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azamat
3 years ago
7

To practice Problem-Solving Strategy 15.1 Mechanical Waves. Waves on a string are described by the following general equation y(

x,t)=Acos(kx−ωt). A transverse wave on a string is traveling in the +x direction with a wave speed of 8.25 m/s , an amplitude of 5.50×10−2 m , and a wavelength of 0.540 m . At time t=0, the x=0 end of the string has its maximum upward displacement. Find the transverse displacement y of a particle at x = 1.51 m and t = 0.150 s .
Physics
1 answer:
NeX [460]3 years ago
4 0

Answer:

The transverse displacement is   y(1.51 , 0.150) = 0.055 m    

Explanation:

 From the question we are told that

     The generally equation for the mechanical wave is

                    y(x,t) = Acos (kx -wt)

     The speed of the transverse wave is v = 8.25 \ m/s

     The amplitude of the transverse wave is A = 5.50 *10^{-2} m

     The wavelength of the transverse wave is \lambda = 0540 m

      At t= 0.150s , x = 1.51 m

 The angular frequency of the wave is mathematically represented as

          w = vk

Substituting values  

         w = 8.25 * 11.64

        w = 96.03 \ rad/s

The propagation constant k is mathematically represented as

                  k = \frac{2 \pi}{\lambda}

Substituting values

                  k = \frac{2 * 3.142}{0. 540}

                   k =11.64 m^{-1}

Substituting values into the equation for mechanical waves

      y(1.51 , 0.150) = (5.50*10^{-2} ) cos ((11.64 * 1.151 ) - (96.03  * 0.150))

       y(1.51 , 0.150) = 0.055 m    

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Answer:

Titan takes 11.634 times longer to orbit Saturn as compared to Enceladus.

Explanation:

We have been given that the average distance of Enceladus from Saturn is 238,000 km; the average distance of Titan from Saturn is 1,222,000 km.

We will use Kepler's Law to solve our given problem.

\frac{(T_1)^2}{(T_2)^2}=\frac{(r_1)^3}{(r_2)^3}

Upon substituting our given values, we will get:

\frac{(T_1)^2}{(T_2)^2}=\frac{(238,000)^3}{(1,222,000)^3}

\frac{(T_1)^2}{(T_2)^2}=\frac{13481272000000000}{1824793048000000000}

\frac{(T_1)^2}{(T_2)^2}=\frac{13481272}{1824793048}

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Taking square root of both sides, we will get:

\frac{T_1}{T_2}=\sqrt{0.0073878361246365}

\frac{T_1}{T_2}=0.0859525225030452495

\frac{T_2}{T_1}=\frac{1}{0.0859525225030452495}

\frac{T_2}{T_1}=11.634329870476699\approx 11.634

T_2=11.634\cdot T_1

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What should you do if you discover food is being stored at the wrong temperatures?
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A rock is thrown upward from a bridge into a river below. The function f(t)=−16t2+44t+138 determines the height of the rock abov
Likurg_2 [28]

Answer:

a) 138 ft

b) 4.62 s

c) 1.375 s

d) 168.25 ft

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The height of a rock (thrown from the top of a bridge) above the level of water surface as it varies with time when thrown is given in the question as

h = f(t) = -16t² + 44t + 138

with t in seconds, and h in feet

a) The bridge's height above the water.

At t=0 s, the rock is at the level of the Bridge's height.

At t = 0,

h = 0 + 0 + 138 = 138 ft

b) How many seconds after being thrown does the rock hit the water?

The rock hits the water surface when h = 0 ft. Solving,

h = f(t) = -16t² + 44t + 138 = 0

-16t² + 44t + 138 = 0

Solving this quadratic equation,

t = 4.62 s or t = -1.87 s

Since time cannot be negative,

t = 4.62 s

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At maximum height or at the maximum of any function, the derivative of that function with respect to the independent variable is equal to 0.

At maximum height,

(dh/dt) = f'(t) = (df/dt) = 0

h = f(t) = -16t² + 44t + 138

(dh/dt) = (df/dt) = -32t + 44 = 0

32t = 44

t = (44/32)

t = 1.375 s

d) What is the rock's maximum height above the water?

The maximum height occurs at t = 1.375 s,

Substituting this for t in the height equation,

h = f(t) = -16t² + 44t + 138

At t = 1.375 s, h = maximum height = H

H = f(1.375) = -16(1.375²) + 44(1.375) + 138

H = 168.25 ft

Hope this Helps!!!

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The mean free path of a helium atom in helium gas at standard temperature and pressure is 0.2 um.What is the radius of the heliu
ivolga24 [154]

Answer: 0.10233nm

Explanation:

The mean free path \lambda   of an atom is given by the following formula:

\lambda=\frac{RT}{\sqrt{2} \pi d^{2}N_{A}P}    (1)

Where:

\lambda=0.2\mu m=0.2(10)^{-6}m

R=8.3145J/mol.K is the Universal gas constant

T=0\°C=273.115K is the absolute standard temperature

d is the diameter of the helium atom

N_{A}=6.0221(10)^{23}/mol is the Avogadro's number

P=1atm=101.3kPa=101.3(10)^{3}Pa=101.3(10)^{3}J/m^{3} absolute standard pressure

Knowing this, let's find d from (1), in order to find the radius r of the helium atom:

d=\sqrt{\frac{RT}{\sqrt{2}\pi\lambda N_{A}P}}    (2)

d=\sqrt{\frac{(8.3145J/mol.K)(273.115K)}{\sqrt{2}\pi(0.2(10)^{-6}m)(6.0221(10)^{23}/mol)(101.3(10)^{3}J/m^{3})}}    (3)

d=2.0467(10)^{-10}m    (4)

If the radius is half the diameter:

r=\frac{d}{2}  (5)

Then:

r=\frac{2.0467(10)^{-10}m}{2}  (6)

r=1.0233(10)^{-10}m  (7)

However, we were asked to find this radius in nanometers. Knowing 1nm=(10)^{-9}m:

r=1.0233(10)^{-10}m.\frac{1nm}{(10)^{-9}m}=0.10233nm  (8)

Finally:

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