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Greeley [361]
3 years ago
12

you will get brainless if you win you get trapped in a house there is two doors one is a lion that hasn't eaten in 2 years the o

ther door has pirannas which will you pick
SAT
2 answers:
Bezzdna [24]3 years ago
5 0
You would pick the door with the piranhas, because the lion hasn’t eaten in two years so that lion, is hungrier than ever, and the piranha has eaten, hopefully, so that means the piranha is not very hungry so I would say you take the chance of the door with the piranha... I so hope this helps
navik [9.2K]3 years ago
3 0
The answer is the room with the lion. if the lion hasn’t eaten in 2 years it will be dead.
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What can be concluded from the audio? the prince believed the story to be true. the prince’s troops overtook the tower. the prin
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There are different kinds of audio message. What can be concluded from the audio is that  the prince’s troops overtook the tower.

<h3>What is an audio message?</h3>

There are a lot of audio messages that are often used in communications. Audio message entails speaking through a phone so as to send a voice message to one's friends or family.

One can also use a recorder to record messages.  The option that could be concluded from the audio is that  the prince’s troops overtook the tower.

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The drama states that Earhart was "commander" of the flight in name only. In what ways was Earhart as much in command of the Fli
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Which of the following is not part of Monroe’s Motivated Sequence?
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Soil contains many nutrients. How do nutrients get into the soil?
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A point charge q1 = -4. 00 nc is at the point x = 0. 60 m, y = 0. 80 m , and a second point charge q2 = +6. 00 nc is at the poin
Alekssandra [29.7K]

The net electric field is the vector sum of the components of the electric

field produced by the two charges.

The values of the magnitude and direction of the net electric field at the origin (approximate values) are;

  • 131.6 N/C
  • 12.6 ° above the negative x–axis

<h3>How are the net electric field magnitude and direction calculated?</h3>

The possible questions based on a similar question posted online are;

(a) The net electric field at the origin.

The electric field due to charge q₁ is given as follows;

\vec E_{1x} = \mathbf{ \dfrac{1}{4 \cdot \pi \cdot \epsilon_0} \cdot \dfrac{q_1}{\vec{r}^2_2}}

Which gives;

\vec{E}_{1x} =\mathbf{ \dfrac{\left(9 \times 10^9 \, N \cdot m/C^2 \right) \cdot \left(-4 \, nC \times \dfrac{10^{-9}C}{1 \, nC} \right)}{\left(1 \, m \right)^2}} \cdot cos\left(arctan\left(\dfrac{0.8}{0.6} \right) \right) =-21.6 \, N/C

\vec{E}_{1y} = \mathbf{\dfrac{\left(9 \times 10^9 \, N \cdot m/C^2 \right) \cdot \left(-4 \, nC \times \dfrac{10^{-9}C}{1 \, nC} \right)}{\left(1 \, m \right)^2}} \cdot sin\left(arctan\left(\dfrac{0.8}{0.6} \right) \right) = 28.8 \, N/C

Which gives;

\vec{E}_1 = \mathbf{21.6 \, N/C  \cdot \hat x +  28.8 \, N/C \hat y}

\vec{E}_{2x} = \dfrac{\left(9 \times 10^9 \, N \cdot m/C^2 \right) \cdot \left(6.00 \, nC \times \dfrac{10^{-9}C}{1 \, nC} \right)}{\left(1 \, m \right)^2} = 150 \, N/C

Therefore;

\vec  {E} = \left[ 21.6 \, N/C - 150 \, N/C \right] \left( \hat x \right) + \left(28.8 \, N/C \right) \left( \hat y \right)

\vec  {E} = \mathbf{\left( -128.4 \, N/C  \right) \left( \hat x \right) + \left(28.8 \, N/C \right) \left( \hat y \right)}

The magnitude of the net electric field is therefore;

E = \sqrt{(-128.4^2 + 28.8^2)} ≈ 131.6

  • The magnitude of the net electric field at the origin is E ≈<u> 131.6 N/C</u>

(b) The direction of the net electric field at the origin.

  • The \ direction \ is \ arctan \left(\dfrac{28.8}{-128.4} \right) \approx \underline{ 12.6^{\circ}} \ above \ the \ negative \ x-axis

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