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Murrr4er [49]
2 years ago
14

By what factor is the self-inductance of an air solenoid changed if its length and number of coil turns are both tripled

Physics
1 answer:
fredd [130]2 years ago
4 0

Answer:

The new self inductance is 3 times of the initial self inductance.

Explanation:

The self inductance of a solenoid is given by :

L=\dfrac{\mu_oN^2 A}{L}

Where

N is number of turns per unit length

A is area of cross section

l is length of solenoid

If length and number of coil turns are both tripled,

l' = 3l and N' = 3N

New self inductance is given by :

L'=\dfrac{\mu_oN'^2 A}{L'}\\\\=\dfrac{\mu_o(3N)^2 A}{3L}\\\\=3\dfrac{\mu_oN^2 A}{L}\\\\=3L

So, the new self inductance is 3 times of the initial self inductance.

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A very delicate sample is placed 0.150 cm from the objective lens of a microscope. The focal length of the objective is 0.140 cm
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Answer:

m = 14*26 = 364

Explanation:

overall magnification is given as m

m = m_{o}* m_{e}

mo magnification of objective lens

me magnification of EYE lens

where mo is given as

m_{o} = \frac{v_{o}}{-u _{0}}

and me as

m_{e} = 1+\frac{D}{f_{e}}

d is distant of distinct vision = 25.0 cm for normal eye

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focal length of objective lense is 0.140 cm

we know that

\frac{1}{v_{0}}-\frac{1}{u_{0}}=\frac{1}{f_{0}}

\frac{1}{v_{0}} = \frac{1}{u_{0}} + \frac{1}{f_{0}}

\frac{1}{v_{0}} = \frac{1}{0.150} + \frac{1}{0.14}

\frac{1}{v_{o}} = 2.1 cm

m_{o} = \frac{2.1}{0.150} = 14

m_{e} = 1+\frac{25}{1}

m_{e} =26

m = m_{o}* m_{e}

m = 14*26 = 364

4 0
3 years ago
In the Bronsted-Lowry Theory of acids and bases, an acid is this.
madreJ [45]

Answer: An acid is a substance that donates a proton and produces a conjugate base.

Explanation:

According to Bronsted-Lowry theory, an acid is a substance that donates a proton and produces a conjugate base while a base is a molecule or ion which accepts the proton.

An example of Bronsted-Lowry acid and base is Ethanoic acid, CH3COOH and hydroxide ion, OH- respectively as shown in the reaction below

CH3COOH(aq) + OH-(aq) <---> CH3COO-(aq) + H2O(l)

Thus, ethanoic acid acts as an acid by donating a proton to the hydroxide ion which accepts it, thus producing ethanoate ion, CH3COO- as a conjugate base.

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Explanation:

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