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3 years ago

True or false. Jake and summer are in the north pole. True false.

SAT

2 answers:

Nutka1998 [239]3 years ago

7 0

Jake and summer are in the north pole is false.

Jake and summer are both at Mexico

Both Jake and summer are teen detectives given the assignment of stopping the Egyptian treasure from being stolen. Their car broke down on the way and they had to trade a sheep for a spare tire. They have another 3 hours to stop the Egyptian treasure from being stolen.

<h3>Who is a detective?</h3>

A detective is a law enforcement officer who is employed to find information not otherwise available to the public or looks for evidence as part of solving a crime.

Learn more about detectives:

brainly.com/question/26059088

Artyom0805 [142]3 years ago

3 0

Answer: 1) What do the underlined words in the following sentence mean? En algunos places, puede hacer frio en septiembre

*it might be cold in September*
2) Translate the following words into English. parecen palacios

*they look like palaces*
3) What is the meaning of the following word in English? sin

*without*

4) According to Jake, what look like chocolate kisses?

*the roofs of the palaces*

5) True or False. Jake and Summer are in the North Pole.

*false*

Explanation:

for future reference

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3 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;

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

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