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Sergio039 [100]
3 years ago
12

Which chart correctly identifies the properties of electric and magnetic fields?

Physics
1 answer:
bazaltina [42]3 years ago
6 0
The correct answer for the question that is being presented above is this one: "D.
Property Electric Field Magnetic Field
<span>Can be produced by moving electric charge X </span>
Have two sides with opposite characteristics X X
Can make metals, such as nickel, iron, and cobalt, into magnets X X
<span>Can be turned on or off with a switch X"</span>
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Two long, parallel, current-carrying wires lie in an xy-plane. The first wire lies on the line y = 0.300 m and carries a current
Marrrta [24]

Answer:

Explanation:

Since the wires attract each other , the direction of current will be same in both the wires .

Let I be current in wire which is along x - axis

force of attraction per unit length between the two current carrying wire is given by

\frac{\mu_0}{4\pi}  x \frac{2 I_1\times I_2}{d}

where I₁ and I₂ are currents in the wires and d is distance between the two

Putting the given values

285 x 10⁻⁶ = 10⁻⁷ x \frac{2\times25.5\times I_2}{.3}

I₂ = 16.76 A

Current in the wire along x axis is 16.76 A

To find point where magnetic field is zero due the these wires

The point will lie between the two wires  as current is in the same direction.

Let at y = y , the neutral point lies

k 2 x  \frac{16.76}{y} = k 2 x \frac{25.5}{.3-y}

25.5y = 16.76 x .3 - 16.76y

42.26 y = 5.028

y = .119

= .12 m

3 0
3 years ago
Do anyone k ow the answers?
OLga [1]
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7 0
3 years ago
Atoms in a molecule held together through sharing ______. (points 3)
Wittaler [7]
Covalent and ionic bonds
8 0
3 years ago
Read 2 more answers
Two teams are playing tug of war. Team A pulls to the right with a force of 450 N. Team B pulls to the left with a force of 415
motikmotik

Answer:

35 N to the right.

Explanation:

450 is going to the right so you subtract what is going against it. Which gives you 35. And because 450 is bigger than 415, it'll be going to the right.

6 0
3 years ago
A plane travels at a speed of 205mph in still air. Flying with a tailwind, the plane is clocked over a distance of 1000 miles. F
vaieri [72.5K]

While plane is moving under tailwind condition it took time "t"

so here we will have

t = \frac{d}{v_{net}}

here net speed of the plane will be given as

v_{net} = v + v_w

t = \frac{1000}{205 + v_w}

similarly when it moves under the condition of headwind its net speed is given as

v_{net} = v - v_w

now time taken to cover the distance is 2 hours more

t + 2 = \frac{1000}{205 - v_w}

now solving two equations

\frac{1000}{205 + v_w} + 2 = \frac{1000}{205 - v_w}

solving above for v_w we got

v_w = 40.4 mph

6 0
3 years ago
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