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3 years ago

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

Physics

2 answers:

likoan [24]3 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]3 years ago

3 0

the answer is a because they start on the north side

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Two spaceships leave Earth in opposite directions, each with a speed of 0.60c with respect to Earth. (a) What is the SPEED of sp

Studentka2010 [4]

Answer:

The relative speed of 1 relative to 2 is 0.88c

Explanation:

In relativistic mechanics the relative speed between 2 objects moving in different direction is given by

$v_{ab}=\frac{v_{a}+v_{b}}{1+\frac{v_{a}v_{b}}{c^{2}}}$

Since it is given that

$v_{a}=0.6c\\\\v_{b}=0.6c$

Applying values in the formula we get

$v_{ab}=\frac{0.6c+0.6c}{1+\frac{(0.6c)^{2}}{c^{2}}}\\\\v_{ab}=0.88c$

8 0

3 years ago

Can you find the magnetic field based on force? a straight segment of wire 35.0 cm long carrying a current of 1.40 a is in a uni

Lostsunrise [7]

The force exerted by a magnetic field on a wire carrying current is:
$F=ILB \sin \theta$
where I is the current, L the length of the wire, B the magnetic field intensity, and $\theta$ the angle between the wire and the direction of B.

In our problem, the force is F=0.20 N. The current is I=1.40 A, while the length of the wire is L=35.0 cm=0.35 m. The angle between the wire and the magnetic field is $53 ^{\circ}$ , so we can re-arrange the formula and substitute the numbers to find B:
$B= \frac{F}{IL \sin \theta}= \frac{0.20 N}{(1.40 A)(0.35 m)(\sin 53^{\circ})}=0.51 T$

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3 years ago

A battery-operated car utilizes a 120.0 V battery with negligible internal resistance. Find the charge, in coulombs, the batteri

fgiga [73]

Answer:

4.29×10⁵ C

Explanation:

From the question,

The energy stored in the battery = Kinetic energy of the car+ Energy needed to make the car climbed the hill+Energy required to exert a force.

E = 1/2mv²+mgh+Fd.................... Equation 1

Where E = Energy stored in the battery, m = mass of the car, v = velocity of the car, h = height of the hill, F = force exerted on the car, d = distance traveled by the car.

But,

d = vt.................... Equation 2

Where v = velocity, t = time.

Substitute equation 2 into equation 1

E = 1/2mv²+mgh+F(vt)................... Equation 3

Given: m = 770 kg, v = 26 m/s, h = 2.15×10² m = 215 m, F = 5.3×10² N = 530 N, t = 1 hour = 3600 s, g = 9.8 m/s²

Substitute into equation 1

E = 1/2(770)(26²)+(770)(9.8)(215)+(530)(26)(3600)

E = 260260+1622390+49608000

E = 51490650 J

Using,

E = qV................. Equation 4

Where q = charge of the battery, V = Voltage.

make q the subject of the equation

q = E/V............... Equation 5

Given: E = 51490650 J, V = 120 V

Substitute into equation 5

q = 51490650/120

q = 429088.75 C

q = 4.29×10⁵ C

7 0

3 years ago

Jane walked 5.00 meters on a road that inclines 13.0 degrees. How much distance did she cover horizontally?

Paladinen [302]

Any options available?

8 0

3 years ago

Two disks of identical mass but different radii (r and 2r) are spinning on frictionless bearings at the same angular speed ?0, b

cestrela7 [59]

Answer:

Explanation:

Moment of inertia of a disc = 1/2 M R²

Since mass is same for both and radius are r and 2r, their moment of inertia can be in the ratio of 1: 4 . Let them be I and 4I . Angular speed are ω₀ and - ω₀ .

We shall apply law of conservation of angular momentum .

initial total angular momentum

I x ω₀ - 4I x ω₀ = - 3Iω₀

Let final common angular momentum be ω

total final angular momentum = ( I + 4I ) ω

Applying law of conservation of angular momentum

( I + 4I ) ω = - 3Iω₀

ω = - 3 / 5 ω₀ .

b )

Initial total rotational K E

= 1/2 I ω₀² + 1/2 4I ω₀²

= 1/2 x5I ω₀²

Final total rotational K E

= 1/2 ( I + 4I ) ( - 3 / 5 ω₀ )²

= 1/2 x 9 / 5 I ω₀²

= 9 / 10I ω₀²

change in rotational kinetic energy = 9 / 10I ω₀² - 1/2 x5I ω₀²

(9/10 - 5/2) xI ω₀²

=( .9 - 2.5 )I ω₀²

= - 1.6 I ω₀² Ans

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3 years ago