All categories

3 years ago

During a total lunar eclipse, the moon remains visible because ________.

1 answer:

3 0

During a total lunar eclipse, the Moon is still visible to the naked eye due to the fact that the atmosphere of our planet, which is Earth bends, which results to the scattering of some sunlight, which actually reaches the Moon during a Lunar Eclipse, which is why the Moon remains visible despite the occurrence of a full lunar eclipse.

You might be interested in

If you were to choose one, which of Earth's spheres is the most important, and why?

Molodets [167]

Earth's land makes up the geosphere.

Earth's water makes up the hydrosphere. ...

Earth's air makes up the atmosphere. ...

Earth's living things make up the biosphere. ...

The four spheres interact. ...

Humans can have major impacts on all the spheres.

pick one

7 0

2 years ago

The innermost satellite of jupiter orbits the planet with a radius of 422 × 103 km and a period of 1.77 days. what is the mass o

FinnZ [79.3K]

The mass of Jupitar is obtained from the calculations as 5.8 * 10^-14 Kg.

<h3>What is the mass of Jupitar?</h3>

There are nine planets in the solar system and the sun lies at the enter of our solar system. This is the heliocentric model of the solar system.

Given that;

T^2 = GMr^3/4π

T = period

G = gravitational constant

r = radius

M = mass of Jupitar

Now;

1 day = 86400 seconds

1.77 days = 1.77 days * 86400 seconds/1 day

= 152928 seconds

Making M the subject of the formula;

M =4πT^2/Gr^3

M = 4 * 3.142 * (152928)^2/6.67 × 10^-11 * (422 × 10^9)^3

M = 2.9 * 10^11/5.0 * 10^24

M = 5.8 * 10^-14 Kg

Learn more about mass of a planet:brainly.com/question/13851553

#SPJ1

5 0

2 years ago

Two long, parallel wires are attracted to each other by a force per unit length of 350 µN/m. One wire carries a current of 22.5

pishuonlain [190]

Answer

given,

force per unit length = 350 µN/m

current, I = 22.5 A

y = y = 0.420 m

$\dfrac{F}{L}= \dfrac{KI_1I_2}{d}$

$I_2 = \dfrac{F}{L}\dfrac{d}{KI_1}$

$I_2 = 350\times 10^{-6}\times \dfrac{0.42}{2 \times 10^{-7}\times 22.5}$

I₂ = 32.67 A

distance where the magnetic field is zero

$\dfrac{4\pi \times 10^{-7}\times 32.67}{2\pi y_1}=\dfrac{4\pi \times 10^{-7}\times 22.5}{2\pi (0.42-y_1)}$

$y_1 = 0.248\ m$

there the distance at which the magnetic field is zero in the two wire is at 0.248 m.

3 0

2 years ago

You are riding a bicycle up a gentle hill. It is fairly easy to increase your potential

Dahasolnce [82]

True or false: while riding a bicycle up a gentle hill, it fairly easy to increase your potential energy, but to increase your kinetic energy would ...

3 0

1 year ago

Two 25.0N weights are suspended at opposite ends of a rope that passes over a frictionless pulley. What is the tension in the ro

goldenfox [79]

Answer:

tension in rope = 25.0 N

Explanation:

Two forces act on the suspended weight. The force coming down is the gravitational force and the upward force by the tension in the rope.

Since the suspended weight is not accelerating so that the net force will be zero. Therefore the tension in the rope should be 25 N.

∑F = F - W = 0

F = W

so tension in rope = F = T = 25 N

8 0

2 years ago