All categories

3 years ago

What is the requirement for a core to be used in an electromagnet?

Physics

1 answer:

xxMikexx [17]3 years ago

8 0

D. It must be able to be magnetized

Explanation:

The requirement for a core to be used in an electromagnet is that it must be able to be magnetized.

An electromagnet is a type of magnet produced by electricity.

They typically run and produce magnetic fields in the vicinity of electrical currents.
These magnets are not permanent magnets.
When the electrical current is removed, the magnetic property of the substance is lost.
The core of an electromagnet is usually made up of a material that can easily be magnetized in the presence of magnetic fields.
Some of the materials used are cobalt, iron, nickel.

learn more:

Electromagnet brainly.com/question/2191993

#learnwithBrainly

You might be interested in

A car accelerates from 30 ft/s to 90 ft/ sec in 6 seconds

Softa [21]

Its acceleration is 10ft/ sec

7 0

3 years ago

Two obiect accumulated a charge of

tamaranim1 [39]

Answer:

A. 181.24 N

Explanation:

The magnitude of hte electrostatic force between two charged objects is given by the equation

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1, q2 are the magnitudes of the two charges

r is the separation between the charges

In this problem, we have:

$q_1=4.5\mu C=4.5\cdot 10^{-6}C$ is the magnitude of the 1st charge

$q_2=2.8\mu C=2.8\cdot 10^{-6}C$ is the magnitude of the 2nd charge

r = 2.5 cm = 0.025 m is the separation between the charges

Therefore, the magnitude of the electric force is:

$F=\frac{(9\cdot 10^9)(4.5\cdot 10^{-6})(2.8\cdot 10^{-6})}{(0.025)^2}=181.44 N$

So, the closest answer is

A) 181.24 N

3 0

3 years ago

A 70- kg bicycle rides his 9.8- kg bicycle with a speed of 16 m/ s. What is the magnitude of the braking force of the bicycle co

Rus_ich [418]

Answer:

F = -319.2 N

Explanation:

Given that,

The mass of a bicyclist, m = 70 kg

Mass of the bicycle = 9.8 kg

The speed of a bicycle, v = 16 m/s

We need to find the magnitude of the braking force of the bicycle come to rest in 4.0 m.

The braking force is given by :

$F=ma\\\\=\dfrac{m(v-u)}{t}\\\\=\dfrac{(70+9.8)(0-16)}{4}\\\\=-319.2\ N$

So, the required force is 319.2 N.

3 0

2 years ago

Density of iron in CGS unit is 7.8 g/cm3. Its density is SI is

kumpel [21]

Answer:

7800kg/m³

Explanation:

Density of iron in CGS unit is 7.8 g/cm3. Its density is SI is

Given the density of iron = 7.8 g/cm3.

The SI units must be in kg/m³

7.8g = 7.8/1000 kg

7.8g = 0.0078kg

1cm³ = 0.000001m³

7.8g/cm³

= 0.0078/0.000001 kg/m³

= 7800kg/m³

Hence the density in SI unit is 7800kg/m³

4 0

3 years ago

The drawing shows a large cube (mass = 28.6 kg) being accelerated across a horizontal frictionless surface by a horizontal force

MrRissso [65]

Answer:

P= 454.11 N

Explanation:

Since P is the only horizontal force acting on the system, it can be defined as the product of the acceleration by the total mass of the system (both cubes).

$P= (M+m)*a\\a = \frac{P}{28.6 +4.3}\\a = \frac{P}{32.9}$

The friction force between both cubes (F) is defined as the normal force acting on the smaller cube multiplied by the coefficient of static friction. Since both cubes are subject to the same acceleration:

$F = m * a*\mu \\F= 4.3*0.710*\frac{P}{32.9}\\F=3.053*\frac{P}{32.9}$

In order for the small cube to not slide down, the friction force must equal the weight of the small cube:

$3.053*\frac{P}{32.9} = 4.3 * g\\\\P = \frac{4.3*9.8*32.9}{3.053} \\P= 454.11 N$

The smallest magnitude that P can have in order to keep the small cube from sliding downward is 454.11 N

8 0

3 years ago