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Firlakuza [10]
3 years ago
14

Two parallel wires are separated by 5.60 cm, each carrying 2.65 A of current in the same direction. (a) What is the magnitude of

the force per unit length between the wires? N/m (b) Is the force attractive or repulsive? attractive repulsive
Physics
1 answer:
ololo11 [35]3 years ago
5 0

Explanation:

It is given that,

The separation between two parallel wires, r = 5.6 cm = 0.056 m

Current in both the wires is 2.65 A

(a) We need to find the magnitude of the force per unit length between the wires. It can be given by :

\dfrac{F}{l}=\dfrac{\mu_o I_1I_2}{2\pi r}\\\\\dfrac{F}{l}=\dfrac{4\pi \times 10^{-7}\times 2.65\times 2.65}{2\pi \times 0.056}\\\\\dfrac{F}{l}=2.5\times 10^{-5}\ N/m

(b) As the current is in same direction, the wires will attract each other.

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Explanation:

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If Mrs. Reichelt throws a chromebook, because it won't login correctly, with a force of 8N, and the chromebook accelerates at 5m
suter [353]

Answer:

1.6 kg

Step-by-step Solution:

Since Force = mass × acceleration we have:

F = 8N

a= 5 m/s^2

m = ?

By plugging the values above into F=ma we obtain:

F=ma\\\\8=m(5)\\\\\frac{8}{5}=\frac{m(5)}{5}\\\\m=\frac{8}{5}=1.6

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7 0
3 years ago
A wave oscillates 5.0 times a second and has a speed of 4.0m/s what is the frequency of this wave
Viefleur [7K]

Answer:

The frequency of the wave is 5.0Hz

Explanation:

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7 0
3 years ago
A 500 kg table with 4 legs rests on solid flat ground. I place 3 books on the center of the table with masses of 10 kg, 20 kg an
BARSIC [14]

Answer:

1,373.4 N

Explanation:

The mass of the table acts at the centre in addition to the books since that is the centre of gravity of the table.

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Total mass of table and books will be 500kg+60kg=560 kg

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3 years ago
A coil of wire containing N turns is in an external magnetic field that is perpendicular to the plane of the coil and it steadil
krok68 [10]

Answer:

The Resultant Induced Emf in coil is 4∈.

Explanation:

Given that,

A coil of wire containing having N turns in an External magnetic Field that is perpendicular to the plane of the coil which is steadily changing. An Emf (∈) is induced in the coil.

To find :-

find the induced Emf if rate of change of the magnetic field and the number of turns in the coil are Doubled (but nothing else changes).

So,

   Emf induced in the coil represented by formula

                          ∈  =   -N\frac{d\phi}{dt}                                  ...................(1)

                                          Where:

                                                    .   \phi = BAcos\theta     { B is magnetic field }

                                                                                 {A is cross-sectional area}

                                                    .  N = No. of turns in coil.

                                                    .  \frac{d\phi}{dt} = Rate change of induced Emf.

Here,

Considering the case :-

                                    N1 = 2N  &      \frac{d\phi1}{dt} = 2\frac{d\phi}{dt}

Putting these value in the equation (1) and finding the  new emf induced (∈1)

                           

                                      ∈1 =-N1\times\frac{d\phi1}{dt}

                                      ∈1 =-2N\times2\frac{d\phi}{dt}

                                       ∈1 =4 [-N\times\frac{d\phi}{dt}]

                                        ∈1 = 4∈             ...............{from Equation (1)}      

Hence,

The Resultant Induced Emf in coil is 4∈.        

                           

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