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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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534.9 N

Explanation:

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Answer:

0.092 m

Explanation:

A charged moving particle immersed in a region with magnetic field follows a circular trajectory at constant speed (uniform circular motion), since the magnetic forces acts perpendicular to the direction of motion of the particle.

Since the magnetic force acts as centripetal force, we can write:

qvB=m\frac{v^2}{r}

where

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v is its velocity

B is the strength of the magnetic field

m is the mass of the particle

r is the radius of the orbit

Solving the equation for r,

r=\frac{mv}{qB}

For the ion of oxygen-16, we have:

m_A=2.66\cdot 10^{-26}kg

q_A = 1.6\cdot 10^{-19}C (it is singly charged)

v_A=2.90\cdot 10^6 m/s

B_A=1.30 T

So the radius of its orbit is

r_A=\frac{m_A v_A}{q_A B_A}=\frac{(2.66\cdot 10^{-26})(2.90\cdot 10^6)}{(1.6\cdot 10^{-19})(1.30)}=0.371 m

For the ion of oxygen-18, we have:

m_B = \frac{18}{16}m_A = 2.99\cdot 10^{-26}kg

q_B = 1.6\cdot 10^{-19}C (it is singly charged)

v_B=2.90\cdot 10^6 m/s

B_B=1.30 T

So the radius of its orbit is

r_B=\frac{m_B v_B}{q_B B_B}=\frac{(2.99\cdot 10^{-26})(2.90\cdot 10^6)}{(1.6\cdot 10^{-19})(1.30)}=0.417 m

After each ion has travelled a semicircle, the separation between the two ions will be twice the difference in their radius, so:

d=2(r_B-r_A)=2(0.417-0.371)=0.092 m

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3 years ago
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Answer:

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Explanation:

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