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Nitella [24]
3 years ago
9

Felipe believes that whenever the Moon is in the position that is shown from above (top view) in Diagram A, the Moon always look

s completely bright in the view from Earth, as shown in Diagram B.
Is Felipe correct? If he is correct, explain why the Moon always looks bright from Earth in that position, and explain how he should show light on the Moon in Diagram A. If he is incorrect, explain how else the Moon can look when it is in that position.

Physics
1 answer:
taurus [48]3 years ago
8 0

Answer:he is incorrect

Explanation:he is wrong because if the sunlight is on the opposite side of the earth than the moon, then the sunlight would be blocked by the earth, making the moon dark, because the moon gets its light from the light that is reflected of the surface from the sun

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A passenger walks from one side of a ferry to the other as it approaches a dock.Passenger's velocity is 1.55 m/s due north relat
victus00 [196]

Answer:\theta =33.22^{\circ}

Explanation:

Given

Velocity of Passenger w.r.t to Ferry

v_{pf}=v_p-v_f=1.55\hat{j}-------1

Velocity of Passenger w.r.t to water

v_{pw}=v_p-v_w=4.5\left ( -\hat{i}+\hat{j}\right )--------2

Subtract 2 from 1

v_f-v_w=-4.5\hat{i}+2.95\hat{j}

(b)direction

tan\theta =\frac{2.95}{4.5}=0.655

\theta =33.22^{\circ} North of west

5 0
4 years ago
In a ballistic pendulum experiment, projectile 1 results in a maximum height h of the pendulum equal to 2.6 cm. A second project
Trava [24]

Answer:

  Second  projectile is 1.4 times faster than first projectile.

Explanation:

By linear momentum conservation

Pi = Pf

m x U + M x 0 = (m + M) x V

U= \dfrac{(m + M)\times V}{m}

Now Since this projectile + pendulum system rises to height 'h', So using energy conservation:

KEi + PEi = KEf + PEf

PEi = 0, at reference point

KEf = 0, Speed of system zero at height 'h'

KEi = \dfrac{(m + M)\times V^2}{2}

PEf = (m + M) g h

So,

\dfrac{(m + M)\times V^2}{2} + 0 = 0+ (m + M) g h

V =\sqrt {2gh}

So from above value of V

Initial velocity of projectile =U

U=\dfrac{(M+m)\sqrt{2gh}}{m}

Now Since mass of projectile and pendulum are constant, So Initial velocity of projectile is proportional to the square root of height swung by pendulum.

Which means

\dfrac{U_2}{U_1}=\sqrt{\dfrac{h_2}{h_1}}

U_2=\sqrt{\dfrac{h_2}{h_1}}\times U_1

U_2=\sqrt{\dfrac{5.2}{2.6}}\times U_1

U₂ = 1.41 U₁

Therefore we can say that ,Second  projectile is 1.4 times faster than first projectile.

4 0
3 years ago
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slavikrds [6]

Answer:

500 feet

Explanation:

It's stipulated that a minimum horizontal distance of 2000 feet from clouds should be allowed for VFR flights but at a night, the minimum distance below clouds requirement for VFR flight outside controlled airspace at altitudes of more than 1200 feet AGL, but less than 10000 feet MSL should be 500 feet.

3 0
3 years ago
Which of these phrases would go in the overlap? Check all that apply.
Sati [7]

Answer:

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Explanation:

5 0
3 years ago
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Daniel [21]

Answer:

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Explanation: i hope this helps :)

3 0
4 years ago
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