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ivolga24 [154]
4 years ago
5

1.5 Kg ball moves in a circle that is 0.4 m radius at a velocity of 5.40 m/s Calculate its centripetal acceleration. *

Physics
1 answer:
topjm [15]4 years ago
5 0
Hdhshidndhbsubdudbybdjdnjdbd mdudnt
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A light ray incident from medium 1 to medium 2, where n1>n2. When the incident angle exceed the critical angle ac, the refrac
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Explanation:

(a)

Critical angle is the angle at the angle of refraction is 90°. After the critical angle, no refraction takes place.

Using Snell's law as:

n_1\times {sin\theta_i}={n_2}\times{sin\theta_r}

Where,  

{\theta_i}  is the angle of incidence

{\theta_r} is the angle of refraction = 90°

{n_2} is the refractive index of the refraction medium

{n_1} is the refractive index of the incidence medium

Thus,

n_1\times {sin\ \theta_{critical}}={n_2}\times{sin\ 90^0}

The formula for the calculation of critical angle is:

{sin\theta_{critical}}=\frac {n_2}{n_1}

Where,  

{\theta_{critical}} is the critical angle

(b)

No it cannot occur. It only occur when the light ray bends away from the normal which means that when it travels from denser to rarer medium.

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An object start from rest with a constant acceleration of 8.00 m/s^2 along straight line.
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Problem 1 Observer A, who is at rest in the laboratory, is studying a particle that is moving through the laboratory at a speed
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Answer:

markers are 29.76 m far apart in the laboratory

Explanation:

Given the data in the question;

speed of particle = 0.624c

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we know that

distance = vt

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so we substitute

distance = 0.624c × 1.59 × 10⁻⁷ s

distance = 0.624(3 × 10⁸ m/s) × 1.59 × 10⁻⁷ s

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distance =  29.76 m

Therefore, markers are 29.76 m far apart in the laboratory

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