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3 years ago

In the diagram, the Sun, Earth, and Moon are in perfect alignment. Which two conclusions can be drawn based on the diagram?

Physics

2 answers:

mihalych1998 [28]3 years ago

8 0

Answer:An observer on the Moon sees a solar eclipse. and An observer on Earth sees a lunar eclipse.

Explanation:from the moon the sun is completly if not just mostly invisible(solar eclipse) and from earth the moon would be in lunar eclipse from the earth blocking the sun.

AysviL [449]3 years ago

4 0

Answer:

I know one answer is an observer on earth sees a lunar eclipse

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What happens to the light rays when they hit the specimen?

pogonyaev

Answer:

Explanation:

Ray of light when hits any specimen or object. The light is partially reflected, partially reflected and partially absorbed. It is never completed reflected, refracted or absorbed. Hence, the correct answer would be c.

4 0

2 years ago

2. Tomas is hanging from a tree limb, that is inclined at a 65° angle. The force

LUCKY_DIMON [66]

Answer:

57 N

Explanation:

Were are told that the force

of gravity on Tomas is 57 N.

And it acts at an inclined angle of 65°

Thus;

The vertical component of the velocity is; F_y = 57 sin 65

While the horizontal component is;

F_x = 57 cos 65

Thus;

F_y = 51.66 N

F_x = 24.09 N

The net force will be;

F_net = √((F_y)² + (F_x)²)

F_net = √(51.66² + 24.09²)

F_net = √3249.0837

F_net = 57 N

4 0

3 years ago

kaheart [24]

Answer: el pepe

Explanation:

8 0

2 years ago

Read 2 more answers

You are holding a positive charge and there are positive charges of equal magnitude 1 mm to your north and 1 mm to your east. Wh

lara [203]

If I hold a positive charge in my hand and there are positive charges of equal magnitude 1 mm to your north and 1 mm to your east then the direction of the force on the charge I am holding is towards the north-east direction.

Reasoning:

It is given that there is a positive charge in my hand. There are two more positive charges with the same magnitude. One is 1 mm far towards the east, and the other one is 1 mm far towards the north. It is required to find the direction of the force acting on the charge in my hand.

Let the magnitude of the charge in my hand is Q, and the magnitude of the other charges is q.

Thus the electric force applied on the charge in my hand due to each other is,

$F=\frac{kQq}{r^2}$

Here k is the Coulomb constant, and r is the distance between the charges.

It is also known that the force on a positive charge due to another positive charge is acted outwards.

Thus, the force on the charge due to the charge on the east is,

$\vec{F_1}=\frac{kQq}{( 10^{-3}\text{ m})^2}\hat{i}$

And the force on the charge due to the charge on the north is,

$\vec{F_2}=\frac{kQq}{( 10^{-3}\text{ m})^2}\hat{j}$

As the forces are equal in magnitude and one is perpendicular to the other, thus the net force will be acted at an angle of $45^\circ$ from the north or from the north direction.