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STALIN [3.7K]
3 years ago
14

What causes the lunar phases we see on Earth? Give a few examples of different phases of the moon and what each looks like.

Physics
1 answer:
Aliun [14]3 years ago
5 0
What causes us to see the moon is the Sun. The moon is usually above or below the Earth. So light from the Sun hits the moon and the moon reflects the light back. Some of it to Earth. For example, when a new moon happens the moon is in between the Sun and the Earth

SUN            --------LIGHT----------->    MOON       EARTH

What happens there is that none of the light gets reflected off the part of the moon that we can see from the ground, giving the illusion that it is not there.

During a full moon the Earth is in between the Sun and moon


SUN            -------LIGHT--------->       EARTH        MOON

Now, nearly all the light from the Sun gets reflected off the surface of the moon and to us on Earth

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A horse running at 3 m/s speeds up with a constant acceleration of 5 m/s2. How fast is the
Elan Coil [88]

Answer:

The horse is going at 12.72 m/s speed.

Explanation:

The initial speed of the horse (u) = 3 m/s

The acceleration of the horse (a)= 5 m/s^{2}

The displacement( it is assumed it is moving in a straight line)(s)= 15.3 m

Applying the second equation of motion to find out the time,

s=ut+\frac{1}{2}at^{2}

15.3=3t+2.5t^{2}

2.5t^{2}+3t-15.3=0

Solving this quadratic equation, we get time(t)=1.945 s, the other negative time is neglected.

Now applying first equation of motion, to find out the final velocity,

v=u+at

v=3+1.945*5

v=3+9.72

v=12.72 m/s

The horse travels at a speed of 12.72 m/s after covering the given distance.

7 0
3 years ago
A Venturi tube may be used as a fluid flowmeter. Suppose the device is used at a service station to measure the flow rate of gas
leonid [27]

Answer

given,

flow rate = p = 660 kg/m³

outer radius = 2.8 cm

P₁ - P₂ = 1.20 k Pa

inlet radius = 1.40 cm

using continuity equation

 A₁ v₁ = A₂ v₂

 π r₁² v₁ = π r₁² v₂

 v_1= \dfrac{r_1^2}{r_2^2} v_2

 v_1= \dfrac{1.4^2}{2.8^2} v_2

 v_1= 0.25 v_2

Applying Bernoulli's equation

 \Delta P = \dfrac{1}{2}\rho (v_2^2-v_1^2)

 \Delta P = \dfrac{1}{2}\rho (v_2^2-(0.25 v_2)^2)

 \Delta P = \dfrac{1}{2}\rho v_2^2 (1 - 0.0625)

 v_2=\sqrt{\dfrac{2\Delta P}{\rho(1 - 0.0625)}}

 v_2=\sqrt{\dfrac{2\times 1200}{660 \times(1 - 0.0625)}}

       v₂ = 1.97 m/s

b) fluid flow rate

Q = A₂ V₂

Q = π (0.014)²  x 1.97

Q = 1.21 x 10⁻³ m³/s

5 0
3 years ago
An archer wants to hit a target that is dropped from a tower. At the sound of a horn, the archer is to shoot an arrow; at the sa
bija089 [108]

It has to be D because the arrow will drop as it moves, if it were a gun, you'd lead the target so fire below it, but due to it being an arrow, you aim high not low. Also, they didnt specify how fast anything is, so you'd probably miss if you actually did it.


6 0
3 years ago
Read 2 more answers
What's the diameter of a dish antenna that will receive 10−20W of power from Voyager at this time? Assume that the radio transmi
Murrr4er [49]

Complete Question:

The Voyager 1 spacecraft is now beyond the outer reaches of our solar system, but earthbound scientists still receive data from the spacecraft s 20-W radio transmitter. Voyager is expected to continue transmitting until about 2025, when it will be some 25 billion km from Earth.

What s the diameter of a dish antenna that will receive 10−20W of power from Voyager at this time? Assume that the radio transmitter on Voyager transmits equally in all directions(isotropically).  In fact, the antenna on Voyager focuses the signal in a beam aimed at the earth, so this problem over-estimates the size of the receiving dish needed.

Answer:

d = 2,236 m.

Explanation:

The received power on Earth, can be calculated as the product of the intensity (or power density) times the area that intercepts the power radiated.

As we assume that  the transmitter antenna is ominidirectional, power is spreading out over a sphere with a radius equal to the distance to the source.

So, we can get the power density as follows:

I = P /A = P / 4*π*r², where P = 20 W, and r= 25 billion km = 25*10¹² m.

⇒ I = 20 W / 4*π* (25*10¹²)² m²

The received power, is just the product of this value times the area of the receiver antenna, which we assumed be a circle of diameter d:

Pr = I. Ar =( 20W / 4*π*(25*10¹²)² m²) * π * (d²/4) = 10⁻²⁰ W

Simplifying common terms, we can solve for d:

d= √(16*(25)²*10⁴/20) = 2,236 m.

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3 years ago
What is the equivalet tempreture of 0 kelvin in celcius scale <br> HELP NEED IT IN 5 MIN
Alona [7]

Answer:

-273.15 degree celcius

Explanation:

7 0
3 years ago
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