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OLga [1]
2 years ago
7

How would the self-inductance, l, of a coil change if you would increase its radius by a factor of two and increase its length b

y a factor of two?
Physics
1 answer:
pochemuha2 years ago
3 0

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: -

" brainly.com/question/15293029 "

#SPJ4

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Two objects with masses m and 3m are connected by a compressed spring so that the energy stored in the spring is 375 joules. If
cluponka [151]

Answer:

281.25 J

Explanation:

We are told that the two objects with masses m and 3m.

Also that energy stored in the spring is 375 joules.

Now, initially the centre of mass of the system took place at rest, it means v1 = v and v2 = v/3

Thus, from principle of conservation of energy, we have;

½mv² + ½(3m)(v/3)² = 375J

(m + 3m/9)½v² = 375

(4/3)m × ½v² = 375

Multiply both sides by ¾ to get;

½mv² = 375 × ¾

½mv² = 281.25 J

Therefore, energy of lighter body is 281.25 J

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The force F required to compress a spring a distance x is given by F 2 F0 5 kx where k is the spring constant and F0 is the prel
IrinaVladis [17]

Answer:

a)W=8.333lbf.ft

b)W=0.0107 Btu.

Explanation:

<u>Complete question</u>

The force F required to compress a spring a distance x is given by F– F0 = kx where k is the spring constant and F0 is the preload. Determine the work required to compress a spring whose spring constant is k= 200 lbf/in a distance of one inch starting from its free length where F0 = 0 lbf. Express your answer in both lbf-ft and Btu.

Solution

Preload = F₀=0 lbf

Spring constant k= 200 lbf/in

Initial length of spring x₁=0

Final length of spring x₂= 1 in

At any point, the force during deflection of a spring is given by;

F= F₀× kx  where F₀ initial force, k is spring constant and x is the deflection from original point of the spring.

W=\int\limits^2_1 {} \, Fds \\\\\\W=\int\limits^2_1( {F_0+kx} \,) dx \\\\\\W=\int\limits^a_b {kx} \, dx ; F_0=0\\\\\\W=k\int\limits^2_1 {x} \, dx \\\\\\W=k*\frac{1}{2} (x_2^{2}-x_1^{2}  )\\\\\\W=200*\frac{1}{2} (1^2-0)\\\\\\W=100.lbf.in\\\\

Change to lbf.ft by dividing the value by 12 because 1ft=12 in

100/12 = 8.333 lbf.ft

work required to compress the spring, W=8.333lbf.ft

The work required to compress the spring in Btu will be;

1 Btu= 778 lbf.ft

?= 8.333 lbf.ft----------------cross multiply

(8.333*1)/ 778 =0.0107 Btu.

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