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Lyrx [107]
3 years ago
13

A thin coil has 17 rectangular turns of wire. When a current of 4 A runs through the coil, there is a total flux of 5 ✕ 10−3 T ·

m2 enclosed by one turn of the coil (note that
Φ = kI,

and you can calculate the proportionality constant k). Determine the inductance in henries.
Physics
1 answer:
Alex73 [517]3 years ago
7 0

Answer:

Inductance, L = 0.0212 Henries

Explanation:

It is given that,

Number of turns, N = 17

Current through the coil, I = 4 A

The total flux enclosed by the one turn of the coil, \phi=5\times 10^{-3}\ Tm^2

The relation between the self inductance and the magnetic flux is given by :

L=\dfrac{N\phi}{I}

L=\dfrac{17\times 5\times 10^{-3}}{4}

L = 0.0212 Henries

So, the inductance of the coil is 0.0212 Henries. Hence, this is the required solution.

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Ball X of mass 1.0 kg and ball Y of mass 0.5kg travel toward each other on a horizontal surface. Both balls travel with a consta
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Answer:

B. Ball Y will travel at a speed less than 5 m/s in the opposite direction of travel as before the collision.

Explanation:

Impulse created by ball Y  on ball X = 40 x 1/6 Ns

Ball X will also create impulse 40 / 6 on ball Y .

impulse = change in momentum .

impulse in Y = change in momentum in Y .

Initial momentum of Y  = .5 x 5 = 2.5

Let final velocity of Y after collision be v  in opposite direction .

change in momentum of Y = v - (-2.5 )

so,

v + 2.5 = 40 / 6 =  6.67

v =  4.17 m / s .

Option B is correct .

B. Ball Y will travel at a speed less than 5 m/s in the opposite direction of travel as before the collision.

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I don't know the choices but...

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Sedimentary rock is a rock formed underground, usually under a source of water too. The rock is made up of smaller rock that, over the years, fused together and solidified.

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when their center-to-center separation is 50 cm. The spheres are then connected by a thin conducting wire. When the wire is remo
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Answer:

q1 = 7.6uC , -2.3 uC

q2 = 7.6uC , -2.3 uC

( q1 , q2 ) = ( 7.6 uC , -2.3 uC ) OR ( -2.3 uC , 7.6 uC )

Explanation:

Solution:-

- We have two stationary identical conducting spheres with initial charges ( q1 and q2 ). Such that the force of attraction between them was F = 0.6286 N.

- To model the electrostatic force ( F ) between two stationary charged objects we can apply the Coulomb's Law, which states:

                              F = k\frac{|q_1|.|q_2|}{r^2}

Where,

                     k: The coulomb's constant = 8.99*10^9

- Coulomb's law assume the objects as point charges with separation or ( r ) from center to center.  

- We can apply the assumption and approximate the spheres as point charges under the basis that charge is uniformly distributed over and inside the sphere.

- Therefore, the force of attraction between the spheres would be:

                             \frac{F}{k}*r^2 =| q_1|.|q_2| \\\\\frac{0.6286}{8.99*10^9}*(0.5)^2 = | q_1|.|q_2| \\\\ | q_1|.|q_2| = 1.74805 * 10^-^1^1 ... Eq 1

- Once, we connect the two spheres with a conducting wire the charges redistribute themselves until the charges on both sphere are equal ( q' ). This is the point when the re-distribution is complete ( current stops in the wire).

- We will apply the principle of conservation of charges. As charge is neither destroyed nor created. Therefore,

                             q' + q' = q_1 + q_2\\\\q' = \frac{q_1 + q_2}{2}

- Once the conducting wire is connected. The spheres at the same distance of ( r = 0.5m) repel one another. We will again apply the Coulombs Law as follows for the force of repulsion (F = 0.2525 N ) as follows:

                          \frac{F}{k}*r^2 = (\frac{q_1 + q_2}{2})^2\\\\\sqrt{\frac{0.2525}{8.99*10^9}*0.5^2}  = \frac{q_1 + q_2}{2}\\\\2.64985*10^-^6 =   \frac{q_1 + q_2}{2}\\\\q_1 + q_2 = 5.29969*10^-^6  .. Eq2

- We have two equations with two unknowns. We can solve them simultaneously to solve for initial charges ( q1 and q2 ) as follows:

                         -\frac{1.74805*10^-^1^1}{q_2} + q_2 = 5.29969*10^-^6 \\\\q^2_2 - (5.29969*10^-^6)q_2 - 1.74805*10^-^1^1 = 0\\\\q_2 = 0.0000075998, -0.000002300123

                         

                          q_1 = -\frac{1.74805*10^-^1^1}{-0.0000075998} = -2.3001uC\\\\q_1 = \frac{1.74805*10^-^1^1}{0.000002300123} = 7.59982uC\\

 

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