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uranmaximum [27]
3 years ago
15

Suppose you have an unknown clear substance immersed in water, and you wish to identify it by finding its index of refraction. Y

ou arrange to have a beam of light enter it at an angle of 45.7∘, and you observe the angle of refraction to be 30.5∘. What is the index of refraction of the substance? Water has an index of refraction equal to 1.333.
Physics
1 answer:
s344n2d4d5 [400]3 years ago
3 0

Answer: 1.65

Explanation: Snell's law says that index of refraction of the first material times sine theta one equals index of refraction of the second material, which we don't know, multiplied by sine theta two, . So, we'll solve this for N two by dividing both sides by sine theta two. So, N two is N one sine theta one over sine theta two . So, that's 1.33 index of refraction of water times sine of 45.7degrees incidence angle divided by sine of 30.5 degrees angle of refraction, giving us 1.65.

The formula for snell's law is N1*sintheta1=N2*sinetheta2

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If a 51kg snowboarder falls of a cliff, and is falling 15 m/s when they impact the snow, what is the average force of the snow o
storchak [24]

Answer:

Average force is F = mass times change in V/ change in time so..

1 366.07143 N

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5 0
3 years ago
A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft)
Phantasy [73]

Complete Question

A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft), where V0=110 V, f=60 Hz. The power supplied to the bulb is P=V2R J/s (joules per second) and the total energy expended over a time period [0,T] (in seconds) is U  =  \int\limits^T_0 {P(t)} \, dt

Compute U if the bulb remains on for 5h

Answer:

The value is  U  =  7.563 *10^{5} \  J

Explanation:

From the question we are told that

   The power rating of the bulb is P  =  100 \  W

   The resistance is   R =  143 \ \Omega

   The  voltage is  V  =  V_o  sin [2 \pi ft]

   The  energy expanded is U  =  \int\limits^T_0 {P(t)} \, dt

   The  voltage  V_o  =  110 \  V

   The frequency is  f =  60 \  Hz

    The  time considered is  t =  5 \  h  =  18000 \  s

Generally power is mathematically represented as

             P =  \frac{V^2}{ R}

=>          P =  \frac{( 110  sin [2 \pi * 60t])^2}{ 144}

=>           P =  \frac{ 110^2 [ sin [120 \pi t])^2}{ 144}

So  

     U  =  \int\limits^T_0 { \frac{ 110^2*  [sin [120 \pi t])^2}{ 144}} \, dt

=>  U  =  \frac{110^2}{144} \int\limits^T_0 { (   sin^2 [120 \pi t]} \, dt

=>  U =  \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 2 (120\pi t)}{2} } \, dt

=>  U =  \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 240 \pi t)}{2} } \, dt

=>  U =  \frac{110^2}{144} [\frac{t}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi t)}{240 \pi} ] ]\left  | T} \atop {0}} \right.

=>  U =  \frac{110^2}{144} [\frac{t}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi t)}{240 \pi} ] ]\left  | 18000} \atop {0}} \right.

U =  \frac{110^2}{144} [\frac{18000}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi (18000))}{240 \pi} ] ]

=>   U  =  7.563 *10^{5} \  J

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so in conclusion metal is stronger and thicker than non modal things
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