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Ludmilka [50]
3 years ago
5

What are the effect of propagation of light​

Physics
1 answer:
ollegr [7]3 years ago
7 0

≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡ HI ≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡

                               ∵∴∵∴∵∴∵∴∵∴∵∴∵∴∵∴∵∴∵∴∵∴∵∴

Due to Rectilinear propagation of light  

  • A shadow is formed
  • Formation of Day and Night
  • An Image in the pinhole camera is formed

∞║║ HOPE THIS HELPS PLEASE MARK AS BRAINLIEST║║∞

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2 years ago
The balance Lenght of a potent ometer wire for a Cell of emf 1.62v is 90cm. if the Cell is replaced by another one of emf 1.08v.
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Answer:

answer is 3.05v

Explanation:

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6 0
2 years ago
15 points! An atomic nucleus initially moving at 420 m/s emits an alpha particle in the direction of its velocity, and the remai
alexandr1967 [171]

The alpha particle is emitted at 4235 m/s

Explanation:

We can use the law of conservation of momentum to solve the problem: the total momentum of the original nucleus must be equal to the total momentum after the alpha particle has been emitted. Therefore:

p_i = p_f\\ Mu=m_1 v_1 + m_2 v_2 =  

where:  

M =222u is the mass of the original nucleus

v=420 m/s is the initial velocity of the nucleus

m_1 = 4 u is the mass of the alpha particle

v_1 is the final velocity of the alpha particle

m_2 = 222u-4u = 218 u is the mass of the daughter nucleus

v_2 = 350 m/s is the final velocity of the nucleus

Solving for v_1, we  find the final velocity of the alpha particle:

v_1 = \frac{Mu-m_2 v_2}{m_1}=\frac{(222)(420)-(218)(350)}{4}=4235 m/s

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

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4 0
3 years ago
The International Space Station has a mass of 1.8 × 105 kg. A 70.0-kg astronaut inside the station pushes off one wall of the st
Aleonysh [2.5K]

Answer:

a = 5.83 \times 10^{-4} m/s^2

Explanation:

Since the system is in international space station

so here we can say that net force on the system is zero here

so Force by the astronaut on the space station = Force due to space station on boy

so here we know that

mass of boy = 70 kg

acceleration of boy = 1.50 m/s^2

now we know that

F = ma

F = 70(1.50) = 105 N

now for the space station will be same as above force

F = ma

105 = 1.8 \times 10^5 (a)

a = \frac{105}{1.8 \times 10^5}

a = 5.83 \times 10^{-4} m/s^2

3 0
3 years ago
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