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Oliga [24]
2 years ago
8

A student moves a horseshoe magnet by hand across a tabletop to varying distances from an electromagnet, as shown in the diagram

above.
The student records observations of how much force it takes to hold the horseshoe magnet in place at each distance. Observations

• 20 cm: no force required
• 10 cm: small force required; easy to hold in place
• 5 cm: large force required; very difficult to hold in place

What do the student's observations demonstrate?

A. The electromagnet's magnetic field becomes stronger as more magnetic material is added to it.

B. The electromagnet's magnetic field is stronger as more turns are added to the wire on the nail.

C. The electromagnet's magnetic field becomes stronger as a small amount of strength is added to the system by the battery.

D. The electromagnet's magnetic field is stronger close to it than far from it.​

Physics
2 answers:
miv72 [106K]2 years ago
8 0

Answer: D

The electromagnet's magnetic field is stronger close to it than far from it.

Explanation:

Hope this helps!

Sloan [31]2 years ago
4 0

The closer together they are the harder it is to hold on to because the magnetic field is stronger as it gets closer.


Answer:

D. The electromagnet's magnetic field is stronger close to it than far from it.​

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Size and Direction

Explanation:

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SP 15The magnetic field in a region of space is measured to be:This field is known to be caused by a cluster of long-straight wi
tino4ka555 [31]

Answer:

 i = 0.477 10⁴ B

the current flows in the  counterclockwise

Explanation:

For this exercise let's use the Ampere law

                    ∫ B . ds = μ₀ I

Where the path is closed

Let's start by locating the current vines that are parallel to the z-axis, so it must be exterminated along the x-axis and as the specific direction is not indicated, suppose it extends along the y-axis.

From BiotSavart's law, the field must be perpendicular to the direction of the current, so the magnetic field must go in the x direction.

We apply the law of Ampere the segment parallel to the x-axis is the one that contributes to the integral, since the other two have an angle of 90º with the magnetic field

Segment on the y axis

        L₀ = (y2-y1)

        L₀ = 3-0 = 3 cm

Segment on the point x = 2 cm

        L₁ = 3-0

        L₁ = 3cm

       B L = μ₀ I

       B 2L = μ₀ I

        i = 2 L B /μ₀

        i= 2 0.03 / 4π 10⁻⁷   B

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The current is perpendicular to the magnetic field whereby the current flows in the  counterclockwise

8 0
3 years ago
Suppose that a charged particle of diameter 1.00 micrometer moves with constant speed in an electric field of magnitude 1.00×105
Dovator [93]
It's a bit of a trick question, had the same one on my homework. You're given an electric field strength (1*10^5 N/C for mine), a drag force (7.25*10^-11 N) and the critical info is that it's moving with constant velocity(the particle is in equilibrium/not accelerating). 
<span>All you need is F=(K*Q1*Q2)/r^2 </span>
<span>Just set F=the drag force and the electric field strength is (K*Q2)/r^2, plugging those values in gives you </span>
<span>(7.25*10^-11 N) = (1*10^5 N/C)*Q1 ---> Q1 = 7.25*10^-16 C </span>
3 0
3 years ago
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What factors might affect how fast a balloon falls to the ground?
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Answer:

air resistance, gravitational force

5 0
3 years ago
A horizontal spring with stiffness 0.4 N/m has a relaxed length of 11 cm (0.11 m). A mass of 21 grams (0.021 kg) is attached and
riadik2000 [5.3K]

Answer:

0.6983 m/s

Explanation:

k = spring constant of the spring = 0.4 N/m

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x = stretch in the spring = L - L₀ = 0.27 - 0.11 = 0.16 m

m = mass of the mass attached = 0.021 kg

v = speed of the mass

Using conservation of energy

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m v² = k x²

(0.021) v² = (0.4) (0.16)²

v = 0.6983 m/s

5 0
4 years ago
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