The self-inductance of a coil will change by 8 times its original value by increasing its radius value by 2 and increasing the length of the coil by 2.
Self-Inductance: -
The definition of self-inductance is the induction of a voltage in a wire that carries current when the current in the wire is changing. In the instance of self-inductance, the circuit itself induces a voltage through the magnetic field produced by a changing current.
We know that the self-inductance of the coil is denoted by: -
L= µ *π*(r)^2*(N)^2*l
Where
L= Self-Inductance of the coil
µ= Magnetic Permeability Constant
r= Radius of the coil
l= Length of the coil
N= Number of turns of the coil
Here Self-inductance of the coil is directly proportional to the length of the coil and the square of the radius of the coil.
So,
On increasing the radius of the coil by a factor of 2 and the length of the coil by 2 the self-inductance of the coil increases by 8 times its original value.
Learn more about Self-Inductance here: -
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Answer:
-6112.26 J
Explanation:
The initial kinetic energy, 
is given by
} where m is the mass of a body and 
is the initial velocity
The final kinetic energy, 
is given by
where 
is the final velocity
Change in kinetic energy, 
is given by
Since the skater finally comes to rest, the final velocity is zero. Substituting 0 for and 12.6 m/s for and 77 Kg for m we obtain
From work energy theorem, work done by a force is equal to the change in kinetic energy hence for this case work done equals <u>-6112.26 J</u>
Answer:
a hamster resulting from sexual reproduction
Answer:
They will not meet
h-hX=1.2*g*t²
hX=v0*t-(1/2*g*t²)
Explanation:
fall h=1/2*g*t²
elevation time if v0=20 m/s te=v0/g=20 m/s /9.81 m/s²=2.0387s
hmax=v0²/(2*g)=(400 m²/s²)/19.62 m/s²2=20.387 m
free fall
t=2.0387s yields hX=1/2*g*t²=20.387 m
h-hX=200m - 20.387 m=179,613 m.
so, the second body has not enough initianoal speed to reach a meeting point
