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3 years ago

I dnt know how to do it

Physics

1 answer:

NNADVOKAT [17]3 years ago

4 0

Here's what you need to know about waves:

Wavelength = (speed) / (frequency)

Now ... The question gives you the speed and the frequency,
but they're stated in unusual ways, with complicated numbers.

Frequency: How many each second ?
The thing that's making the waves is vibrating 47 times in 26.9 seconds.
Frequency = (47) / (46.9 s) = 1.747... per second. (1.747... Hz)

Speed: How far a point on a wave travels in 1 second.
The crest of one wave travels 4.16 meters in 13.7 seconds.
Speed = (4.16 m / 13.7 sec) = 0.304... m/s

Wavelength = (speed) / (frequency)

Wavelength = (0.304 m/s) / (1.747 Hz) = 0.174 meter per second

You might be interested in

Two forces,

serg [7]

First compute the resultant force F:

$\mathbf F_1=(5.90\,\mathbf i-5.60\,\mathbf j)\,\mathrm N$

$\mathbf F_2=(4.65\,\mathbf i-5.55\,\mathbf j)\,\mathrm N$

$\implies\mathbf F=\mathbf F_1+\mathbf F_2=(10.55\,\mathbf i-11.15\,\mathbf j)\,\mathrm N$

Then use Newton's second law to determine the acceleration vector $\mathbf a$ for the particle:

$\mathbf F=m\mathbf a$

$(10.55\,\mathbf i-11.15\,\mathbf j)\,\mathrm N=(2.10\,\mathrm{kg})\mathbf a$

$\mathbf a\approx(5.02\,\mathbf i-5.31\,\mathbf j)\dfrac{\rm m}{\mathrm s^2}$

Let $\mathbf x(t)$ and $\mathbf v(t)$ denote the particle's position and velocity vectors, respectively.

(a) Use the fundamental theorem of calculus. The particle starts at rest, so $\mathbf v(0)=0$ . Then the particle's velocity vector at t = 10.4 s is

$\mathbf v(10.4\,\mathrm s)=\mathbf v(0)+\displaystyle\int_0^{10}\mathbf a(u)\,\mathrm du$

$\mathbf v(10.4\,\mathrm s)=\left((5.02\,\mathbf i-5.31\,\mathbf j)u\,\dfrac{\rm m}{\mathrm s^2}\right)\bigg|_{u=0}^{u=10.4}$

$\mathbf v(10.4\,\mathrm s)\approx(52.2\,\mathbf i-55.2\,\mathbf j)\dfrac{\rm m}{\rm s}$

If you don't know calculus, then just use the formula,

$v_f=v_i+at$

So, for instance, the velocity vector at t = 10.4 s has x-component

$v_{f,x}=0+\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)(10.4\,\mathrm s)=52.2\dfrac{\rm m}{\mathrm s^2}$

(b) Compute the angle $\theta$ for $\mathbf v(10.4\,\mathrm s)$ :

$\tan\theta=\dfrac{-55.2}{52.2}\implies\theta\approx-46.6^\circ$

so that the particle is moving at an angle of about 313º counterclockwise from the positive x axis.

(c) We can find the velocity at any time t by generalizing the integral in part (a):

$\mathbf v(t)=\mathbf v(0)+\displaystyle\int_0^t\mathbf a\,\mathrm du$

$\implies\mathbf v(t)=\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf i+\left(-5.31\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf j$

Then using the fundamental theorem of calculus again, we have

$\mathbf x(10.4\,\mathrm s)=\mathbf x(0)+\displaystyle\int_0^{10.4}\mathbf v(u)\,\mathrm du$

where $\mathbf x(0)=(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m$ is the particle's initial position. So we get

$\mathbf x(10.4\,\mathrm s)=(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m+\displaystyle\int_0^{10.4}\left(\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf i+\left(-5.31\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\right)\,\mathrm du$

$\mathbf x(10.4\,\mathrm s)=(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m+\dfrac12\left(\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf i+\left(-5.31\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf j\right)\bigg|_{u=0}^{u=10.4}$

$\mathbf x(10.4\,\mathrm s)\approx(542\,\mathbf i-570\,\mathbf j)\,\mathrm m$

So over the first 10.4 s, the particle is displaced by the vector

$\mathbf x(10.4\,\mathrm s)-\mathbf x(0)\approx(270\,\mathbf i-283\,\mathbf j)\,\mathrm m-(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m\approx(272\,\mathbf i-287\,\mathbf j)\,\mathrm m$

or a net distance of about 395 m away from its starting position, in the same direction as found in part (b).

(d) See part (c).

3 0

3 years ago

Which of these materials is permeable?

kogti [31]

Correct answer is letter B. sandstone

8 0

3 years ago

Fl-19 in florida, if your pwc is equipped with an engine cut-off lanyard, what must you do with it?

melamori03 [73]

Answer: Attach it to clothing, personal flotation device or the person's person.

Explanation:

This particular Question or problem has to do with the rules and regulations or laws governing the use of Personal Watercraft(PWC) in the state of Florida in the United States of America. Hence, one of the rules that is applicable to the use of boats and waterways or the use of personal watercraft in Florida is that in florida, if ones pwc is equipped with an engine cut-off lanyard the person operating it must ATTACH IT TO THE CLOTHING, PERSONAL FLOATATION DEVICE IR THE OPERATORS' PERSON.

Another rule band the use of flammable personal flotation device with Personal Watercraft(PWC) in the state of Florida. All this rules and guildlines are made to limit or minimize hazards.

3 0

3 years ago

100 grados a revoluciones

djyliett [7]

No. Quatro cientos grads.

6 0

3 years ago

5. A 10 kg ball is traveling at the same speed as a 1 kg ball. Compared to the 10 kg ball, the 1 kg ball has (2 points)

suter [353]

What’s the rest of the question or is that it?

4 0

3 years ago