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4 years ago

Which figure represents the electric field lines of two negative point charges? Help

Physics

1 answer:

Yuliya22 [10]4 years ago

3 0

Answer:

The question is incomplete because the options are not given. We'll solve this question with the options.

There are two properties of negative charges which we have to consider to find the figure of its electrical field.

Firstly, for a negative charge, the electrical field lines are always directed radially outwards and they don't intersect.

Secondly, we know that similar charges repel each other, so there will be no electrical field present directly between these two negative charges. Electrical field line will be present between two charges only when there is a force of attraction.

Taking both of these facts into consideration, the electrical field line between two negatives is represented by the figure below.

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A ball filled with an unknown material starts from rest at the top of a 2 m high incline that makes a 28o with respect to the ho

Lady_Fox [76]

Answer:

Searching in google I found the total mass and the radius of the ball (m = 1.5 kg and r = 10 cm) which are needed to solve the problem!

The ball rotates 6.78 revolutions.

Explanation:

Searching in google I found the total mass and the radius of the ball (m = 1.5 kg and r = 10 cm) which are needed to solve the problem!

At the bottom the ball has the following angular speed:

$\omega_{f} = \frac{v_{f}}{r} = \frac{4.9 m/s}{0.10 m} = 49 rad/s$

Now, we need to find the distance traveled by the ball (L) by using θ=28° and h(height) = 2 m:

$sin(\theta) = \frac{h}{L} \rightarrow L = \frac{h}{sin(\theta)} = \frac{2 m}{sin(28)} = 4.26 m$

To find the revolutions we need the time, which can be found using the following equation:

$v_{f} = v_{0} + at$

$t = \frac{v_{f} - v_{0}}{a}$ (1)

So first, we need to find the acceleration:

$v_{f}^{2} = v_{0}^{2} + 2aL \rightarrow a = \frac{v_{f}^{2} - v_{0}^{2}}{2L}$ (2)

By entering equation (2) into (1) we have:

$t = \frac{v_{f} - v_{0}}{\frac{v_{f}^{2} - v_{0}^{2}}{2L}}$

Since it starts from rest (v₀ = 0):

$t = \frac{2L}{v_{f}} = \frac{2*4.26 m}{4.9 m/s} = 1.74 s$

Finally, we can find the revolutions:

$\theta_{f} = \frac{1}{2} \omega_{f}*t = \frac{1}{2}*49 rad/s*1.74 s = 42.63 rad*\frac{1 rev}{2\pi rad} = 6.78 rev$

Therefore, the ball rotates 6.78 revolutions.

I hope it helps you!

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3 years ago

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Answer:

The answer is "0.91238 and 744.8"

Explanation:

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$=0.076 \times 1.225 \times 9.8 \\\\ =0.91238 \ N\\\\$

$F_B \ in \ water = v \& water \ g \\\\$

$=0.076 \times 1000 \times 9.8\\\\= 744.8 \ N\\$

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Hey there,

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~Top

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