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3 years ago

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An unknown additional charge q3q3q_3 is now placed at point B, located at coordinates (0 mm, 15.0 mm ). Find the magnitude and s

ign of q3q3q_3 needed to make the total electric field at point A equal to zero.

1 answer:

Yanka [14]3 years ago

7 0

Answer:

0.3nanocouloumb

Explanation:

Pls see attached file

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Which of the following would most likely happen if water did not form hydrogen bonds?

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Option C is the correct answer

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3 years ago

Read 2 more answers

A green light is submerged 2.70 m beneath the surface of a liquid with an index of refraction 1.31. What is the radius of the ci

Anettt [7]

Answer:

The radius is $r = 3.1905 \ m$

Explanation:

From the question we are told that

The distance beneath the liquid is $d = 2.70 \ m$

The refractive index of the liquid is $n_i = 1.31$

Now the critical value is mathematically represented as

$\theta = sin ^{-1} [\frac{1}{n_i} ]$

substituting values

$\theta = sin ^{-1} [\frac{1}{131} ]$

$\theta = 49.76^o$

Using SOHCAHTOA rule we have that

$tan \theta = \frac{ r}{d}$

=> $r = d * tan \theta$

substituting values

$r = 2.7 * tan (49.76)$

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5 0

3 years ago

What is the weight of a box with a mass of 150 kg on Earth?

Dmitry [639]

Answer:

W = M g = 150 kg * 9.81 m/s^2 = 1470 N

You were only given 3 significant figures in the question.

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2 years ago

A 1.55-kg object hangs in equilibrium at the end of a rope (taken as massless) while a wind pushes the object with a 13.3-n hori

yarga [219]

<span>To answer this problem, we use balancing of forces: x and y components to determine the tension of the rope.

First, the vertical component of tension (Tsin theta) is equal to the weight of the object.
T * sin θ = mg =</span> 1.55 * 9.81 <span>
T * sin θ = 15.2055

Second, the horizontal component of tension (t cos theta) is equal to the force of the wind.
T * cos θ = 13.3

Tan θ = sin </span>θ / cos θ = 15.2055/13.3 = 1.143
we can find θ that is equal to 48.82.

T then is equal to 20.20 N

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3 years ago