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3 years ago

How much farther away from a current-carrying wire would you

1 answer:

4 0

Answer: Your question is open ended therefore i will provide a general answer based on the general concept of your question.

Answer : At half distance between the center of the Magnetic field the magnetic pole the effect of the magnetic field generated will be half as strong

Explanation:

Magnetic force ∝ 1 / d^2

This simply means increase in the squared distance of an object leads to the decrease in the effect of the magnetic field generated by a current carrying wire.

Hence ;

At half distance between the center of the Magnetic field the magnetic pole, the effect of the magnetic field generated will be half as strong

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Randomly assigning students to two different groups in an experiment on

Mars2501 [29]

Answer:c

Explanation:

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4 years ago

Read 2 more answers

A dinner plate falls vertically to the floor and breaks up into three pieces, which slide horizontally along the floor. Immediat

N76 [4]

Answer:

p_k=\sqrt{p_x^2+p_y^2}}

Explanation:

Apply the momentum in each direction knowing that the impact is at the same time for the pieces so

$p_x=m_1*v_1$

$p_x=200g*2.0m/s=0.4kgm/s$

$p_y=m_2*v_2$

$p_y=235g*1.5m/s=0.3525kgm/s$

So the momentum in the other piece can be find knowing that

$p_x^2+p_y^2=p_k^2$

So:

$p_k=\sqrt{p_x^2+p_y^2}}$

$p_k=\sqrt{0.4^2+0.3525^2}}=\sqrt{0.2842 kg^2*m^2/s^2}$

$p_k=0.5331kg*m/s$

To find the velocity knowing the mass

$p_k=m_k*v_k$

$v_k=\frac{p_k}{m_k}=\frac{0.5331 kg*m/s}{0.10kg}$

$v_k=5.331 m/s$

4 0

3 years ago

When a ball increases in speed by the same amount each second its acceleration?

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Its acceleration is constant.

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4 years ago

At a beach the light is generallypartially polarized owing to reflections off sand and water. At a particular beach on a particu

Ne4ueva [31]

Answer:

a) 0.159

b) 0.84

Explanation:

The Horizontal component is 2.3 times the vertical component

Let the horizontal electric field component = $E_{h}$

Let the vertical electric field component = $E_{v}$

The formula for light intensity is given by:

$I = \frac{E_{m} ^{2} }{2c \mu}$ ..............................(1)

$E_{m}$ is the resolution of the vertical and horizontal components, $E_{h} and E_{v}$

$E_{m} ^{2} = E_{h} ^{2} + E_{v} ^{2}$ ..................(2)

Light intensity before the glasses were put on:

$I_{1} = \frac{E_{m} ^{2} }{2c \mu_{1} }$ .............................(3)

Put equation (2) into equation (3)

$I_{1} = \frac{E_{h} ^{2} + E_{v} ^{2}}{2c \mu_{1} }$ .............................(4)

After the glasses were put on the horizontal component vanishes, i.e. $E_{h} = 0$

$I_{2} = \frac{ E_{v} ^{2}}{2c \mu_{2} }$ ...................................(5)

Divide equation (5) by equation (4)

$\frac{I_{2} }{I_{1} } = \frac{E_{v} ^{2} }{E_{h} ^{2} + E_{v} ^{2}}$ ...............................(6)

But $E_{h} = 2.3E_{v}$ ......................(7)

Insert equation (7) into (6)

$\frac{I_{2} }{I_{1} } = \frac{E_{v}^{2} }{(2.3E_{v})^{2} + E_{v} ^{2} } \\\frac{I_{2} }{I_{1} } = \frac{E_{v}^{2} }{5.29E_{v}^{2} + E_{v} ^{2} }\\\frac{I_{2} }{I_{1} } = \frac{E_{v}^{2} }{6.29E_{v}^{2} }\\\frac{I_{2} }{I_{1} } =\frac{1}{6.29} \\$