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Vedmedyk [2.9K]
2 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]2 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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projectile motion of a particle of mass M with charge Q is projected with an initial speed V in a driection opposite to a unifor
SIZIF [17.4K]

Answer:

Range, R = MV²/2QE

Explanation:

The question deals with the projectile motion of a particle mass M with charge Q, having an initial speed V in a direction opposite to that of a uniform electric field.

Since we are dealing with projectile motion in an electric field, the unknown variable here, would be the range, R of the projectile. We note that the electric field opposes the motion of the particle thereby reducing its kinetic energy. The particle stops when it loses  all its kinetic energy due to the work done on it in opposing its motion by the electric field. From work-kinetic energy principles, work done on charge by electric field = loss in kinetic energy of mass.

So, [tex]QER = MV²/2{/tex} where R is the distance (range) the mass moves before it stops

Therefore {tex}R = MV²/2QE{/tex}

5 0
3 years ago
A 6,000N is applied to a formula one car that weighs 500kg. What is the car's acceleration?
Vesna [10]

Answer:

<h2>12 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

f is the force

m is the mass

From the question

f = 6000 N

m = 500 kg

We have

a =  \frac{6000}{500}  =  \frac{60}{5}  = 12 \\

We have the final answer as

<h3>12 m/s²</h3>

Hope this helps you

8 0
2 years ago
The nine ring wraiths want to fly from barad-dur to rivendell. rivendell is directly north of barad-dur. the dark tower reports
Katarina [22]
The west constituent of their sequence needs to cancel out 58 mph crosswind. Subsequently a northwest direction is a 45-degree angle up to even with the destination. That is the third point out of the triangle and the right angle is at the destination. The top side is the west constituent of their flight the vertical side is their resultant travel and the hypotenuse is their definite distance flown. Since the 58 mph crosswind was negated by flying northwest, the distance from the beginning to the destination must be the same distance as the west component of their travel. The hypotenuse is square root of twice the side since it has 2 identical sides.

c = sqrt (58^2 + 58^2) = sqrt (6728) = 82.02 
 
Alternative solution:

c = sqrt (2) * 58 = 1.414 * 58 = 82.02

Therefore, they have to fly 82.02 mph
5 0
3 years ago
Read 2 more answers
Find the angle of depression from the top of a lighthouse 250 feet above water level to the water line of a ship 2.5 miles offsh
Rudik [331]
Height of the lighthouse = 250 feet
Distance of the ship from the shore = 2.5 miles
We already know that
1 mile = 5280 feet
Then
2.5 miles = 5280 * 2.5
               = 13200 feet
Now, to find the angle of depression, we can use the formula
<span>θ = <span>tan<span>−1 </span></span>(13200/250)
</span>   = 1.50 degrees
I hope that this is the answer that you were looking for and the answer has come to your desired help. 
4 0
3 years ago
The average human has a density of 945 kg/m3 after inhalation, and 1020 kg/m3 after exhalation. Fresh water has a density of 100
Montano1993 [528]

Answer:

a. The human body has nearly the same density as salt water after exhaling.

b. The human body will always float in the Dead Sea.

Explanation:

According to the concept of floating on the basis of  density, any body that is put in a fluid of density greater than its own density will always float due to the force of buoyancy from the liquid.

  • The portion of the object submerged  while the object is floating depends upon the density of the object as compared to the density of the fluid. This is governed by the equation:

\rho_f.V_s=\rho_o.V_o

where:

\rho_f= density of the fluid

\rho_o= density of the object

V_s= volume of the object submerged in the fluid

V_o= total volume of the object

6 0
3 years ago
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