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Vedmedyk [2.9K]
3 years ago
11

Select the Moon and use the Info view to determine which of the following statements is correct. The Last Quarter Moon.... rises

near noon and sets near midnight. rises at about 6am and sets at about 6pm. rises near midnight and sets near midday. rises and sets at the same time as the Sun. Submit Your Answer
Physics
1 answer:
Ann [662]3 years ago
5 0

Answer: The correct statement is that the Last Quarter Moon (rises near midnight and sets near midday).

Explanation:

Phases of the moon also called the LUNAR PHASE can be defined as the different shades of illumination on the moon as seen from the earth. The moon is the natural satellite of the earth that illuminates upon reflection of light from the sun. This means it doesn't have power to shine on its own. When carefully observed, there are times it gets dark and beings to glow brighter over a period of time. This occurs because as the moon completes its four weeks lunar cycle round the earth, how much of its face we see illuminated by sunlight depends on the angle the Sun makes with the Moon.

There are 8 main types of the moon phases these includes:

--> New moon: This is when the moon is not visible to the earth because it's between the earth and the sun. It rises at sunrise and sets at sunset.

--> The waxing crescent: At this phase the moon gets brighter and illuminated from the sun that a crescent shape is seen.

--> First quarter: this occurs one week after the new moon. The moon rises at noon and sets at midnight.

--> The waxing gibbous: This occurs after the first quarter phase where more than half of the lit part of the moon is seen.

--> Full moon: This occurs when the moon and the sun are opposite each other. That is way it is said to rise at sunset and sets at sun rise.

--> The waning gibbous: this occurs when more than half of the lit part of the moon gradually becomes darker

--> Third quarter ( Last Quarter): The moon rises at midnight and sets at noon. This occurs a week after the full moon.

--> The waning crescent: This occurs after the last quarter phase where a very thin fading crescent shaped moon is seen, just before the Moon is invisible again at the start of the cycle, the new moon.

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What is the force applied to a baseball that has a mass of 142kg and has a acceleration of 30m/s to the power of 2
aev [14]
  • Mass=m=142kg
  • Acceleration=a=30m/s
  • Force=F

Using Newton's second law

\\ \sf\Rrightarrow F=ma

\\ \sf\Rrightarrow F=142(30)

\\ \sf\Rrightarrow F=4260N

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3 years ago
A projectile is launched at an angle of 30 and lands 20 s later at the same height as it was launched. (a) What is the initial s
Pavlova-9 [17]

Answer:

(a) 196 m/s

(b) 490 m

(c) 3394.82 m

(d) 2572.5 m

Explanation:

First of all, let us know one thing. When an object is thrown in the air, it experiences two forces acting in two different directions, one in the horizontal direction called air resistance and the second in the vertically downward direction due to its weight. In most of the cases, while solving numerical problems, air resistance is neglected unless stated in the numerical problem. This means we can assume zero acceleration along the horizontal direction.

Now, while solving our numerical problem, we will discuss motion along two axes according to our convenience in the course of solving this problem.

<u>Given:</u>

  • Time of flight = t = 20 s
  • Angle of the initial velocity of projectile with the horizontal = \theta = 30^\circ

<u>Assume:</u>

  • Initial velocity of the projectile = u
  • R = Range of the projectile during the time of flight
  • H = maximum height of the projectile
  • D = displacement of the projectile from the initial position at t = 15 s

Let us assume that the position from where the projectile was projected lies at origin.

  • Initial horizontal velocity of the projectile = u\cos \theta
  • Initial horizontal velocity of the projectile = u\sin \theta

Part (a):

During the time of flight the displacement of the projectile along the vertical is zero as it comes to the same vertical height from where it was projected.

\therefore u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow u\sin \theta t=\dfrac{1}{2}(g)t^2\\\Rightarrow u=\dfrac{gt^2}{2\sin \theta t}\\\Rightarrow u=\dfrac{9.8\times 20^2}{2\sin 30^\circ \times 20}\\\Rightarrow u=196\ m/s

Hence, the initial speed  of the projectile is 196 m/s.

Part (b):

For a projectile, the time take by it to reach its maximum height is equal to return from the maximum height to its initial height is the same.

So, time taken to reach its maximum height will be equal to 10 s.

And during the upward motion of this time interval, the distance travel along the vertical will give us maximum height.

\therefore H = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow H = 196\times \sin 30^\circ \times 10 + \dfrac{1}{2}\times(-9.8)\times 10^2\\ \Rightarrow H =490\ m

Hence, the maximum altitude is 490 m.

Part (c):

Range is the horizontal displacement of the projectile from the initial position. As acceleration is zero along the horizontal, the projectile is in uniform motion along the horizontal direction.

So, the range is given by:

R = u\cos \theta t\\\Rightarrow R = 196\times \cos 30^\circ \times 20\\\Rightarrow R =3394.82\ m

Hence, the range of the projectile is 3394.82 m.

Part (d):

In order to calculate the displacement of the projectile from its initial position, we first will have to find out the height of the projectile and its range during 15 s.

\therefore h = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow h = 196\times \sin 30^\circ \times 15 + \dfrac{1}{2}\times(-9.8)\times 15^2\\ \Rightarrow h =367.5\ m\\r = u\cos \theta t\\\Rightarrow r = 196\times \cos 30^\circ \times 15\\\Rightarrow r =2546.11\ m\\\therefore D = \sqrt{r^2+h^2}\\\Rightarrow D = \sqrt{2546.11^2+367.5^2}\\\Rightarrow D =2572.5\ m

Hence, the displacement from the point of launch to the position on its trajectory at 15 s is 2572.5 m.

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Answer:

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Explanation:

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The answer is "36 grams".

Explanation:

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