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2 years ago

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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Which object is least likely to allow light to pass through it?

jeka57 [31]

a tree truck

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3 years ago

Read 2 more answers

You push a desk with 245 N, but the desk doesn't move due to its friction with the ground. What is the magnitude of the friction

const2013 [10]

If the desk doesn't move, then it's not accelerating.

If it's not accelerating, then the net force on it is zero.

If the net force on it is zero, then any forces on it are balanced.

If there are only two forces on it and they're balanced, then they have equal strengths, and they point in opposite directions.

So the friction on the desk must be equal to your<em> 245N</em> .

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3 years ago

If a ball is thrown straight up into the air with an initial velocity of 65 ft/s, its height in feet after t seconds is given by

fgiga [73]

Answer:

a) $v_{1}=\frac{(62.5-66)ft}{(2.5-2)s}=-7ft/s$

$v_{2}=\frac{(65.94-66)ft}{(2.1-2)s}=-0.6ft/s$

$v_{3}=\frac{(66.0084-66)ft}{(2.01-2)s}=0.84ft/s$

$v_{4}=\frac{(66.001-66)ft}{(2.001-2)s}=1ft/s$

b) $v=65-32(2)=1ft/s$

Explanation:

From the exercise we got the ball's equation of position:

$y=65t-16t^{2}$

a) To find the average velocity at the given time we need to use the following formula:

$v=\frac{y_{2}-y_{1} }{t_{2}-t_{1} }$

Being said that, we need to find the ball's position at t=2, t=2.5, t=2.1, t=2.01, t=2.001

$y_{t=2}=65(2)-16(2)^{2} =66ft$

$y_{t=2.5}=65(2.5)-16(2.5)^{2} =62.5ft$

$v_{1}=\frac{(62.5-66)ft}{(2.5-2)s}=-7ft/s$

$y_{t=2.1}=65(2.1)-16(2.1)^{2} =65.94ft$

$v_{2}=\frac{(65.94-66)ft}{(2.1-2)s}=-0.6ft/s$

$y_{t=2.01}=65(2.01)-16(2.01)^{2} =66.0084ft$

$v_{3}=\frac{(66.0084-66)ft}{(2.01-2)s}=0.84ft/s$

$y_{t=2.001}=65(2.001)-16(2.001)^{2} =66.001ft$

$v_{4}=\frac{(66.001-66)ft}{(2.001-2)s}=1ft/s$

b) To find the instantaneous velocity we need to derivate the equation

$v=\frac{df}{dt}=65-32t$

$v=65-32(2)=1ft/s$

7 0

3 years ago

I’m sorry this is a quiz and we’re correcting it and I still don’t get it anyone know ?

kotegsom [21]

Answer:

3.0M

Explanation:

Thats two wavelengths,not one.

Pretty honest mistake I would've made the same if I was rushing

5 0

3 years ago

A diode vacuum tube consists of a cathode and an anode spaced 5-mm apart. If 300 V are applied across the plates. What is the ve

MAXImum [283]

Answer:

Explanation:

There is electric field between the plates whose value is given by the following expression

electric field E = V /d where V is potential between the plates and d is distance between them

E = 300 / 5 x 10⁻³

= 60 x 10³ N/c

Force on electron = q E where q is charge on the electron

F = 1.6 X 10⁻¹⁹ X 60 X 10³ = 96 X 10⁻¹⁶ N.

Acceleration a = force / mass

a = 96 x 10⁻¹⁶/ mass = 96 x 10⁻¹⁶ / 9.1 x 10⁻³¹

= 10.55 x 10¹⁵ m / s²

For midway , distance travelled

s = 2.5 x 10⁻³ m

s = 1\2 a t²

t = $\sqrt{\frac{2s}{a\\ } }$

= $\sqrt{\frac{2\times2.5\times10^{-3}}{ 10.55\times10^{15}}$

t = .474 x 10⁻¹⁸ s

For striking the plate time is calculated as follows

t = $[tex]\sqrt{\frac{2\times5\times10^{-3}}{ 10.55\times10^{15}}$ [/tex]

t = 0.67 x 10⁻¹⁸ s

3 0

3 years ago