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3 years ago

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The speed of a given wave is 8.96 x10x10x10 m/s, if the frequency of this wave is 254 Hz, what is the wavelength of this wave fr

om compression to compression?

1 answer:

alexdok [17]3 years ago

8 0

Answer: 35.3m (3sigfig)

Explanation:

I would use the equation wave speed= frequency x wavelength

Rearrange to get wavelength=wave speed/ frequency

Plug in your numbers and that's the answer you get

Hope this helps :)

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Ashley decides to enter her pet turtle in a race. She knows her turtle can travel at a rate of 2 meters per hour. The race track

Genrish500 [490]

It would probably be B

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3 years ago

Two point charges each carrying a charge of + 4.5 E - 6 C are located 4.5 meters away from each other. How strong is the electro

BARSIC [14]

Answer:

0.009 N, repulsive

Explanation:

The electrostatic force between two electric charges is given by:

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1 and q2 are the two charges

r is the separation between the two charges

In this problem, we have

$q_1 =q_2 = +4.5\cdot 10^{-6}C$ are the two charges

r = 4.5 m is their separation

Substituting into the equation, we find

$F=(9\cdot 10^9 Nm^2 C^{-2})\frac{(+4.5\cdot 10^{-6} C)(4.5\cdot 10^{-6} C)}{(4.5 m)^2}=0.009 N$

Moreover, the force is repulsive. In fact, the following rules apply:

- When two charges have same sign, they repel each other

- When two charges have opposite signs, they attract each other

7 0

2 years ago

When a carpenter shuts off his circular saw, the 10.0 inch diameter blade slows from 4250 rpm to 0.00 in 4.00 s. (a) What is the

MaRussiya [10]

Answer:

(a) $\alpha=-111.26rad/s$

(b) $s=4450.6in$

(c) $8.66in$

Explanation:

First change the units of the velocity, using these equivalents $1rev=2\pi rad$ and $1 min =60s$

$4250rpm(\frac{2\pi rad}{1rev})(\frac{1 min}{60 s} )=445.06rad/s$

The angular acceleration $\alpha$ the time rate of change of the angular speed $\omega$ according to:

$\alpha=\frac{\Delta \omega}{\Delta t}$

$\Delta \omega=\omega_i-\omega_f$

Where $\omega_i$ is the original velocity, in the case the velocity before starting the deceleration, and $\omega_f$ is the final velocity, equal to zero because it has stopped.

$\alpha=\frac{\Delta \omega}{\Delta t} =\frac{\omega_i-\omega_f}{4}\frac{0-445.06}{4} =\frac{-445.06}{4} =-111.26rad/s$

b) To find the distance traveled in radians use the formula:

$\theta = \omega_i t + \frac{1}{2} \alpha t^2$

$\theta = 445.06 (4) + \frac{1}{2}(-111.26) (4)^2=1780.24-890.12=890.12rad$

To change this result to inches, solve the angular displacement $\theta$ for the distance traveled $s$ ( $r$ is the radius).

$\theta=\frac{s}{r} \\s=\theta r$

$s=890.12(5)=4450.6in$

c) The displacement is the difference between the original position and the final. But in every complete rotation of the rim, the point returns to its original position. so is needed to know how many rotations did the point in the 890.16 rad of distant traveled:

$\frac{890.12}{2\pi}=141.6667$

The real difference is in the 0.6667 (or 2/3) of the rotation. To find the distance between these positions imagine a triangle formed with the center of the blade (point C), the initial position (point A) and the final position (point B). The angle $\gamma=\frac{2\pi}{3}=\frac{360^o}{3}=120$ is between the two sides known. Using the theorem of the cosine we can find the missing side of the the triangle(which is also the net displacement):

$c^2=a^2+b^2-2abcos(\gamma)$

$c^2=5^2+5^2-2(5)(5)cos(\frac{2\pi}{3} )\\c^2=25+25+25\\c^2=75\\c=5\sqrt{3}=8.66in$

4 0

3 years ago

Using Gauss's law, calculate the electric field at a point distance s from a long wire bearing uniform charge density. i need he

11111nata11111 [884]

Answer:

E = 2k $\frac{\lambda}{ r}$

Explanation:

Gauss's law states that the electric flux equals the wax charge between the dielectric permeability.

We must define a Gaussian surface that takes advantage of the symmetry of the problem, let's use a cylinder with the faces perpendicular to the line of charge. Therefore the angle between the cylinder side area has the same direction of the electric field which is radial.

Ф = ∫ E . dA = E ∫ dA = q_{int} /ε₀

tells us that the linear charge density is

λ = q_ {int} /l

q_ {int} = l λ

we substitute

E A = l λ /ε₀

is area of cylinder is

A = 2π r l

we substitute

E = $\frac{ l \ \lambda}{ \epsilon_o \ 2\pi \ r \ l }$

E = $\frac{\lambda}{ 2\pi \epsilon_o \ r}$

the amount

k = 1 / 4πε₀

E = 2k $\frac{\lambda}{ r}$

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2 years ago

How might the velocity (speed) of wind or water affect the deposition of sediments?

Maksim231197 [3]

Well depending on the speed of both of those things is were the rock will be placed and it also determines how fast can an environment change
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6 0

3 years ago