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Brums [2.3K]
3 years ago
15

Melanie made the following diagram shown to represent some phases of the moon as seen from Earth.

Physics
1 answer:
timofeeve [1]3 years ago
6 0

Answer:

D. The position of the first quarter moon and third quarter moon have been interchanged.    

Explanation:

Phases of moon are the shapes of illuminated portion of the moon that is visible from the earth.

New moon occurs when the moon comes between the earth and the sun. The lit portion faces away from the earth.

After new moon, waxing crescent phase occurs followed by first quarter. In the first quarter, half moon is visible.

Waxing gibbous occurs which is when more than half and less than full face of the moon is visible. This is followed by full moon when earth comes between moon and the sun.

The waning gibbous phase is visible for next few days till the third quarter phase. Then waning crescent phase and finally new moon again.

In the given diagram made by Melanie, the position of first quarter and third quarter are wrong and must be interchanged.

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An electron and a proton are held on an x axis, with the electron at x = + 1.000 m
mixas84 [53]

Answer:

  r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

Explanation:

For this exercise we must use conservation of energy

the electric potential energy is

          U = k \frac{q_1q_2}{r_{12}}

for the proton at x = -1 m

          U₁ =- k \frac{e^2 }{r+1}

for the electron at x = 1 m

          U₂ = k \frac{e^2 }{r-1}

starting point.

        Em₀ = K + U₁ + U₂

        Em₀ = \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1}

final point

         Em_f = k e^2 ( -\frac{1}{r_2 +1} + \frac{1}{r_2 -1})

   

energy is conserved

        Em₀ = Em_f

        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e^2 (- \frac{1}{r_2 +1} + \frac{1}{r_2 -1})              

       

        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e²(  \frac{2}{(r_2+1)(r_2-1)} )

we substitute the values

½ 9.1 10⁻³¹ 450 + 9 10⁹ (1.6 10⁻¹⁹)² [ - \frac{1}{20+1} + \frac{1}{20-1} ) = 9 109 (1.6 10-19) ²( \frac{2}{r_2^2 -1} )

          2.0475 10⁻²⁸ + 2.304 10⁻³⁷ (5.0125 10⁻³) = 4.608 10⁻³⁷ ( \frac{1}{r_2^2 -1} )

          2.0475 10⁻²⁸ + 1.1549 10⁻³⁹ = 4.608 10⁻³⁷     \frac{1}{r_2^2 -1}

          \frac{2.0475 \ 10^{-28} }{1.1549 \ 10^{-37} } = \frac{1}{r_2^2 -1}

          r₂² -1 = (4.443 10⁸)⁻¹

           

          r2 = \sqrt{1 + 2.25 10^{-9}}

          r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

4 0
3 years ago
Help, please. I am not sure what to do.
miss Akunina [59]

Answer:

option D) -3m

Explanation:

if 6m is diplaced by -3m then it would be -3+6=3m

feel free to ask if you are confused

3 0
2 years ago
Two resistors, of R1 = 3.93 Ω and R2 = 5.59 Ω, are connected in series to a battery with an EMF of 24.0 V and negligible interna
son4ous [18]

Answer:

2.521 (A); 14.0924 (V)

Explanation:

more info in the attachment, the answers are marked with red colour.

4 0
2 years ago
A helicopter blade spins at exactly 110 revolutions per minute. Its tip is 4.50 m from the center of rotation. (a) Calculate th
NeX [460]

Answer:

(a). The average speed is 51.83 m/s.

(b). The average velocity over one revolution is zero.

Explanation:

Given that,

Angular velocity = 110 rev/m

Radius = 4.50 m

(a). We need to calculate the average speed

Using formula of average speed

v=r\omega

v = 4.50\times110\times\dfrac{2\pi}{60}

v=51.83\ m/s

(b). The average velocity over one revolution is zero because the net displacement is zero in one revolution.

Hence, (a). The average speed is 51.83 m/s.

(b). The average velocity over one revolution is zero.

8 0
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Describe and explain how the movement and arrangement of the particles in a block of ice change as the ice melts to form liquid
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Answer:

Ice is water in solid phase, in this phase, the particles are very close together and relatively in fixed positions.

As the temperature starts to increase (thermal energy), also does the kinetic energy of the particles (so we have a change from thermal energy to kinetic energy), so they start to move "more", and the position of the particles starts to be less "fixed". There is a point where the particles have enough energy, and this point is where the phase of the water changes from solid to liquid phase (the fusion point). After this point the water can not hold his shape, and takes the shape of the container where it is.

4 0
3 years ago
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