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melamori03 [73]

3 years ago

9

What is the threshold velocity vthreshold(water) (i.e., the minimum velocity) for creating Cherenkov light from a charged partic

le as it travels through water (which has an index of refraction of n

1 answer:

VladimirAG [237]3 years ago

4 0

Complete Question

The complete question is shown on the first uploaded image

Answer:

A

$v_w = 2.256 *10^{8} \ m/s$

B

$v_e = 2.21 *10^{8} \ m/s$

C

The correct option is B

Explanation:

From the question we are told that

The refractive index of water is $n_w = 1.33$

The refractive index of ethanol is $n_e = 1.36$

Generally the threshold velocity for creating Cherenkov light from a charged particle as it travels through water is mathematically evaluated as

$v_w = \frac{c}{n_w }$

Where c is the speed of light with value $c = 3.0 *10^{8} \ m/s$

$v_w = \frac{3.0 *10^{8}}{1.33 }$

$v_w = 2.256 *10^{8} \ m/s$

Generally the threshold velocity for creating Cherenkov light from a charged particle as it travels through water is mathematically evaluated as

$v_e = \frac{ c}{n_e }$

=> $v_e = \frac{3.0 *10^{8}}{1.36 }$

=> $v_e = 2.21 *10^{8} \ m/s$

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3 years ago

Two ice skaters, Daniel (mass 70.0 kg) and Rebecca (mass 45.0 kg), are practicing. Daniel stops to tie his shoelace and, while a

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a) $v=7.32m/s$

b) $\alpha =-35$ º

c) $ΔK=-1094.62J$

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From the exercise we know that the collision between Daniel and Rebecca is elastic which means they do not stick together

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Now, lets analyze the movement in the vertical direction

$p_{1y}=p_{2y}$

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$0=m_{D}v_{D2y}+m_{R}v_{R2y}$

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