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3 years ago

B) Explain the method of preparing electromagnet. How do you test the

Physics

1 answer:

kap26 [50]3 years ago

5 0

Answer:

An electromagnet is made by forming a coil around a soft iron bar (known here as the metal) such as a nail or screw and connect with an insulated copper wire (known here as the electric current conductor) the ends of the wound copper is then connected separately to the positive and negative terminals of a battery (known here as the source of electric current)

The north seeking needle of the magnetic compass will move away when brought close to the north pole of the formed electromagnet which can then be labelled N

The magnetic compass needle will be attracted to the south pole of the electromagnet which can then be labelled S

Explanation:

An electromagnet is an electric powered magnet that is formed (temporarily) by the perpendicular movement of electric current with respect to a metal core

The magnitude and the poles of an electromagnet can be changed by changing the magnitude and the direction of flow of the electric current respectively.

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If a girl does 1,350 Joules of work to pedal her bike using a force of 75 N, how far did she ride her bike?

shepuryov [24]

Answer:

D = 18m

Explanation:

We use this formula:

Work = Force * Distance

Now, we replace it:

W = F*D

1,350J = 75N * D

D = 1,350/75

D = 18m

Hope it was helpful ;)

7 0

2 years ago

An experiment is conducted on a long straight wire of diameter d. A constant current is sent through the wire and the magnetic f

Anton [14]

Answer:

$B_2 = \frac{B_1}{2}$

Explanation:

As per Ampere's law the magnetic field at the surface of the wire is given as

$\int B. dL = \mu_0 i$

here we have

$B . (\pi d) = \mu_0 i$

so we will have

$B_1 = \frac{\mu_0 i}{\pi d}$

now again we use same value of current but wire with double the diameter

so the magnetic field at the surface is given as

$B_2 = \frac{\mu_0 i}{2\pi d}$

so we have

$B_2 = \frac{B_1}{2}$

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3 years ago

12 POINTS PLEASE HELP.

bazaltina [42]

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3 years ago

A parallel-plate capacitor initially has a potential difference of 400 V and is then disconnected from the charging battery. If

Damm [24]

The capacitance of a capacitor is the ratio of the stored charge to its potential difference, i.e.

C = Q/ΔV

C is the capacitance

Q is the stored charge

ΔV is the potential difference

Rearrange the equation:

ΔV = Q/C

We also know the capacitance of a parallel-plate capacitor is given by:

C = κε₀A/d

C is the capacitance

κ is the capacitor's dielectric constant

ε₀ is the electric constant

A is the area of the plates

d is the plate separation

If we substitute C:

ΔV = Qd/(κε₀A)

We assume the stored charge and the area of the plates don't change. Then if we double the plate spacing, i.e. we double the value of d, then the potential difference ΔV is also doubled.

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2 years ago

The net force (unbalanced force) applied on an object is the product of_________and___________

RSB [31]

Answer:

of its mass and its acceleration

Explanation:

based on Newton's second law of motion

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1 year ago