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11111nata11111 [884]

2 years ago

8

Suppose the sphere is positively charged. Is it attracted to, repelled by, or not affected by the magnet?.

1 answer:

Nutka1998 [239]2 years ago

4 0

When a positively charged sphere is brought near the north pole of a magnet, the positively charged sphere will be attracted to the magnet.

<h3>Positively charged object</h3>

When a positively charged object is brought near the north pole of a magnet, the positively charged object will be attracted to the magnet beacuse of polarity.

Positively charged metals have the tendency to show the polarization of charges.

Thus, when a positively charged sphere is brought near the north pole of a magnet, the positively charged sphere will be attracted to the magnet. Also, if the south pole is brought near the sphere, the positively charged sphere will repel the magnet.

Learn more about attraction of magnet here: brainly.com/question/14749231

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Raindrops fall vertically at 7.5 m/s relative to the Earth. What does an observer in a car moving at 20.2 m/s in a straight line

Vilka [71]

Answer:

vDP = 21.7454 m/s

θ = 200.3693°

Explanation:

Given

vDE = 7.5 m/s

vPE = 20.2 m/s

Required: vDP

Assume that

vDE to be in direction of - j

vPE to be in direction of i

According to relative motion concept the velocity vDP is given by

vDP = vDE - vPE (I)

Substitute in (I) to get that

vDP = - 7.5 j - 20.2 i

The magnitude of vDP is given by

vDP = √((- 7.5)²+(- 20.2)²) m/s = 21.7454 m/s

θ = Arctan (- 7.5/- 20.2) = 20.3693°

θ is in 3rd quadrant so add 180°

θ = 20.3693° + 180° = 200.3693°

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3 years ago

4. A car accelerates, from rest, at 2.4 m/s?. How fast is the car traveling

pantera1 [17]

Answer:

19.2m/s

Explanation:

Assuming that 2.4m/s^2 was the acceleration and not a typo, we can use the equation v=at, where v=velocity, a=acceleration, and t=time,

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v=2.4*8

v=19.2m/s

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10. How far does a transverse pulse travel in 1.23 ms on a string with a density of 5.47 × 10−3 kg/m under tension of 47.8 ?????

KATRIN_1 [288]

Answer: Tension = 47.8N, Δx = 11.5× $10^{-6}$ m.

Tension = 95.6N, Δx = 15.4× $10^{-5}$ m

Explanation: A speed of wave on a string under a tension force can be calculated as:

$|v| = \sqrt{\frac{F_{T}}{\mu} }$

$F_{T}$ is tension force (N)

μ is linear density (kg/m)

Determining velocity:

$|v| = \sqrt{\frac{47.8}{5.47.10^{-3}} }$

$|v| = \sqrt{0.00874 }$

$|v| =$ 0.0935 m/s

The displacement a pulse traveled in 1.23ms:

$\Delta x = |v|.t$

$\Delta x = 9.35.10^{-2}*1.23.10^{-3}$

Δx = 11.5× $10^{-6}$

With tension of 47.8N, a pulse will travel Δx = 11.5× $10^{-6}$ m.

Doubling Tension:

$|v| = \sqrt{\frac{2*47.8}{5.47.10^{-3}} }$

$|v| = \sqrt{2.0.00874 }$

$|v| = \sqrt{0.01568}$

|v| = 0.1252 m/s

Displacement for same time:

$\Delta x = |v|.t$

$\Delta x = 12.52.10^{-2}*1.23.10^{-3}$

$\Delta x =$ 15.4× $10^{-5}$

With doubled tension, it travels $\Delta x =$ 15.4× $10^{-5}$ m

4 0

3 years ago