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2 years ago

What are the three elements that magnets are made of?

Physics

1 answer:

tatuchka [14]2 years ago

4 0

Answer:

iron (Fe), cobalt

(Co), and nickel (Ni) .

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Briefly explain why arterial injuries are more dangerous than damage to veins

MArishka [77]

Answer:

Injury to a vein increases the risk of forming a blood clot.

Explanation:

hoped this helped

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3 years ago

This glass of lemonade is sitting in the hot summer sun. As time passes, in which direction will heat transfer take place?

Ad libitum [116K]

Answer:

ice → lemonade it is the correct answer of this question

4 0

3 years ago

Plssss help Do you guys have any other more interesting, funny or creative ideas of incline problems than something sliding down

VashaNatasha [74]

U can always just do the classic roller coaster going up an incline and create some sort of story from that.

8 0

3 years ago

Can you answer the b. thanks

levacccp [35]

Answer:

current going into a junction in a circuit is EQUAL TO the current comming out of the junction.

Explanation:

Krichhoff's Current Law

Kirchhoff's current law (1st Law) states that current flowing into a node (or a junction) must be equal to current flowing out of it.

4 0

2 years ago

A hollow cylinder with an inner radius of 5 mm and an outer radius of 26 mm conducts a 4-A current flowing parallel to the axis

bearhunter [10]

Answer:

B = 38.2μT

Explanation:

By the Ampere's law you have that the magnetic field generated by a current, in a wire, is given by:

$B=\frac{\mu_o I_r}{2\pi r}$ (1)

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

r: distance from the center of the cylinder, in which B is calculated

Ir: current for the distance r

In this case, you first calculate the current Ir, by using the following relation:

$I_r=JA_r$

J: current density

Ar: cross sectional area for r in the hollow cylinder

Ar is given by $A_r=\pi(r^2-R_1^2)$

The current density is given by the total area and the total current:

$J=\frac{I_T}{A_T}=\frac{I_T}{\pi(R_2^2-R_1^2)}$

R2: outer radius = 26mm = 26*10^-3 m

R1: inner radius = 5 mm = 5*10^-3 m

IT: total current = 4 A

Then, the current in the wire for a distance r is:

$I_r=JA_r=\frac{I_T}{\pi(R_2^2-R_1^2)}\pi(r^2-R_1^2)\\\\I_r=I_T\frac{r^2-R_1^2}{R_2^2-R_1^2}$ (2)

You replace the last result of equation (2) into the equation (1):

$B=\frac{\mu_oI_T}{2\pi r}(\frac{r^2-R_1^2}{R_2^2-R_1^2})$

Finally. you replace the values of all parameters:

$B=\frac{(4\pi*10^{-7}T/A)(4A)}{2\PI (12*10^{-3}m)}(\frac{(12*10^{-3})^2-(5*10^{-3}m)^2}{(26*10^{-3}m)^2-(5*10^{-3}m)^2})\\\\B=3.82*10^{-5}T=38.2\mu T$

hence, the magnitude of the magnetic field at a point 12 mm from the center of the hollow cylinder, is 38.2μT

8 0

2 years ago