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Over [174]
4 years ago
9

What is hidden during a lunar eclipse? Isn't the Moon hidden?

Physics
2 answers:
shtirl [24]4 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 Moon are exactly or very closely aligned (in syzygy), with Earth between the other two. A lunar eclipse can occur only on the night of a full moon.

So yes the moon is hidden during a Lunar eclipse.

Setler [38]4 years ago
4 0

Here you go:

This light gets reflected onto the moon. ... The moon doesn't always hide completely behind Earth's shadow. During partial lunar eclipses, the sun, Earth and moon are slightly off in their alignment and so our planet's shadow engulfs just part of the moon.


Hope it helps you

Please mark me as Brainsliest


You might be interested in
1. In what year did India (our second most populous country) begin to produce more energy than Spain?
choli [55]

Answers:

<h2>1) Sometime in 2014 </h2>

As we can see in the graph in 2014 there is an intersection between the lines that represent the wind energy production of Spain and India, then the production of India is increased, while that of Spain remains constant.

<h2>2) Germany </h2>

As we can see in the graph in 2000 Germany was the number 1 producer of gigawatts of wind energy. In addition it is observed a rapid growth and increase in production compared to the other countries.

<h2>3) In 2006, America was producing over 10 GW and began a steep climb in wind energy production.</h2>

In 1997, America began with a constant and very small energy production, which was growing gradually until 2006, when its rapid growth in production began to exceed the other countries shown in the graph.

<h2>4) Years</h2>

When a function is plotted, generally the independent variable, whose value is previously set, is represented on the X axis.

In this sense, the variable found on the X axis of this graph is Years.

<h2>5) Wind Energy Capacity </h2>

When a function is plotted, generally the dependent variable (which is generated from the independent variable) is represented on the Y axis.

In this sense, the variable found on the Y axis of this graph is Wind Energy Capacity.

<h2>6) Gigawatts (GW) </h2>

We already know the dependent variable in this <u>Wind Energy Capacity vs Years</u> graph is Wind Energy Capacity, and its Unit is Gigawatts (GW).

Where 1GW=10^{6} W

<h2>7) Both onshore and offshore wind sources </h2>

When we talk about scientific experimentation and representing data, the Control Variable is the one that remains constant (it does not change during the investigation).  

According to the explanation above, the option that fits with this characteristic is: Both onshore and offshore wind sources

<h2>8) 55 GW </h2>

If we want to make a linear interpolation to estimate how much wind energy capacity was there in the United States in the middle of 2011, firstly we need to <u>find two points where the value we want to find is in the middle.</u>

These points (X,Y) are:

P1: (2011,48) and P2: (2012,60)

Where X1:2011, Y1:48, X2:2012, Y2:60

Now we find the <u>slope</u> m of the line with the following equation:

m=\frac{Y2 - Y1}{X2 - X1}

m=\frac{60 - 48}{2012 - 2011}=12

Then, with this value of the slope we can write the <u>equation of the line</u> and find the <u>intersection point</u> b with one of the given points (we will use P1):

Y=mX+b

48=12(2011)+b

b=-24084

With this value and the slope, we can find Y for X=2011.5 (we need to know the wind capacity in the middle of 2011):

Y=(12)(2011.5)-24084

Y=54 This is the e<u>stimated result</u> 54 GW, and the option that is near this value is 55 GW

<h2>9) 36 GW </h2>

In this case we will apply a similar method as the previous answer, but estimating a prediction:

Let's find two points near the value we want to find. These points (X,Y) are:

P1: (2014,22) and P2: (2016,29)

Where X1:2014, Y1:22, X2:2016, Y2:29

Now we find the <u>slope</u> m of the line with the following equation:

m=\frac{Y2 - Y1}{X2 - X1}

m=\frac{29 - 22}{2016 - 2014}=\frac{7}{2}

Then, with this value of the slope we can write the <u>equation of the line</u> and find the <u>intersection point</u> b with one of the given points (we will use P2):

Y=mX+b

29=\frac{7}{2}(2016)+b

b=-7027

With this value and the slope, we can find Y for X=2020 (we need to estimate the expected wind capacity in 2020):

Y=\frac{7}{2}(2020)-7027

Y=43 This is the e<u>stimated result</u> 43 GW (note linear extrapolation is not as accurate as other methods), and the option that is near this value is 36 GW

<h2>10) Additional 7 GW/Year</h2>

In this case we will use the equation of the slope:

m=\frac{Y2 - Y1}{X2 - X1}

With the following points:

P1: (2009,35) and P2: (2016,82)

m=\frac{82 - 35}{2016 - 2009}

m=6.7 \approx 7

Therefore, the correct option is Additional 35 GW/Year

4 0
3 years ago
8.0kg rocket fired horizontally encounters a force of air resistance of 4.9 N. The force supplied by the rocket's engine is 60.9
Gelneren [198K]
60.9 - 4.9= 56
56 is the net force
using the formula F = ma
56= 8a
a = 7
acceleration is 7ms^-2
6 0
3 years ago
A boat crossing a 153.0 m wide river is directed so that it will cross the river as quickly as possible. The boat has a speed of
Lynna [10]

We have the relation

\vec v_{B \mid E} = \vec v_{B \mid R} + \vec v_{R \mid E}

where v_{A \mid B} denotes the velocity of a body A relative to another body B; here I use B for boat, E for Earth, and R for river.

We're given speeds

v_{B \mid R} = 5.10 \dfrac{\rm m}{\rm s}

v_{R \mid E} = 3.70 \dfrac{\rm m}{\rm s}

Let's assume the river flows South-to-North, so that

\vec v_{R \mid E} = v_{R \mid E} \, \vec\jmath

and let -90^\circ < \theta < 90^\circ be the angle made by the boat relative to East (i.e. -90° corresponds to due South, 0° to due East, and +90° to due North), so that

\vec v_{B \mid R} = v_{B \mid R} \left(\cos(\theta) \,\vec\imath + \sin(\theta) \, \vec\jmath\right)

Then the velocity of the boat relative to the Earth is

\vec v_{B\mid E} = v_{B \mid R} \cos(\theta) \, \vec\imath + \left(v_{B \mid R} \sin(\theta) + v_{R \mid E}\right) \,\vec\jmath

The crossing is 153.0 m wide, so that for some time t we have

153.0\,\mathrm m = v_{B\mid R} \cos(\theta) t \implies t = \dfrac{153.0\,\rm m}{\left(5.10\frac{\rm m}{\rm s}\right) \cos(\theta)} = 30.0 \sec(\theta) \, \mathrm s

which is minimized when \theta=0^\circ so the crossing takes the minimum 30.0 s when the boat is pointing due East.

It follows that

\vec v_{B \mid E} = v_{B \mid R} \,\vec\imath + \vec v_{R \mid E} \,\vec\jmath \\\\ \implies v_{B \mid E} = \sqrt{\left(5.10\dfrac{\rm m}{\rm s}\right)^2 + \left(3.70\dfrac{\rm m}{\rm s}\right)^2} \approx 6.30 \dfrac{\rm m}{\rm s}

The boat's position \vec x at time t is

\vec x = \vec v_{B\mid E} t

so that after 30.0 s, the boat's final position on the other side of the river is

\vec x(30.0\,\mathrm s) = (153\,\mathrm m) \,\vec\imath + (111\,\mathrm m)\,\vec\jmath

and the boat would have traveled a total distance of

\|\vec x(30.0\,\mathrm s)\| = \sqrt{(153\,\mathrm m)^2 + (111\,\mathrm m)^2} \approx \boxed{189\,\mathrm m}

3 0
2 years ago
Which of the following properties did Rutherford use in his experiment?
Ronch [10]

<u>Answer</u>:

(C) The positive charge of the alpha particles and the negative charge of the electrons  are the properties Rutherford used in his experiment

<u>Explanation</u>:

In his scattering experiment, scientist Ernest Rutherford used the property of positively charged alpha particles and negatively charge electrons. He performed this experiment by passing some of the alpha particles through a gold foil. The result was that the some alpha particles scattered while some passed through the gold foil without collision.

He concluded that the alpha particles are centrally positively charged and needed a large amount repelling force. This experiment of Rutherford is also known as Rutherford model of atom. This experiment helped him in doing so many other discoveries.

3 0
4 years ago
PLEASE HELP BRAINLIEST
Leni [432]

B is the answer!!!!!!!!!!!!!

7 0
3 years ago
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