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Citrus2011 [14]
3 years ago
7

Please answer the following question

Physics
2 answers:
Elden [556K]3 years ago
6 0

Answer:

Where is the resistance value?

Question is incomplete

Liula [17]3 years ago
6 0
Can you show the rest of the problems?
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It would take roughly 3 seconds.
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When the two surfaces of a thin film are parallel and the light is incident at some angle what fringes are formed? A. Circular B
lord [1]

Answer:

The correct option is;

A. Circular

Explanation:

Some of the light that impinges on the surface are reflected and the rest are transmitted to a different medium

At the surface of the next medium also, some of the light are transmitted while the others are reflected and refracted through the first medium

The speed of light (and hence the wavelength and color) refracted through the thin film is changed as the distance the refracted light travels through the thin film is increased as we move away from the point directly in the front view to some distance as the reflected light path from those distance to the eye is increased due to their inclination giving them a different wavelength which are all equal at a radial distance from the eye hence forming a circular fringes.

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A ball is thrown horizontally with a velocity of 12 m/s. How
miskamm [114]

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6 0
3 years ago
Read 2 more answers
A 1.0 meter long rod is held horizontally in the east-west direction on a distant planet and is dropped from a height of 1.23 m.
enyata [817]

Answer:

The induced emf between two end is 34.02 \times 10^{-5} V

Explanation:

Given:

Length of rod l = 1 m

Height h = 1.23 m

Magnetic field B = 6.93 \times 10^{-5} T

For finding induced emf,

  \epsilon = Blv

Where v = velocity of rod,

For finding the velocity of rod.

From kinematics equation,

 v^{2} = v_{o} ^{2} + 2gh

Where v_{o} = initial velocity, g = 9.8 \frac{m}{s^{2} }

   v = \sqrt{2gh}

   v = \sqrt{2 \times 9.8 \times 1.23}

   v = 4.91 \frac{m}{s}

Put the velocity in above equation,

   \epsilon = 6.93 \times 10^{-5} \times 1 \times 4.91

   \epsilon = 34.02 \times 10^{-5} V

Therefore, the induced emf between two end is 34.02 \times 10^{-5} V

6 0
3 years ago
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