28.04.2020

You’re in a mall and you need some money to buy a nose ring, but you’re broke. you decide to scoop some quarters out of the fountain. the water in the fountain is one foot deep. how far below the water do the quarters appear to be? the index of refraction of water is 4/3.

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Step-by-step answer

30.05.2023, solved by verified expert
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0.75 ft

Explanation:

This situation is due to Refraction, a phenomenon in which a wave (the light in this case) bends or changes its direction when passing through a medium with an index of refraction different from the other medium.  In this context, the index of refraction is a number that describes how fast light propagates through a medium or material.  

In addition, we have the following equation that states a relationship between the apparent depth You’re in a mall and you need some money to buy, №16483577, 28.04.2020 09:24 and the actual depth You’re in a mall and you need some money to buy, №16483577, 28.04.2020 09:24:  

You’re in a mall and you need some money to buy, №16483577, 28.04.2020 09:24(1)  

Where:

You’re in a mall and you need some money to buy, №16483577, 28.04.2020 09:24 is the air's index of refraction  

You’re in a mall and you need some money to buy, №16483577, 28.04.2020 09:24 water's index of refraction.

You’re in a mall and you need some money to buy, №16483577, 28.04.2020 09:24 is the actual depth of the quarters

Now. when You’re in a mall and you need some money to buy, №16483577, 28.04.2020 09:24 is smaller than You’re in a mall and you need some money to buy, №16483577, 28.04.2020 09:24 the apparent depth is smaller than the actual depth. And, when You’re in a mall and you need some money to buy, №16483577, 28.04.2020 09:24 is greater than You’re in a mall and you need some money to buy, №16483577, 28.04.2020 09:24 the apparent depth is greater than the actual depth.  

Let's prove it:

You’re in a mall and you need some money to buy, №16483577, 28.04.2020 09:24(2)  

Finally we find the aparent depth of the quarters, which is smaller than the actual depth:

You’re in a mall and you need some money to buy, №16483577, 28.04.2020 09:24

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Physics
Step-by-step answer
P Answered by PhD

0.75 ft

Explanation:

This situation is due to Refraction, a phenomenon in which a wave (the light in this case) bends or changes its direction when passing through a medium with an index of refraction different from the other medium.  In this context, the index of refraction is a number that describes how fast light propagates through a medium or material.  

In addition, we have the following equation that states a relationship between the apparent depth {d}^{*} and the actual depth d:  

{d}^{*}=d\frac{{n}_{1}}{{n}_{2}}(1)  

Where:

n_{1}=1 is the air's index of refraction  

n_{2}=\frac{4}{3}=1.33 water's index of refraction.

d=1 ft is the actual depth of the quarters

Now. when n_{1} is smaller than n_{2} the apparent depth is smaller than the actual depth. And, when n_{1} is greater than n_{2} the apparent depth is greater than the actual depth.  

Let's prove it:

{d}^{*}=1 ft\frac{1}{1.33}(2)  

Finally we find the aparent depth of the quarters, which is smaller than the actual depth:

{d}^{*}=0.75 ft

Physics
Step-by-step answer
P Answered by Specialist
Answer: Option B and C are True.

Explanation:
The weight of the two blocks acts downwards.
Let the weight of the two blocks be W. Solving for T₁ and T₂:
w = T₁/cos 60° -----(1);
w = T₂/cos 30° ----(2);
equating (1) and (2)
T₁/cos 60° = T₂/cos 30°;
T₁ cos 30° = T₂ cos 60°;
T₂/T₁ = cos 30°/cos 60°;
T₂/T₁ =1.73.
Therefore, option a is false since T₂ > T₁.
Option B is true since T₁ cos 30° = T₂ cos 60°.
Option C is true because the T₃ is due to the weight of the two blocks while T₄ is only due to one block.
Option D is wrong because T₁ + T₂ > T₃ by simple summation of the two forces, except by vector addition.
Answer: Option B and C are True.

Explanation:  
The weight of the two blocks acts downwards.
Le
Physics
Step-by-step answer
P Answered by PhD
Answer:
7.25 secs.

Explanation:
First find the distance it takes to stop
s = [v^2-u^2]/2a = 0^2 - 8.7^2/2[-2.4] = 8.7^2/4.8
Next find the time it takes to go that distance , s = ut +[1/2] at^2
8.7^2/4.8 = 8.7t +[1/2] [ -2.4]t^2 , rearrange and
t^2 -[8.7/1.2]+ 8.7^2/[(1.2)(4.8)]=0 complete the square
[t - (8.7/2.4)]^2=0
t = 8.7/2.4 = 3.625 secs
At this stage the deceleration will push the object back in the direction it came from for another 3.625 secs when it will be 8.7 m/s again
Total time , T =2t = 7.25 secs.

Note:
The term differential is used in calculus to refer to an infinitesimal (infinitely small) change in some varying quantity. For example, if x is a variable, then a change in the value of x is often denoted Δx (pronounced delta x). The differential dx represents an infinitely small change in the variable x.
Physics
Step-by-step answer
P Answered by PhD
The change in temperature is 9.52°CExplanation:Since, the heat supplied by the electric kettle is totally used to increase the temperature of the water.Thus, from the law of conservation of energy can be stated as:Heat Supplied by Electric Kettle = Heat Absorbed by WaterHeat Supplied by Electric Kettle = m C ΔTwhere,Heat Supplied by Electric Kettle = 20,000 JMass of water = m = 0.5 kgSpecific Heat Capacity of Water = C = 4200 J/kg.°CChange in Temperature of Water = ΔTTherefore,20,000 J = (0.5 kg)(4200 J/kg.°C) ΔTΔT = 20,000 J/(2100 J/°C)ΔT = 9.52°C
Physics
Step-by-step answer
P Answered by PhD
Weight of barbell (m) = 100 kg
Uplifted to height (h) = 2m
Time taken= 1.5 s
Work done by Jordan = potential energy stored in barbell = mgh
= 100×2×9.8
= 1960J
Power = energy/time
= 1960/1.5
1306.67watts
Physics
Step-by-step answer
P Answered by PhD
Weight of jasmine (m) = 400 N
Height climbed on wall (h) = 5m
Total time taken in climbing = 5 sec
Work done in climbing the wall = rise in potential energy = mgh
= 400×9.8×51
= 19600J
Power generated by Jasmine = potential energy / time
= 19600/5
= 3920Watts
Physics
Step-by-step answer
P Answered by PhD
The horizontal and vertical motions of balloons are independent from each other.
Let vertical component of initial velocity U' horizontal component of initial velocity U"
Time of landing (t) is found with the help of vertical motion.
Since vertical component of initial velocity of balloon is zero(U' = 0)
From equation h = U't + 1/2gt^2
h = 1/2gt^2
t = √(2h/g)
t = √( 2×150/9.8)
t = 5.53 sec
Horizontal velocity = 50m/s
Horizontal range of balloon, R = U"t
= 50× 5.53
= 27.65m
So the balloon will go 27.65 metre away from the bridge

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