Based on the trend in valence electrons across periods for main-group

Anit [1.1K]

A, C, and D all happen at different stages
of a total lunar eclipse.

I'll describe the stages of the eclipse, but before I do, I just need
to clarify: The Earth doesn't have an umbra or a penumbra, but
its shadow does.

-- the eclipse begins when the first edge of the moon
   moves into the penumbra of Earth's shadow; ( C )
   this part of the moon grows steadily.

-- After a while, the first edge of the moon begins to move
   into the umbra of Earth's shadow ( A ), and gets very dark.

-- The total phase of the eclipse begins when the ENTIRE
    moon is in the umbra of Earth's shadow.

Then everything happens in reverse.

-- Eventually, the leading edge of the moon moves out
   of the shadow's umbra, into the penumbra. This part
   steadily grows.

-- After a while, none of the moon is in the umbra, and
   the whole thing is in the penumbra. The moon is
   fully illuminated, but not quite as bright as it should be.

-- Soon, the leading edge of the moon leaves the penumbra
of Earth's shadow, and gets brighter. This portion of the moon
    steadily grows, until ...

-- the moon completely leaves the penumbra, all of it is as bright
as it's supposed to be. The eclipse is completely over. ( B )

==> The whole process lasts several hours.

==> Everybody on the night side of the Earth sees the same thing
   at the same time. It doesn't matter WHERE you are on the night
   side ... if you can see the moon in the sky, you see the present
   phase of the eclipse.

==> The lunar eclipse can only happen at the Full Moon. In fact, the
   mid-point of the total phase is the exact moment of Full Moon.

8 0

2 years ago

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Technician A says that a proper U-joint inspection can be done with the driveshaft in place on the vehicle. Technician B says th

Julli [10]

Answer:

TECHNICIAN B

Explanation: Drive shaft is an elongated part of the drive system made of steel of a vehicle that transmits the ENGINE TORQUE to the WHEELS of a vehicle. For proper inspection a technician is expected to completely remove the driveshaft to check all its compactment. It is Cylindrical and usually used for REAR-WHEEL-VEHICLES. Drive shafts are known by different names like propeller shaft,Cardan shaft etc it has different types which range from

One-piece drive shaft ,Two-piece drive shaft to Slip-in-tube drive shaft .

6 0

2 years ago

. A water balloon is thrown horizontally at a speed of 2.00 m/s from the roof of a building that is 6.00m above the ground. At t

dedylja [7]

Answer:

Explanation:

Height of building

H = 6m

Horizontal speed of first balloon

U1x = 2m/s

Second ballot is thrown straight downward at a speed of

U2y = 2m/s

Time each gallon hits the ground

Balloon 1.

Using equation of free fall

H = Uoy•t + ½gt²

Uox = 0 since the body does not have vertical component of velocity

6 = ½ × 9.8t²

6 = 4.9t²

t² = 6 / 4.9

t² = 1.224

t = √1.224

t = 1.11 seconds

For second balloon

H = Uoy•t + ½gt²

6 = 2t + ½ × 9.8t²

6 = 2t + 4.9t²

4.9t² + 2t —6 = 0

Using formula method to solve the quadratic equation

Check attachment

From the solution we see that,

t = 0.9211 and t = -1.329

We will discard the negative value of time since time can't be negative here

So the second balloon get to the ground after t ≈ 0.92 seconds

Conclusion

The water ballon that was thrown straight down at 2.00 m/s hits the ground first by 1.11 s - 0.92s = 0.19 s.

4 0

2 years ago

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A 20 KeV electron emits two bremsstrahlung photons as it is being brought to rest in two successive decelerations. The wavelengt

Degger [83]

Answer:

λ₁ = 87.5 10⁻¹² m , λ₂ = 2.175 10⁻¹⁰ m, E₂ = 5.8 10³ eV

Explanation:

In this case you can use the law of conservation of energy, all the energy of the electron is converted into energized emitted photons

Let's reduce to the SI system

E₀ = 20 10³ eV (1.6 10⁻¹⁹ J / 1eV) = 3.2 10⁻¹⁵ J

Δλ = 1.30 A = 0.13 nm = 0.13 10⁻⁹ m

Ef = E₁ + E₂

E₀ = Ef

E₀ = E₁ + E₂

The energy can be found with the Planck equation

E = h f

c = λ f

f = c / λ

E = hc / λ

They indicate that the wavelength of the second photon is

λ₂ = λ₁ +0.130 10⁻⁹

We replace

E₀ = hv / λ₁ + hc / ( λ₁ + 0.130 10⁺⁹)

E₀ / hv = 1 / λ₁ + 1 / ( λ₁ + 0.13 10⁻⁹)

3.2 10⁻¹⁵ / (6.63 10⁻³⁴ 3 10⁸) = ( λ₁ + 0.13 10⁻⁹ + λ₁) / λ₁ ( λ₁ + 0.13 10⁻⁹)

1.6 10¹⁰ ( λ₁² +0.13 10⁻⁹ λ₁) = 2 λ₁ + 0.13 10⁻⁹

λ₁² + 0.13 10⁻⁹ λ₁ = 1.25 10⁻¹⁰ λ₁ + 8.125 10⁻²¹

λ₁² + 0.005 10⁻⁹ λ₁ = 8.125 10⁻²¹

λ₁² + 5 10⁻¹² λ₁ - 8.125 10⁻²¹ = 0

Let's solve the second degree equation

λ₁ = [-5 10⁻¹² ±√((5 10⁻¹²)² + 4 8.125 10⁻²¹)] / 2

λ₁ = [-5 10⁻¹² ±√(25 10⁻²⁴ +32.5 10⁻²¹)] / 2 = [-5 10⁻¹² ±√ (32525 10⁻²⁴)] / 2

λ₁ = [-5 10⁻¹² ± 180 10⁻¹²] / 2

λ₁ = 87.5 10⁻¹² m

λ₂ = -92.5 10⁻¹² m

We take the positive wavelength

The wavelength of the photons is

λ₁ = 87.5 10⁻¹² m

λ₂ = λ₁ + 0.13 10⁻⁹

λ₂ = 87.5 10⁻¹² + 0.13 10⁻⁹

λ₂ = 0.2175 10⁻⁹ m = 2.175 10⁻¹⁰ m

The energy after the first deceleration is

E₂ = E₀ –E₁

E₂ = E₀ –hc / λ₁

E₂ = 3.2 10⁻¹⁵ - 6.63 10⁺³⁴ 3 10⁸ / 87.5 10⁻¹²

E₂ = 3.2 10⁻¹⁵ - 2.27 10⁻¹⁵

E₂ = 0.93 10⁻¹⁵ J

E₂ = 0.93 10⁻¹⁵ J (1 eV / 1.6 10⁻¹⁹ J)

E₂ = 5.8 10³ eV

7 0

2 years ago

A ball is falling from the sky down to the left. What two forces are being used?

suter [353]

Answer:

push and pull

Explanation:

i beleve good luck❤️❤️

4 0

2 years ago

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