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4 years ago

This diagram shows the magnetic field lines near the ends of two magnets. There is an error in the diagram.

Physics

2 answers:

ss7ja [257]4 years ago

8 0

Answer:

C-Changing the S to N

Explanation:

As we know that magnetic field lines originated from North pole of magnet and then terminate into the south pole of magnet.

Now if we put similar poles of magnet in front of each other then we can say that the magnetic field lines of two poles will repel each other and due to this two similar poles have repulsion force.

Similarly if two different poles of magnet will place in front of each other then the magnetic field lines will originate from north pole and terminate at south pole.

This will show attraction force between the poles of magnet.

Now we can see the figure that magnetic field lines are originating from both poles here as well as these lines are repelling each other.

So here both poles must have North pole

So correction here must be

C)-Changing the S to N

hoa [83]4 years ago

4 0

I believe the answer is c

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An emf of 22.0 mV is induced in a 519-turn coil when the current is changing at the rate of 10.0 A/s. What is the magnetic flux

zhenek [66]

Answer:

$\phi=1.56\times 10^{-5}\ Wb$

Explanation:

Given that,

Emf, V = 22 mV

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Rate of change of current is 10 A/s.

We need to find the magnetic flux through each turn of the coil at an instant when the current is 3.70 A.

Let's find the inductance first. So,

$L=\dfrac{\epsilon}{(dI/dt)}\\\\L=\dfrac{0.022}{10}\\\\L=0.0022\ H$

We have,

$L=\dfrac{N\phi}{I}$ , $\phi$ is magnetic flux

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So, the magnetic flux is $1.56\times 10^{-5}\ Wb$ .

8 0

3 years ago

A cat dozes on a stationary merry-go-round, at a radius of 4.4 m from the center of the ride. The operator turns on the ride and

monitta

Answer:

The coefficient of static friction is 0.29

Explanation:

Given that,

Radius of the merry-go-round, r = 4.4 m

The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 7.7 s.

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v is the speed of cat, $v=\dfrac{2\pi r}{t}$

$\mu=\dfrac{4\pi^2r}{gt^2}\\\\\mu=\dfrac{4\pi^2\times 4.4}{9.8\times (7.7)^2}\\\\\mu=0.29$

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3 years ago

When the cylinder is displaced slightly along its vertical axis it will oscillate about its equilibrium position with a frequenc

Nesterboy [21]

Answer:

w = √[g /L (½ r²/L2 + 2/3 ) ]

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Explanation:

We can simulate this system as a physical pendulum, which is a pendulum with a distributed mass, in this case the angular velocity is

w² = mg d / I

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d = L / 2

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I = ¼ m r2 + ⅓ m L2

I = m (¼ r2 + ⅓ L2)

now let's use the concept of density to calculate the mass of the system

ρ = m / V

m = ρ V

the volume of a cylinder is

V = π r² L

m = ρ π r² L

let's substitute

w² = m g (L / 2) / m (¼ r² + ⅓ L²)

w² = g L / (½ r² + 2/3 L²)

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Vsevolod [243]

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Leviafan [203]

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3 years ago

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