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4 years ago

Which of the Moon’s positions will cause the highest or spring tides?

Physics

2 answers:

olya-2409 [2.1K]4 years ago

7 0

The answer is B. A and C. Spring tides occur during a new moon and full moon

Contact [7]4 years ago

5 0

Answer:

Positions A and C will cause the highest tides, called spring tides, because the gravity of the Moon and Sun will work together to create the two tidal bulges on opposite sides of Earth. So B.

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A passengers train travels from reno to Las Vegas a distance of 706 km in 6.5 h. what is the change average speed?

12345 [234]

Answer: learn how to do it on your own idiot

Explanation: try hard in school

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3 years ago

The original meaning of the word "essay' is which of these? treatise dialogue attempt thought

Elina [12.6K]

The answer to this is, attempt.

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4 years ago

Read 2 more answers

A mail carrier leaves the post ofﬁce and drives 2.00 km to the north. He then drives in a direction 60.0° south of east for 7.00

Oxana [17]

Answer:

12.97 km

Explanation:

In order to find the resultant displacement, we have to resolve each of the 3 displacements along the x and y direction.

Taking north as positive y direction and east as positive x-direction, we have:

- Displacement 1: 2.00 km to the north

$A_x = 0\\A_y = +2.00 km$

- Displacement 2: 60.0° south of east for 7.00 km

$B_x=(7.00)(cos (-45^{\circ}))=4.95 km\\B_y = (7.00)(sin (-45^{\circ}))=-4.95 km$

- Displacement 3: 9.50 km 35.0° north of east

$C_x=(9.50)(cos 35^{\circ})=7.78 km\\C_y=(9.50)(sin 35^{\circ})=5.45 km$

So the net displacement along the two directions is:

$R_x=A_x+B_x+C_x=0+4.95+7.78=12.73 km\\R_y=A_y+B_y+C_y=2.00+(-4.95)+5.45=2.50 km$

So, the distance between the initial and final position is equal to the magnitude of the net displacement:

$R=\sqrt{R_x^2+R_y^2}=\sqrt{12.73^2+2.50^2}=12.97 km$

7 0

3 years ago

Two identical ions are each missing two electrons, each ion has a charge of 2e. What is the magnitude of the force between the i

xz_007 [3.2K]

Answer:

|F| = 2.09 × 10⁻⁸ assuming that the two ions are point charges.

Explanation:

What's the charge on each ion?

The symbol $e$ here stands for fundamental charge. Each electron carries a negative fundamental charge of -e. Each proton carry a positive fundamental charge of +e.

Molecules and atoms are neutral. They contain an equal number of electrons and protons. Remove one electron from a molecule or atom, and that particle will end up with more protons (which are positive) than electrons. That particle will carry a positive charge of +e become an ion (a cation to be precise.) Remove another electron and the ion will carry a charge of +2e.

For each ion $q = +2 \;e = 2\times 1.60\times 10^{-19}\;\text{C} = 3.2\times 10^{-19}\;\text{C}$ .

What's the size of the electrostatic force between the two ions?

Consider Coulomb's Law for the electrostatic force $F$ between two point charges:

$\displaystyle F = -\frac{k\cdot q_1\cdot q_2}{r^{2}}$ ,

where

$k$ is Coulomb's constant,
$q_1$ and $q_2$ are the charge on the two point charges, and
$r$ is the separation between the two charges.

Make sure that all values are in SI units. Assume that the two ions are small enough that they act like point charges:

$\displaystyle \begin{aligned}F &= -\frac{k\cdot q_1 \cdot q_2}{r^{2}}\\&=-\frac{8.99\times 10^{9}\cdot(3.2\times 10^{-19}) \cdot (3.2\times 10^{-19})}{(2.1\times 10^{-10})^{2}}\\ &= -2.09\times 10^{-8}\;\text{N}\end{aligned}$ .

The value of $F$ is negative, meaning that the two charges will repel each other because they are both positive. The question is asking for the magnitude of this force. Thus drop the sign in front of $F$ to obtain $2.09\times 10^{-8}\;\text{N}$ , which is the magnitude of $F$ .