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4 years ago

What is the direction of magnetic field lines inside any magnet?

Physics

2 answers:

likoan [24]4 years ago

7 0

<h2>Answer:</h2>

<u>The right option is </u><u>(C)They travel in a loop.</u>

<h2>Explanation:</h2>

The magnetic field lines point from the north pole to the south pole. Magnetic lines of force always forms closed loops. However, the magnetic field lines do not just end at the tip of the magnet. It is equally strong at the north pole when compared with the south pole.They go right through it, so that inside the magnet the magnetic field points from the south pole to the north pole.

Kruka [31]4 years ago

3 0

the answer is a because they start on the north side

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suter [353]

Answer:

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2 years ago

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A point charge (–5.0 µC) is placed on the x axis at x = 4.0 cm, and a second charge (+5.0 µC) is placed on the x axis at x = –4.

AveGali [126]

Answer:

The magnitude of electric force is $7.2\times10^{-3} N$

Explanation:

Coulomb's Law:

The force of attraction or repletion is

directly proportional to the products of charges i.e $F\propto q_1q_2$

inversely proportional to the square of distance i.e $F\propto \frac{1}{r^2}$

$\therefore F\propto \frac{q_1q_2}{r^2}$

$\Rightarrow F=k \frac{q_1q_2}{r^2}$ [ k is proportional constant=9×10⁹N m²/C²]

There are two types of force applied on Q=+2.5 μC=2.5×10⁻⁶ C

Let F₁ force be applied on Q =+2.5 μC by q₁= -5.0 μC = - 5.0×10⁻⁶ C

and F₂ force be applied on Q=+2.5 μC by q₂= 5.0 μC= 5.0×10⁻⁶ C

Since the magnitude of F₁ and F₂ are same. Therefore their y component cancel.

If we draw a line from q₁ to Q .

The it forms a triangle whose base = 4.0 cm and altitude =3.0 cm.

Let hypotenuse = r

Therefore, $r=\sqrt{altitude^2+base^2} =\sqrt{3^2+4^2} =5$

we know,

$cos \theta = \frac{base }{hypotenuse}$

$\Rightarrow cos \theta = \frac{4 }{r}$

Total force $F_Q = 2.F_1 cos\theta \hat{i}$

$=2k\frac{Qq_1}{r^2} cos\theta \hat i$

$=2\ \frac{9\times1 0^9\times2.5 \times 5\times 10^{-12}}{r^2} \frac{4}{r} \hat i$

$=8\ \frac{9\times10^9\times2.5 \times 5\times 10^{-12}}{5^3} \hat i$ [ r=5]

$=7.2\times10^{-3}\hat i$ N

The magnitude of electric force is $7.2\times10^{-3} N$

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3 years ago

Do you think the inner planets should be explored or should the money be spent on other things? Justify your opinion.

frosja888 [35]

Answer:

yes i think that inner planets should be explored because if we ever find new life or are able to live there then i think its worth takeing a risk.

Explanation:

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3 years ago

A crate on a motorized cart starts from rest and moves with a constant eastward acceleration ofa= 2.60 m/s^2. A worker assists t

Leona [35]

Answer:

To obtain the power, we first need to find the work made by the force.

1) To calculate the work, we need the next equation:

$\int\limits {F} \, dx$

So the force is given by the problem so our mission is to find 'dx' in terms of 't'

2) we know that:

$\frac{dV}{dt} = a = 2.6$

So we have:

$v = 2.6t$

Then:

$\frac{dx}{dt} = V = 2.6t\\ \\dx = 2.6t*dt$

3) Finally, we replace everything:

$\int\limits^{4.7}_{0} {5.4t*2.6t} \, dt$

After some calculation, we have as a result that the work is:

161.9638 J.

4) To calculate the power we need the next equation:

$P = \frac{W}{t}$

P = 161.9638/4.7 = 34.46 W

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3 years ago

A superhero swings a magic hammer over her head in a horizontal plane. The end of the hammer moves around a circular path of rad

Korolek [52]

Answer:

9.21954 m/s

54 m/s²

Angle is zero

Explanation:

r = Radius of arm = 1.5 m

$\omega$ = Angular velocity = 6 rad/s

The horizontal component of speed is given by

$v_h=\omega r\\\Rightarrow v_h=6\times 1.5\\\Rightarrow v_h=9\ m/s$

The vertical component of speed is given by

$v_v=2\ m/s$

The resultant of the two components will give us the velocity of hammer with respect to the ground

$v=\sqrt{v_h^2+v_v^2}\\\Rightarrow v=\sqrt{9^2+2^2}\\\Rightarrow v=9.21954\ m/s$

The velocity of hammer relative to the ground is 9.21954 m/s

Acceleration in the vertical component is zero

Net acceleration is given by

$a_n=a_h=\omega^2r\\\Rightarrow a_n=6^2\times 1.5\\\Rightarrow a_n=54\ m/s^2$

Net acceleration is 54 m/s²

As the acceleration is towards the center the angle is zero.

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3 years ago