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-Dominant- [34]
3 years ago
13

If the sun is 400 times bigger than the moon, how couild the moon possibly cover the sun during a solar eclipse?​

Physics
2 answers:
Vilka [71]3 years ago
6 0

Answer:

It's all about perspective. The moon is far closer to the Earth than the Sun is, so they appear roughly the same size. If they were closer to each other, then obviously the moon wouldn't be large enough to cover a substantial amount of the sun's light. But given the huge distance both between them and the moon and the earth, to us they look relatively the same.

AlekseyPX3 years ago
3 0

Explanation:

the Moon passes between Earth and the Sun Even though the Moon is much smaller than the Sun, because it is just the right distance away from Earth, the Moon can fully block the Sun's light from Earth's perspective This completely blocks out the Sun's light

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5. A cable is attached 32.0 m from the base of a flagpole that is about to
soldi70 [24.7K]

Answer:

The length of the flagpole is approximately 87.43 m

Explanation:

The given parameters of the cable attached to the flagpole are;

The point along the flagpole at which the cable is attached = 32.0 m

The angle with respect to the ground at which the raising of the flagpole is halted = 60.0°

The downward force exerted by the cable, F_v = 1.233 × 10⁴ N

The force exerted by the cable to the left = 1.233 × 10⁴ N

Let 'W' represent the weight of the flagpole, at equilibrium, we have;

The sum of vertical forces = 0

Therefore;

F_v + W - R = 0

W - R = -1.233 × 10⁴ N

Taking moment about the support at the base of the pole, we get;

1.233 × 10⁴ × d × cos(60.0°) - 1.233 × 10⁴ ×d× sin(60.0°) + W × d/2 ×cos(60.0°) = 0

∴ W × d/2 ×cos(60.0°) ≈  4513.093·d  

W = 2 × 4513.093/(cos(60.0°)) ≈ 18,052.373 N

R = 18,052.373 + 1.233 × 10⁴ ≈ 30,382.373

R ≈ 30,382.373 N

Taking moment about the point of attachment of the cable to the ground, we have;

W × ((d/2) × cos(60.0°) + 32) = R × 32

∴ (d/2) = ((30,382.373 × 32/18,052.373) - 32)/(cos(60.0°)) ≈ 43.71281

d = 2 × 43.71281 ≈ 87.43

The length of the flagpole, d ≈ 87.43 m

7 0
3 years ago
A spaceship, at rest in some inertial frame in space, suddenly needs to accelerate. The ship forcibly expels 103 kg of fuel from
Cerrena [4.2K]

Answer:

V_s = 1.8*10^5m/s

Explanation:

There is no external force applied, therefore there is a moment's preservation throughout the trajectory.

<em>Initial momentum  = Final momentum. </em>

The total mass is equal to

m_T= m_1 +m_2

Where,

m_1 = mass of ship

m_2 = mass of fuell expeled.

As the moment is conserved we have,

0=V_fm_2+V_sm_1

Where,

V_f = Velocity of fuel

V_s =Velocity of Space Ship

Solving and re-arrange to V_swe have,

V_s = \frac{V_f m_2 }{m_1}

V_s = \frac{3/5c}{10^6}

V_s = 3.5*10^{-3}c

Where c is the speed of light.

Therefore the ship be moving with speed

V_s = \frac{3}{5}*10^{-3}*3*10^8m/s

V_s = 1.8*10^5m/s

5 0
3 years ago
Which is a product of photosynthesis?
evablogger [386]

<u>Answer</u>:-

The main product is glucose despite as the matter of fact it produces several products such as:

•Oxygen

&

•Water too.

Explanation:

The reaction of photosynthesis is given below:

6CO2 + 12H2O + (in the presence of sunlight) → C6H12O6 + 6O2 + 6H2O

Carbon dioxide + Water + Sunlight → Glucose + Oxygen + Water.

8 0
3 years ago
Number the steps for balancing equations: Use coefficients to increase the atoms on each side. Check to make sure you have the s
Sav [38]

Answer:

1.Identify the atoms on each side,

2. Count the atoms on each side,

3.Use coefficients to increase the number of atoms on each side,

4.Check to make sure you have same number of each type of atom on each side

Explanation:

THEY ALL RIGHT!!!!!!!!!

6 0
3 years ago
Read 2 more answers
Help!!!
snow_lady [41]

Answer:

Explanation:

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5 0
3 years ago
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