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2 years ago

What causes a material to be classified as ferromagnetic?

Physics

1 answer:

denpristay [2]2 years ago

4 0

Answer:

A material displaying ferromagnetism, such as the various forms of iron, steel, cobalt, nickel, and their alloys.

Explanation:

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23. Two spheres, with radii of R, are in contact with each other and attract each other with a force of F. If the radii of both

Mama L [17]

Answer:

Drop by a factor of 64

Explanation:

By Newtons law of gravity,

$F=\frac{Gm₁m₂}{r^{2} }$ where;

G = Universal gravitaional constant

r = distance between center of gravities of two objects

m₁,m₂ = masses of thee objects.

F = Gravitational force.

m = $\frac{4}{3}π r³ρ$ (mass = volume into density.

So when radius is halved, mass drops by a factor of 8,

m' = $\frac{4}{3}π (r/2)³ρ$

= $\frac{1}{6}π r³ρ$

So in substitution to the equation,

$F'=\frac{G(m₁/8)(m₂/8)}{r^{2} }$

$F' = F/64$

6 0

3 years ago

Which of the following is not an essential feature of scientific explanations?

sineoko [7]

Answer:

D I think I might be wrong its been a while scense I did something like that

4 0

2 years ago

A train 400 m long is moving on a straight track with a speed of 81.4 km/h. The engineer applies the brakes at a crossing, and l

Luda [366]

Answer:

86.5m

Explanation:

first convert km/h

then

81.4*1000/60*60=22.6

17.6*1000/60*60=4.89

then, x1/t1=x2/t2

we get

x2=400*4.89/22.6=86.5//

3 0

2 years ago

____ [38]

Answer:

Explanation:

D = 8.27 m ⇒ R = D / 2 = 8.27 m / 2 = 4.135 m

ω = 0.66 rev/sec = (0.66 rev/sec)*(2π rad/1 rev) = 4.1469 rad/s

We can apply the equation

Ff = W ⇒ μ*N = m*g (I)

then we have

N = Fc = m*ac = m*(ω²*R)

Returning to the equation I

μ*N = m*g ⇒ μ*m*ω²*R = m*g ⇒ μ = g / (ω²*R)

Finally

μ = (9.81 m/s²) / ((4.1469 rad/s)²*4.135 m) = 0.1379

3 0

2 years ago

Given that an electric field of 3×106V/m3×106V/m is required to produce an electrical spark within a volume of air, estimate the

Andre45 [30]

Answer:

Length, l = 33.4 m

Explanation:

Given that,

Electrical field, $E=3\times 10^6\ V/m$

Let the electrical potential is, $V=10^8\ V$

We need to find the length of a thundercloud lightning bolt. The relation between electric field and the electric potential is given by :

$V=E\times d\\\\d=\dfrac{V}{E}\\\\d=\dfrac{10^8}{3\times 10^6}\\\\d=33.4\ m$

So, the length of a thundercloud lightning bolt is 33.4 meters. Hence, this is the required solution.

5 0

3 years ago