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2 years ago

Which image represents the force on a positively charged particle caused by an approaching magnet?

Physics

1 answer:

Amanda [17]2 years ago

4 0

Answer:

Image B represents the force on a positively charged particle caused by an approaching magnet.

Explanation:

The most fundamental law of magnetism is that like shafts repulse each other and dissimilar to posts pull in one another; this can without much of a stretch be seen by endeavoring to put like posts of two magnets together. Further attractive impacts additionally exist. On the off chance that a bar magnet is cut into two pieces, the pieces become singular magnets with inverse shafts. Also, pounding, warming or winding of the magnets can demagnetize them, on the grounds that such dealing with separates the direct game plan of the particles. A last law of magnetism alludes to maintenance; a long bar magnet will hold its magnetism longer than a short bar magnet. The domain theory of magnetism expresses that every single enormous magnet involve littler attractive districts, or domains. The attractive character of domains originates from the nearness of significantly littler units, called dipoles. Iotas are masterminded in such a manner in many materials that the attractive direction of one electron counteracts the direction of another; in any case, ferromagnetic substances, for example, iron are unique. The nuclear cosmetics of these substances is with the end goal that littler gatherings of particles unite as one into zones called domains; in these, all the electrons have the equivalent attractive direction.

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A positively charged particle 1 is at the origin of a Cartesian coordinate system, and there are no other charged objects nearby

8090 [49]

Answer:

P=(2 nm, 8mn)

Explanation:

Given :

Position of positively charged particle at origin, $O=(0\ nm,0\ nm)$

Position of desired magnetic field, $D\equiv(1\ nm,8\ nm)$

Magnitude of desired magnetic field, $E=0\ N.C^{-1}$

Let q be the positive charge magnitude placed at origin.

<u>We know the distance between the two Cartesian points is given as:</u>

$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

<u>For the electric field effect to be zero at point D we need equal and opposite field at the point.</u>

$\frac{1}{4\pi.\epsilon_0} \times \frac{q}{r^2 } =\frac{1}{4\pi.\epsilon_0} \times \frac{q}{r^2 }$

$\therefore (1-0)^2+(8-0)^2=r^2$

$r^2=65\ nm$

$r=\sqrt{65}$

as we know that the electric field lines emerge radially outward of a positive charge so the second charge will be at equally opposite side of the given point.

assuming that the second charge is placed at (x,y) nano-meters.

Therefore,

$x=2\times 1=2\ nm$

and

$y=2\times 8=16\ nm$

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3 years ago

A garden hose attached with a nozzle is used to fill a 10‐ gal bucket. The inner diameter of the hose is 2 cm, and it reduces to

nadya68 [22]

Answer:

Explanation:

Given

Volume of bucket $V=10\ gallon$

Time taken to fill the bucket $t=50\ s$

so volume flow rate is $\dot{V}=\frac{10}{50}=0.2\ gal/s$

1 gal is equivalent to $0.133\ ft^3$

$\dot{V}=0.0267\ ft^3/s$

mass flow rate $\dot{m}=\rho \times \dot{V}$

$\dot{m}=62.4\times 0.0267$

$\dot{m}=1.668\ lbs$

(b)Average velocity through nozzle exit

$\dot{V}=Av_{avg}$

$v_{avg}=\dfrac{0.0267}{\frac{\pi}{4}\times (0.0262)^2}$

$v_{avg}=49.51\ ft/s$

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3 years ago

A bowling (mass = 7.2 kg, radius = 0.11 m) and a billiard ball (mass = 0.38 kg, radius = 0.028 m) may each be treated as uniform

cestrela7 [59]

Hope this helps you!

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2 years ago

A rock is thrown straight up. What is the net external force acting on the rock when it is at the top of its trajectory (ignorin

Nat2105 [25]

Answer:

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Explanation:

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andrew-mc [135]

Answer:

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