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cestrela7 [59]
3 years ago
12

The phases of the moon are the changing appearances of the moon, as seen from Earth. Which phase happens immediately after a thi

rd quarter moon?
Physics
2 answers:
Drupady [299]3 years ago
6 0

The phases of the moon are the changing appearances of the moon, as seen from Earth. Which phase happens immediately after a third quarter moon are the following

Explanation:

  • After the full moon (maximum illumination), the light continually decreases. So the waning gibbous phase occurs next. Following the third quarter is the waning crescent, which wanes until the light is completely gone -- a new moon.

waning gibbous phase

  • The waning gibbous phase occurs between the full moon and third quarter phases. The last quarter moon (or a half moon) is when half of the lit portion of the Moon is visible after the waning gibbous phase.

Time takes by the moon to go through all the phases

about 29.5 days

  • It takes 27 days, 7 hours, and 43 minutes for our Moon to complete one full orbit around Earth. This is called the sidereal month, and is measured by our Moon's position relative to distant “fixed” stars. However, it takes our Moon about 29.5 days to complete one cycle of phases (from new Moon to new Moon).
  • At 3rd quarter, the moon rises at midnight and sets at noon. Then we see only a crescent. At new, the moon rises at sunrise and sets at sunset, and we don't see any of the illuminated side!
lisov135 [29]3 years ago
3 0

Answer:

Waning Gibbous

Explanation:

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Q1. After three half-lives of an isotope, 1 billion of the original isotope's atoms still remain in a certain amount of this ele
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The original parent isotope count was 125 + 875 = 1000. The remaining parent isotope is 125/1000 or 1/8. 1/8 = (1/2)^h, where h is the number of half-lives. 

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3 years ago
A child on a sled slides (starting from rest) down an icy slope that makes an angle of 15◦ with the horizontal. After sliding 20
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Answer:

A) v₁ = 10.1 m/s t₁= 4.0 s

B) x₂= 17.2 m

C) v₂=7.1 m/s

D) x₂=7.5 m

Explanation:

A)

  • Assuming no friction, total mechanical energy must keep constant, so the following is always true:

       \Delta K + \Delta U = (K_{f} - K_{o}) +( U_{f} - U_{o}) = 0  (1)

  • Choosing the ground level as our zero reference level, Uf =0.
  • Since the child starts from rest, K₀ = 0.
  • From (1), ΔU becomes:
  • \Delta U = 0- m*g*h = -m*g*h (2)  
  • In the same way, ΔK becomes:
  • \Delta K = \frac{1}{2}*m*v_{1}^{2}  (3)      
  • Replacing (2) and (3) in (1), and simplifying, we get:

       \frac{1}{2}*v_{1}^{2}  = g*h  (4)

  • In order to find v₁, we need first to find h, the height of the slide.
  • From the definition of sine of an angle, taking the slide as a right triangle, we can find the height h, knowing the distance that the child slides down the slope, x₁, as follows:

       h = x_{1} * sin \theta_{1} = 20.0 m * sin 15 = 5.2 m (5)

       Replacing (5) in (4) and solving for v₁, we get:

      v_{1} = \sqrt{2*g*h} = \sqrt{2*9.8m/s2*5.2m} = 10.1 m/s  (6)

  • As this speed is achieved when all the energy is kinetic, i.e. at the bottom of the first slide, this is the answer we were looking for.
  • Now, in order to finish A) we need to find the time that the child used to reach to that point, since she started to slide at the its top.
  • We can do this in more than one way, but a very simple one is using kinematic equations.
  • If we assume that the acceleration is constant (which is true due the child is only accelerated by gravity), we can use the following equation:

       v_{1}^{2} - v_{o}^{2} = 2*a* x_{1}  (7)

  • Since v₀ = 0 (the child starts from rest) we can solve for a:

       a = \frac{v_{1}^{2}}{2*x_{1} } = \frac{(10.1m/s)^{2}}{2* 20.0m} = 2.6 m/s2  (8)

  • Since v₀ = 0, applying the definition of acceleration, if we choose t₀=0, we can find t as follows:

       t_{1} =\frac{v_{1} }{a} =\frac{10.1m/s}{2.6m/s2} = 4.0 s  (9)

B)

  • Since we know the initial speed for this part, the acceleration, and the time, we can use the kinematic equation for displacement, as follows:

       x_{2} = v_{1} * t_{2} + \frac{1}{2} *a_{2}*t_{2}^{2}  (10)

  • Replacing the values of v₁ = 10.1 m/s, t₂= 2.0s and a₂=-1.5m/s2 in (10):

       x_{2} = 10.1m/s * 2.0s + \frac{1}{2} *(-1.5m/s2)*(2.0s)^{2}  = 17.2 m (11)

C)  

  • From (6) and (8), applying the definition for acceleration, we can find the speed of the child whem she started up the second slope, as follows:

       v_{2} = v_{1} + a_{2} *t_{2} = 10.1m/s - 1.5m/s2*2.0s = 7.1 m/s (12)

D)

  • Assuming no friction, all the kinetic energy when she started to go up the second slope, becomes gravitational potential energy when she reaches to the maximum height (her speed becomes zero at that point), so we can write the following equation:

       \frac{1}{2}*v_{2}^{2}  = g*h_{2}   (13)

  • Replacing from (12) in (13), we can solve for h₂:

       h_{2} =\frac{v_{2} ^{2}}{2*g} = \frac{(7.1m/s) ^{2}}{2*9.8m/s2} = 2.57 m  (14)

  • Since we know that the slide makes an angle of 20º with the horizontal, we can find the distance traveled up the slope applying the definition of sine of an angle, as follows:

       x_{3} = \frac{h_{2} }{sin 20} = \frac{2.57m}{0.342} = 7.5 m (15)

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