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Zolol [24]
3 years ago
7

Question below in photo!! Please answer! Will mark BRAINLIEST! ⬇⬇⬇⬇⬇⬇⬇

Physics
1 answer:
Jobisdone [24]3 years ago
3 0
Both frequencies and amplitudes are needed to transfer energy through a water wave ..
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This is physics 11th grade and a homework question I don’t understand how to do this or what the question is asking me
Alexxx [7]

a) Frequency is the number of complete oscillations per second. Looking at the graph, there are 9 complete oscillations in 5 seconds. Thus,

Frequency = 9/5 = 1.8 oscillations per second

Frequency = 1.8 Hz

Period = 1/frequency = 1/1.8

Period = 0.056 s

b) When we differenctiate displacement with respect to time, the result is velocity.

Recall, period = 1/f = 5/9 cycles

1/4 cycle behind = 1/4 x 5/9 = 5/36

It is delayed with 5/36 sec with respect to displacement.

5/36 sec = 0.139 sec

Acceleration = first derivative of velocity = second derivative of displacement = 1/4 cycle behind velocity = 1/2 cycle behind displacement =

5/36 = 0.139 sec delayed with respect to velocity

= 5/18 = 0.2777 secs delayed with respect to displacement

Thus, the number of seconds out of phase with the displacements is 0.278 seconds

c) The formula for calculating the period of an ideal pendulum anywhere is

T = 2π√length/local gravity). We would calculate the local gravity.

From the information given,

length = 0.2

T = P = 5/9

Thus,

5/9 = 2π√0.2/local gravity)

(5/9)/2π = √0.2/local gravity

Square both sides. It becomes

[(5/9)/2π]^2 = 0.2/local gravity

local gravity = 0.2/[(5/9)/2π]^2

local gravity = 25.56 m/s^2

Thus,

acceleration due to gravity = 25.56 m/s^2

Recall, earth's gravity = 9.8 m/s^2

number of g forces = 25.56/9.8

number of g forces = 2.61

6 0
1 year ago
What is the voltage drop across the 10.0 2 resistor?
amid [387]

Answer:

the answer is equal to 2.00v

4 0
3 years ago
If a plane has an airspeed of 40 m/s and is experiencing a crosswind of 30 m/s, what is its ground speed in m/s?
Anon25 [30]
<h2>Answer:50ms^{-1}</h2>

Explanation:

Let v_{a} be the airspeed.

Let v_{w} be the cross wind speed.

We know that,ground speed is the vector sum of airspeed and cross wind speed and airspeed is perpendicular to cross wind speed.

If v_{1} and v_{2} are two perpendicular vectors,the resultant vector has the magnitude \sqrt{|v|_{1}^{2}+|v|_{2}^{2}}

Given,

v_{a}=40ms^{-1}\\v_{c}=30ms^{-1}

So,the ground speed is \sqrt{40^{2}+30^{2}}=\sqrt{2500}=50ms^{-1}

6 0
3 years ago
What kind of process can create "rogue" waves?
algol [13]
The correct answer for the question that is being presented above is this one: "a. Constructive interference." The kind of process that can create "rogue" waves is constructive interference. Constructive interference refers to <span>the </span>interference<span> of two or more waves of equal frequency and phase creating a mutual one.</span>
3 0
3 years ago
Read 2 more answers
The refractive index n of transparent acrylic plastic (full name Poly(methyl methacrylate)) depends on the color (wavelength) of
Novosadov [1.4K]

Answer:

The angle between the blue beam and the red beam in the acrylic block is  

 \theta _d  =0.19 ^o

Explanation:

From the question we are told that

     The  refractive index of the transparent acrylic plastic for blue light is  n_F  =  1.497

     The  wavelength of the blue light is F  =  486.1 nm  =  486.1 *10^{-9} \ m

    The  refractive index of the transparent acrylic plastic for red light is  n_C  =  1.488

       The  wavelength of the red light is C =  656.3 nm  = 656.3 *10^{-9} \  m

    The incidence angle is  i  =  45^o

Generally from Snell's law the angle of refraction of the blue light  in the acrylic block  is mathematically represented as

       r_F =  sin ^{-1}[\frac{sin(i) *  n_a }{n_F} ]

Where  n_a is the refractive index of air which have a value ofn_a =  1

So

     r_F =  sin ^{-1}[\frac{sin(45) *  1 }{ 1.497} ]

      r_F  =  28.18^o

Generally from Snell's law the angle of refraction of the red light in the acrylic block is mathematically represented as

       r_C =  sin ^{-1}[\frac{sin(i) *  n_a }{n_C} ]

Where  n_a is the refractive index of air which have a value ofn_a =  1

So

     r_C =  sin ^{-1}[\frac{sin(45) *  1 }{ 1.488} ]

      r_F  =  28.37^o

The angle between the blue beam and the red beam in the acrylic block

     \theta _d  =  r_C  - r_F

substituting values

       \theta _d  = 28.37 -  28.18

       \theta _d  =0.19 ^o

 

4 0
3 years ago
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