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Ludmilka [50]
3 years ago
8

During a total solar eclipse, the Earth, Sun, and Moon are arranged in a line, and light from the Sun is blocked out of view for

a certain part of the Earth by the Moon. Beginning with the sun, in what order do the Earth, Sun, and Moon line up?
Physics
1 answer:
kari74 [83]3 years ago
3 0

Answer:

Sun, Moon, Earth

Explanation:

<em>A solar eclipse occurs when the moon comes in-between the sun and the earth with all the three bodies in alignment. Thus, the moon prevents the rays of the sun to get to the earth and in its stead, cast a shadow of itself on the earth. The casting of the moon's shadow on the earth creates a region of total darkness (umbra) for certain parts of the earth while some other regions at the periphery of the moon's shadow only experience partial darkness (penumbra).</em>

Hence, beginning with the sun, the correct order of the three bodies during a solar eclipse would be <u>the Sun, the Moon, and then the Earth</u>.

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kozerog [31]

Newton's second equation of motion :-

S=ut+1/2at^2 [where, u is the initial velocity, a is the acceleration and t is the time interval]

This Equation simply finds a relation between distance travelled by a particle (classically) under uniform acceleration.

So let's see what pieces of information (bundles of equations) do we have with us, initially.

We have, a very primary equation with us,

dS/dt = v… (I)

(Considering motion in a straight line only)

And we also have the equation

dv/dt = a…(II)

Simply replacing the v in eqn (II) by eqn (I), we find

d2S/dt^2 = a…(III)

This is what we need to solve. It's easy.

You know,

d2S/dt^2 = d/dt(dS/dt) = a

⟹ dS/dt = ∫adt = at+c1

Since, dS/dt is the velocity of the particle,

Therefore, at t = 0, dS/dt|t = 0 = u

⟹ u = a∗0 + c1 = c1

⟹ c1 = u

Therefore, dS/dt = u + at

Thus, S = ∫(udt + atdt)

⟹ S = ut + 1/2at^2 +c^2

If say, the particle is already having a displacement S0 the moment you start measuring it's motion. Then, at t = 0, S = S0

This makes S = S0 +ut + 1/2at^2

Since, in most of the practical cases, we start measuring a motion when the particle starts displacing (i.e., when S0=0 ),

We get

S = ut + 1/2at^2

Hope it helps :)

5 0
3 years ago
Ocean waves pass through two small openings, 20.0 m apart, in a breakwater. You're in a boat 70.0 m from the breakwater and init
Klio2033 [76]

Answer:

λ = 5.65m

Explanation:

The Path Difference Condition is given as:

δ=(m+\frac{1}{2})\frac{lamda}{n}  ;

where lamda is represent by the symbol (λ) and is the wavelength we are meant to calculate.

m = no of openings which is 2

∴δ= \frac{3*lamda}{2}

n is the index of refraction of the medium in which the wave is traveling

To find δ we have;

δ= \sqrt{70^2+(33+\frac{20}{2})^2 }-\sqrt{70^2+(33-\frac{20}{2})^2 }

δ= \sqrt{4900+(\frac{66+20}{2})^2}-\sqrt{4900+(\frac{66-20}{2})^2}

δ= \sqrt{4900+(\frac{86}{2})^2 }-\sqrt{4900+(\frac{46}{2})^2 }

δ= \sqrt{4900+43^2}-\sqrt{4900+23^2}

δ= \sqrt{4900+1849}-\sqrt{4900+529}

δ= \sqrt{6749}-\sqrt{5429}

δ=  82.15 -73.68

δ= 8.47

Again remember; to calculate the wavelength of the ocean waves; we have:

δ= \frac{3*lamda}{2}

δ= 8.47

8.47 = \frac{3*lamda}{2}

λ = \frac{8.47*2}{3}

λ = 5.65m

3 0
4 years ago
According to Newton's law of universal gravitation, which statements are true?
kogti [31]

Answer:

c.  As we gain mass, the force of gravity on us increases

8 0
3 years ago
Read 2 more answers
When a bicycle coasts uphill, it moves slower and slower as it climbs. Why?
astra-53 [7]

Letter B because it is gaining more potential energy as it SLOWLY climbs up the hill.

the less motion the more potential energy there is

6 0
3 years ago
Read 2 more answers
An infinite conducting cylindrical shell of outer radius r1 = 0.10 m and inner radius r2 = 0.08 m initially carries a surface ch
irinina [24]

Answer:

a) \sigma_{\rm in} = -2.18~{\rm \mu C/m^2}

b) \sigma_{\rm out}= 1.12~{\rm \mu C/m^2}

c) E = \frac{\sigma(r_1 + r_2)}{\epsilon_0 r}

Explanation:

Before the wire is inserted, the total charge on the inner and outer surface of the cylindrical shell is as follows:

Q_{\rm in} = \sigma A_{\rm in} = \sigma(2\pi r_1 h) = (-0.35)(2\pi (0.08) h) = -0.175h~{\rm \mu C}

Q_{\rm out} = \sigma A_{\rm out} = \sigma(2\pi r_2 h) = (-0.35)(2\pi (0.1) h) = -0.22h~{\rm \mu C}

Here, 'h' denotes the length of the cylinder. The total charge of the cylindrical shell is -0.395h μC.

When the thin wire is inserted, the positive charge of the wire attracts the same amount of negative charge on the inner surface of the shell.

Q_{\rm wire} = \lambda h = 1.1h~{\rm \mu C}

a) The new charge on the inner shell is -1.1h μC. Therefore, the new surface charge density of the inner shell can be calculated as follows:

\sigma_2 = \frac{Q_{\rm in}}{2\pi r_1h} = \frac{-1.1h}{2\pi r_1 h} = \frac{-1.1}{2\pi(0.08)} = -2.18~{\rm \mu C/m^2}

b) The new charge on the outer shell is equal to the total charge minus the inner charge. Therefore, the new charge on the outer shell is +0.705 μC.

The new surface charge density can be calculated as follows:

\sigma_{\rm out}= \frac{Q_{\rm out}}{2\pi r_2h} = \frac{0.705h}{2\pi r_2 h} = \frac{0.705}{2\pi(0.1)} = 1.12~{\rm \mu C/m^2}

c) The electric field outside the cylinder can be found by Gauss' Law:

\int{\vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}

We will draw an imaginary cylindrical shell with radius r > r2. The integral in the left-hand side will be equal to the area of the imaginary surface multiplied by the E-field.

E(2\pi r h) = \frac{Q_{\rm enc}}{\epsilon_0}\\E2\pi rh = \frac{\sigma 2\pi (r_1 + r_2)h}{\epsilon_0}\\E = \frac{\sigma(r_1 + r_2)}{\epsilon_0 r}

4 0
3 years ago
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