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3 years ago

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What is a lunar eclipse 2 paragraph

1 answer:

konstantin123 [22]3 years ago

5 0

A lunar eclipse occurs when the Moon passes directly behind Earth and into its shadow. This can occur only when the Sun, Earth, and the Moon are aligned (in syzygy) exactly or very closely so, with the planet in between. Hence, a lunar eclipse can occur only on the night of a full moon. The type and length of an eclipse depend on the Moon's proximity to either node of its orbit.

During a total lunar eclipse, Earth completely blocks direct sunlight from reaching the Moon. The only light reflected from the lunar surface has been refracted by Earth's atmosphere. This light appears reddish for the same reason that a sunset or sunrise does: the Rayleigh scattering of bluer light. Due to this reddish color, a totally eclipsed Moon is sometimes called a blood moon.

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A horizontal force of 92.7 N is applied to a 40.5 kg crate on a rough, level surface. If the crate accelerates at 1.13 m/s2, wha

lord [1]

Answer:

The value is $F_f = 46.935 \ N$

Explanation:

From the question we are told that

The magnitude of the horizontal force is $F = 92.7 \ N$

The mass of the crate is $m = 40.5 \ kg$

The acceleration of the crate is $a = 1.13 \ m/s$

Generally the net force acting on the crate is mathematically represented as

$F_{net} = F - F_f = ma$

Here $F_f$ is force of kinetic friction (in N) acting on the crate

So

$92.7 - F_f = 40.5 * 1.13$

=> $F_f = 46.935 \ N$

5 0

3 years ago

Two spheres, 1.00 kg each, whose centers are 2.00 m apart, would have what gravitational force between them? A. 3.14 X 10-17 N

Angelina_Jolie [31]

Answer: B

Explanation: the teacher just told us the answer

4 0

3 years ago

27.0 mL to the nearest milliliter

Mila [183]

Answer:

27.0 milliliters is the nearest mililiter so 27.0 is the answer

Explanation:

8 0

3 years ago

Q= Which one of the following statement is incorrect?

Lunna [17]

Answer:

A.

Explanation:

I think it might be the big number A

8 0

3 years ago

If it requires 2.0 J of work to stretch a particular spring by 2.0 cm from its equilibrium length, how much more work will be re

valina [46]

Answer:

16 J

Explanation:

It is given that,

Work done, W = 2 J

A spring is stretched by 2.0 cm from its equilibrium length

We need to find how much more work will be required to stretch it an additional 4.0 cm.

Let k is the spring constant of the spring. When W = 2J, and x = 2 cm, then energy required to stretch the spring is :

$U=\dfrac{1}{2}kx^2\\\\k=\dfrac{2U}{x^2}\\\\k=\dfrac{2(2)}{(0.02)^2}\\\\k=10000\ N/m$

The energy required to stretch the spring from 2 cm to additional 4 cm i.e. 2+4= 6 cm.

$W=\dfrac{1}{2}k(x_2^2-x_1^2)\\\\=\dfrac{1}{2}\times 10000\times ((0.06)^2-(0.02)^2)\\\\W=16\ J$

So, the required work done is 16 J.

7 0

3 years ago