Physics, 27.07.2019 15:30, hey840

What is the wavelength of a 2.0 mhz ultrasound wave traveling through aluminum? assume that the speed of sound in aluminum is 5100 m/s. express your answer using two significant figures. λ = 2.6 mm submitprevious answers correct part b what frequency of electromagnetic wave would have the same wavelength as the ultrasound wave of part a?

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answered: zayyyyy70

The wavelength and the frequency is mathematically given as

a)w= 2.55 mm

b)f=0.13 MHz

Wavelength and frequency

Question Parameters:

the wavelength of a 2.0 mhz ultrasound wave traveling through aluminum

. λ = 2.6 mm submitprevious

the speed of sound in aluminum is 5100 m/s.

Generally the equation for the wavelength is mathematically given as

w=v/f

Therefore

w= 5100 / 2*10^6

w= 2.55 mm

f=v/w

Therefore

f= 333 / 0.00255

f=0.13 MHz

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answered: izzy123abc

(a) The relationship between wavelength $\lambda$ , frequency f and velocity v of a wave is
$v=\lambda f$
For the sound wave of our problem, the frequency is $f=2.0 MHz=2\cdot 10^6 Hz$ while the velocity is v=5100 m/s

, so its wavelength is
$\lambda = \frac{v}{f} = \frac{5100 m/s}{2 \cdot 10^6 Hz} =2.6 \cdot 10^{-3}m=2.6 mm$

(b) The only difference here is the velocity, because the speed of an electromagnetic wave is equal to the speed of light:
$v=c=3 \cdot 10^8 m/s$
So, if we use the wavelength we found in the previous part, we get the frequency of the electromagnetic wave
$f= \frac{c}{\lambda}= \frac{3\cdot 10^8 m/s}{2.6 \cdot 10^{-3}m}=1.5 \cdot 10^{11}Hz$

answered: Guest

d. air pressure decreases as altitude increases.

answered: Guest

Answer: a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same sign as the -4.0 µc charge.======electricstatic force between two point charges is given by $f = \dfrac{kq_1q_2}{r^2}$ where k is coulomb's constant, 8.99×10⁹ n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ = 6.0 µc, q₂ = -4.0 µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1 µc = 10⁻⁶ c) $f_{01} = \dfrac{kq_0q_1}{(\text{6.0 m})^2}$ the force from q₂ $f_{02} = \dfrac{kq_0q_2}{(\text{15 m})^2}$ the force from q₃ $f_{03} = \dfrac{kq_0q_1}{(\text{26 m})^2}$ the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃ $\begin{aligned} f_{net} & = 0\\ f_{01} + f_{02} + f_{03} & = 0 \\ \dfrac{kq_0q_1}{(\text{6.0 m})^2} + \dfrac{kq_0q_2}{(\text{15 m})^2} + \dfrac{kq_0q_3}{(\text{26 m})^2} & = 0 \\ kq_0 \left( \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2}\right) & = 0 \\ \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2} & = 0 \end{aligned}$ isolating: $\begin{aligned} \dfrac{q_3}{(\text{26 m})^2} & = -\dfrac{q_1}{(\text{6.0 m})^2} - \dfrac{q_2}{(\text{15 m})^2} \\ q_3 & = -(\text{26 m})^2\left(\dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2}\right) \\ q_3 & = -(\text{26 m})^2\left(\dfrac{6.0\times10^{-6} \text{ c}}{(\text{6.0 m})^2} + \dfrac{-4.0\times10^{-6} \text{ c}}{(\text{15 m})^2}\right) \\ q_3 & = -1.0064888 \times 10^{-4}\text{ c} \\ q_3 & = -1.0\times 10^{-4}\text{ c} \end{aligned}$ a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same as the -4.0 µc charge