Lf some body need help you can ask me

SAT

2 answers:

Rufina [12.5K]3 years ago

7 0

Thanks! Its really nice of you to offer!

strojnjashka [21]3 years ago

7 0

Answer:

Okay, thank you so much! I'll friend you just incase I forget :D! DO you do Middle School Work?

Explanation:

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More college <br><br>right answer please

Nastasia [14]

Here’s the answers, down there

5 0

3 years ago

What is an example of a scope three carbon emission brainly.

Sati [7]

An example of a scope three carbon emission is; driving a gasoline-fueled vehicle

<h3>What is Scope three carbon emission?</h3>

Scope 3 emissions are a defined as the consequence of the activities of the company that occur from sources not owned or controlled by the company.

Some examples of scope 3 carbon emission activities are;

Extraction and production of purchased materials.

Transportation of purchased fuels.

Use of products and services.

Thus, when applied in practical, we can say that an example of a scope three carbon emission is; driving a gasoline-fueled vehicle

Read more about carbon emissions at; brainly.com/question/1301348

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7 0

2 years ago

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;

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

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