it becomes fixed . When the plate distance becomes half , its capacitance becomes twice from C to 2C . Let charge stored in it at the time of disconnection from battery be Q . Let plate separation reduces from d to d / 2
So charged stored in it will remain unchanged .
ii )
Potential difference = charge / capacitance
in the first case potential difference = Q / C
in the second case potential difference = Q / 2C
So potential difference becomes half .
iii ) electric field = potential diff / plate separation
in the first case electric field = Q / (d x C )
in the second case electric field = 2 Q / (d x 2C)
= Q / (d x C )
So electric field remains unchanged .
iv)
energy stored in first case = Q² / 2C
In the second case energy stored = Q² / 2x2C
so energy stored becomes half .
