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3 years ago

What does the symbol for Monday mean???

Physics

1 answer:

pantera1 [17]3 years ago

6 0

I believe it means Ice or Freezing temperatures.
Basically just watch for the cold, and have a jacket ready

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What motivates the squire in lines 85- 90?

just olya [345]

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3 years ago

Why is centrifugal force is a fake force?<br>

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2 years ago

Question: How is the energy in the food we eat an example of chemical potential energy?

Elanso [62]

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3 years ago

Which object is most efficient?

Radda [10]

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2 years ago

A transverse sinusoidal wave on a string has a period T = 25.0 ms and travels in the negative x direction with a speed of 30.0 m

lesya [120]

Answer:

a) A =0.021525m

b) $\phi=0.37869rad$

c) $v_{max}=5.4098\frac{m}{s}$

d) $y(x,t)=(0.021525m)cos(\frac{8\pi}{3}x+80\pi t+0.37869)$

Explanation:

1) Notation

A= Amplitude

v= velocity

$\lambda$ = wavelength

k= wave number

$\omega$ = angular frequency

f= frequency

2) Part a and b

The equation of movement for a transverse sinusoidal wave is gyben by (1)

$y(t)=Acos(kx+ \omega t +\phi)$ (1)

At x=0 ,t=0 we have that:

$0.02=Acos(\phi)$

The velocity would be the derivate of the position, so taking the derivate of (1) respect to t we got (2)

$v(t)=-\omega Asin(kx+ \omega t+\phi)$ (2)

And replacing the conditions at x=0, t=0 we got

$-2\frac{m}{s}=-\omega Asin(\phi)$

Now we can find the angular frequency with equation (3)

$\omega =\frac{2\pi}{T}$ (3)

Replacing the values obtained we got:

$\omega =\frac{2\pi}{0.025s}=80\pi \frac{rad}{s}$

From equation (1) we have:

$Acos(\phi)=0.02$ (a)

$-2=-80\pi Asin(\phi)$ (b)

So from condition (b) we have:

$Asin(\phi)=\frac{1}{40\pi}$ (c)

If we divide condition (c) by condition (a) we got:

$\frac{Asin(\phi)}{Acos(\phi)}=tan(\phi)=\frac{1}{0.02x40\pi}=\frac{1}{0.8\pi}=0.39789$

If we solve for $\phi$ we got:

$\phi =tan^{-1}(0.39789)=0.37869$

And now since we have $\phi$ we can find A from equation (a)

$Acos(0.37869)=0.02$

So then Solving for A we got $A=\frac{0.02}{cos(0.37869)}=0.021525$

3) Part c

From equation (2) we can see that the maximum speed occurs when $sin(\omega t+\phi)=1$ , so on this case we have:

$v_{max}=\omega A=80\pi \frac{rad}{s}x0.021525m=5.4098\frac{m}{s}$

4) Part d

On this case we need an equation like (1), and we have everything except the wave number, and we can obtain this from the following expression:

$v=\lambda f=\frac{2\pi}{k}\frac{\omega}{2\pi}=\frac{\omega}{k}$ (4)

And solving for k from equation (4) we got

$k=\frac{\omega}{v}=\frac{80\pi \frac{rad}{s}}{30\frac{m}{s}}=\frac{8\pi}{3}m^{-1}}$

And with the k number we have everythin in order to create the wave function, given by:

$y(x,t)=(0.021525m)cos(\frac{8\pi}{3}x+80\pi t+0.37869)$

7 0

3 years ago