What is the awnser to Blank occurs when a substance changes from a liquid to a gas

Diagnostic ultrasound of frequency 3.45 MHz is used to examine tumors in soft tissue.

Mekhanik [1.2K]

Answer:

4.35×10⁻⁴ m

Explanation:

(a)

From wave,

v = λf...................... Equation 1

Where v = speed of sound in air, λ = wavelength of sound, f = frequency of sound.

Make λ the subject of formula in equation 1

λ = v/f................. Equation 2

Given: v = 343, f = 3.45 MHz = 3.45×10⁶ Hz

Substitute into equation 2

λ = 343/(3.45×10⁶)

λ = 99.42×10⁻⁶ m

λ = 9.942×10⁻⁵ m

(b)

using,

v' = λ'f............... Equation 3

Where v' = speed of sound in tissue, λ' = wavelength of sound in tissue.

make λ' the subject of the equation

λ' = v'/f......................Equation 4

Given: v' = 1500 m/s, f = 3.45 MHz = 3.45×10⁶ Hz

Substitute into equation 4

λ' = 1500/(3.45×10⁶)

λ' = 434.783×10⁻⁶

λ' ≈ 4.35×10⁻⁴ m.

6 0

3 years ago

The volume of water in the Pacific Ocean is about 7.00 × 108 km3. The density of seawater is about 1030 kg/m3. For the sake of t

oee [108]

The concepts used to solve this exercise are given through the calculation of distances (from the Moon to the earth and vice versa) as well as the gravitational potential energy.

By definition the gravitational potential energy is given by,

$PE=\frac{GMm}{r}$

Where,

m = Mass of Moon

G = Gravitational Universal Constant

M = Mass of Ocean

r = Radius

First we calculate the mass through the ratio given by density.

$m = \rho V$

$m = (1030Kg/m^3)(7*10^8m^3)$

$m = 7.210*10^{11}Kg$

PART A) Gravitational potential energy of the Moon–Pacific Ocean system when the Pacific is facing away from the Moon

Now we define the radius at the most distant point

$r_1 = 3.84*10^8 + 6.4*10^6 = 3.904*10^8m$

Then the potential energy at this point would be,

$PE_1 = \frac{GMm}{r_1}$

$PE_1 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.904*10^8}$

$PE_1 = 9.05*10^{15}J$

PART B) when Earth has rotated so that the Pacific Ocean faces toward the Moon.

At the nearest point we perform the same as the previous process, we calculate the radius

$r_2 = 3.84*10^8-6.4*10^6 - 3.776*10^8m$

The we calculate the Potential gravitational energy,

$PE_2 = \frac{GMm}{r_2}$

$PE_2 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.776*10^8}$

$PE_2 = 9.361*10^{15}J$

7 0

4 years ago

A small bulb is rated at 7.5 W when operated at 125 V. The tungsten filament has a temperature coefficient of resistivity α = 0.

alisha [4.7K]

To solve this problem we will apply the concepts related to resistance as a function of temperature, product of the relationship between the squared voltage and the power. Mathematically this is,

$R = \frac{v^2}{P}$