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Anna71 [15]
3 years ago
10

Is the displacement D(x,t)=cx2+dt2, where c and d are constants, a possible traveling wave?

Physics
1 answer:
Ilya [14]3 years ago
3 0
Ok, Brainly doesn't have the required math symbols I need to give you the best looking answer so we'll have to do this the best we can.  In short, yes it could be a wave.  A simple wave,ψ,  is described by the equation
d²ψ/dx² - (1/c²)d²ψ/dt² = 0.  The right side must be a nonzero constant, which is zero here.  This is a basic wave equation.  It says if the second partial derivative of a function with respect to space minus the second partial derivative of that function with respect to time, multiplied by a constant (1/c² here) equals a constant (0 here) then that function describes a wave.  Brainly doesn't have a partial derivative symbol so I used "d". Also the c in my equation is the speed of the wave.  Not necessarily the same c as in your equation.

Now let's look at your function.  The second partial derivative of it with respect to x makes the t term 0 (it's a constant as far as x is concerned) and the x term becomes 2c.  Doing the same thing for the partial derivative of it for t leaves 2d.  These terms represent two constants, which if subtracted, as the wave equation requires, would lead to a constant on the right, which could be made zero if the coefficients c and d are chosen correctly.  
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if the input force on a wheel and axle is 2 N and the output force is 20N what is its mechanical advantage
Nezavi [6.7K]

Mechanical Advantage is always (Output Force) / (Input Force) .

MA = (20 N) / (2 N)

<em>MA = 10</em>

5 0
3 years ago
A car with bad shock absorbers bounces up and down with a period of 1.50 s after hitting a bump. The car has a mass of 1 500 kg
kkurt [141]

Answer:

k =  26319 N/m

Explanation:

given,

Period of bounce = 1.5 s

mass of the car = 1500 kg

four springs                    

to find the value of k                              

T = 2\pi \sqrt{\dfrac{m}{k}}      

1.5 = 2\pi \sqrt{\dfrac{1500}{k}}

\dfrac{2.25}{4\pi^2}=\dfrac{1500}{k}      

k = \dfrac{4\pi^2 \times 1500}{2.25}      

k =  26319 N/m            

Now the value of force constant is k =  26319 N/m      

6 0
2 years ago
A ball is thrown straight up from a bridge at a speed of 11.0 m/s. If it takes 5.5 seconds to hit the water below, what is the v
seropon [69]

Answer:

 v = 42.92 m/s

Explanation:

Given,

initial speed of the ball, v = 11 m/s

time taken to hit the ground = 5.5 m/s

velocity of the ball just before it hit the ground, v = ?

time taken by the ball to reach the maximum height

using equation of motion

 v = u + at

final velocity = 0 m/s  

 0 = 11 - 9.8 t

  t = 1.12 s.

time taken by the ball to reach the water from the maximum height

 t' - 5.5 -1.12 = 4.38 s

using equation of motion for the calculation of speed just before it hit the water.

 v  = u + a t

 v = 0 + 9.8 x 4.38

 v = 42.92 m/s

Velocity of the ball just before it reaches the water is equal to v = 42.92 m/s

3 0
3 years ago
A group of 25 particles have the following speeds:
svp [43]

Answer:

Part a: The average speed is 24.12 m/s

Part b: The rms speed is 25.55 m/s

Part c: The most probable speed is 17 m/s.

Explanation:

Part a

Average Speed

<em>Average speed is given as </em>

<em />v_{avg}=\frac{\sum}{n}

v_{avg}=\frac{(2 \times 11) +(7 \times 17)+(4 \times 19)+(3 \times 27) +(6 \times 32) +(1 \times 33)+(2 \times 40)  }{25}\\v_{avg}=\frac{22+119+76+81+192+33+80}{25}\\v_{avg}=\frac{603}{25}\\v_{avg}=24.12 m/s

So the average speed is 24.12 m/s

Part b

RMS Speed

v_{rms}=\frac{\sum v^2}{n}

v_{rms}=\sqrt{\frac{(2 \times 11^2) +(7 \times 17^2)+(4 \times 19^2)+(3 \times 27^2) +(6 \times 32^2) +(1 \times 33^2)+(2 \times 40^2)  }{25}} \\v_{rms}=\sqrt{\frac{242+2023+1444+2187+6144+1089+3200}{25}}\\\\v_{rms}=\sqrt{\frac{16329}{25}}\\v_{rms}=\sqrt{653.16}\\v_{rms}=25.55 m/s

So the rms speed is 25.55 m/s

Part c

Most Probable Speed

As 7 particles have speed of 17 m/s i.e. 7 is the highest frequency so 17 m/s is the most probable speed.

3 0
3 years ago
A wave has a frequency of 15,500 Hz and a wavelength of 0.20 m. What is the
Hoochie [10]

Answer:

3100 m/s

Explanation:

The relationship between frequency and wavelength of a wave is given by the wave equation:

v=f\lambda

where

v is the speed of the wave

f is its frequency

\lambda is the wavelength

For the wave in this problem,

f = 15,500 Hz

\lambda=0.20 m

Therefore, the wave speed is

v=(15500)(0.20)=3100 m/s

4 0
3 years ago
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