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dusya [7]
3 years ago
5

Could anyone please help me? I really need help, I am not understanding this. I will love the help, please.

SAT
2 answers:
Ratling [72]3 years ago
5 0

Answer:

Problem 1: Even

Problem 2: Even

Problem 3: Neither

Problem 4: Neither

Explanation:

You have to consider the fact that just because it is negative doesn't mean its odd. You have to do this, you take the function and plug –x in for x, and then simplify. If you end up with the exact same function that you started with (that is, if f (–x) = f (x), so all of the signs are the same), then the function is even.

Hope this helps! :)

Serhud [2]3 years ago
4 0
Even
even
neither
neither
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<h3>What is the study about?</h3>

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

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

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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

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

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