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3 years ago

Why is soft iron used for the core of an electromagnet?

Physics

2 answers:

aliina [53]3 years ago

5 0

Answer:

Because of the high permeability and high saturation level of holding magnetism.

Explanation:

Soft iron always used in the core of an electromagnet because of the following reason which are discussed below.

Permeability: These magnets are highly permeable means they can gain magnetism when current is passes through them and instantly loose magnetism after removal of current.
High saturation level: The property of these soft magnets are they have the high value of saturation means they can hold high level of magnetism without easily saturating.

Tanzania [10]3 years ago

3 0

The soft iron inside the coil makes the magnetic field stronger because it becomes a magnet itself when the current is flowing. Soft iron is used because it loses its magne

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What type of convergent boundary is the himalayan mountains formed by

Diano4ka-milaya [45]

The Himalayan Mountains formed at a convergence plate boundary between the Eurasian plate and the Indian plate.

8 0

3 years ago

.A coin rolls off the edge of a table. The coin

geniusboy [140]

Answer:

Apply the following formulae horizontally And get A value for time

Remember horizontal acceleration is zero

$s = ut + \frac{1}{2}a {t}^{2} \\ 0.8 = 1.7 \times t \\ \frac{0.8}{1.7} = t \\ t = 0.47s$

and then to find the height apply the same above equation vertically...remember vertical initial velocity is zero

$s = ut + \frac{1}{2} a {t}^{2} \\ s = \frac{1}{2} \times 10 \times (0.47) ^{2} \\ s = 1.1045m$

5 0

2 years ago

What magnification would be obtained if an eyepiece with a focal length of 0.38 m was placed on telescope?

weqwewe [10]

Answer:

This question is incomplete

Explanation:

This question is incomplete because the telescope's focal length was not provided. The formula to be used here is

Magnification = telescope's focal length/eyepiece's focal length

The eyepiece's focal length was provided in the question as 0.38 m.

NOTE: Magnification can be described as the power of an instrument (in this case telescope) to enlarge an object. It has no unit and thus the two focal lengths mentioned in the formula above must be in the same unit (preferably meters since one of them is in meters already).

7 0

2 years ago

A student of weight 652 N rides a steadily rotating Ferris wheel (the student sits upright). At the highest point, the magnitude

miskamm [114]

Answer:

Explanation:

Given

Weight of person $W=652\ N$

At highest point Magnitude of the normal force $F=585\ N$

net force at highest point

$F_{net}=W+F_c$

where $F_c=$ centripetal force

$F_c=\dfrac{mv^2}{r}$

$F_{net}=F_{n}=$ Normal Force

$585=652+F_c$

$F_c=-67\ N$

Negative sign shows force is in upward direction

At bottom point centripetal force is towards the bottom

$F_n=F_c+W$

$F_n=652+67$

$F_n=719\ N$

8 0

3 years ago

Two Earth satellites, A and B, each of mass m, are to be launched into circular orbits about Earth's center. Satellite A is to o

Pachacha [2.7K]

(a) 0.448

The gravitational potential energy of a satellite in orbit is given by:

$U=-\frac{GMm}{r}$

where

G is the gravitational constant

M is the Earth's mass

m is the satellite's mass

r is the distance of the satellite from the Earth's centre, which is sum of the Earth's radius (R) and the altitude of the satellite (h):

r = R + h

We can therefore write the ratio between the potentially energy of satellite B to that of satellite A as

$\frac{U_B}{U_A}=\frac{-\frac{GMm}{R+h_B}}{-\frac{GMm}{R+h_A}}=\frac{R+h_A}{R+h_B}$

and so, substituting:

$R=6370 km\\h_A = 5970 km\\h_B = 21200 km$

We find

$\frac{U_B}{U_A}=\frac{6370 km+5970 km}{6370 km+21200 km}=0.448$

(b) 0.448

The kinetic energy of a satellite in orbit around the Earth is given by

$K=\frac{1}{2}\frac{GMm}{r}$

So, the ratio between the two kinetic energies is

$\frac{K_B}{K_A}=\frac{\frac{1}{2}\frac{GMm}{R+h_B}}{\frac{1}{2}\frac{GMm}{R+h_A}}=\frac{R+h_A}{R+h_B}$

Which is exactly identical to the ratio of the potential energies. Therefore, this ratio is also equal to 0.448.

(c) B

The total energy of a satellite is given by the sum of the potential energy and the kinetic energy:

$E=U+K=-\frac{GMm}{R+h}+\frac{1}{2}\frac{GMm}{R+h}=-\frac{1}{2}\frac{GMm}{R+h}$

For satellite A, we have

$E_A=-\frac{1}{2}\frac{GMm}{R+h_A}=-\frac{1}{2}\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)(28.8 kg)}{6.37\cdot 10^6 m+5.97\cdot 10^6 m}=-4.65\cdot 10^8 J$

For satellite B, we have

$E_B=-\frac{1}{2}\frac{GMm}{R+h_B}=-\frac{1}{2}\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)(28.8 kg)}{6.37\cdot 10^6 m+21.2\cdot 10^6 m}=-2.08\cdot 10^8 J$

So, satellite B has the greater total energy (since the energy is negative).

(d) $-2.57\cdot 10^8 J$

The difference between the energy of the two satellites is:

$E_B-E_A=-2.08\cdot 10^8 J-(-4.65\cdot 10^8 J)=-2.57\cdot 10^8 J$

4 0

3 years ago