Ultrasonic imaging is made possible due to the fact that a sound wave is partially reflected whenever it hits a boundary between two materials with different densities within the body. the percentage of the wave reflected when traveling from material 1 into material 2 is r=(ρ1−ρ2ρ1+ρ2)2. knowing this, why does the technician apply ultrasound gel to the patient before beginning the examination?

Answers

Answer 1

Answer:

The gel that is applied before ultrasonic imaging is a conducting material. It acts as a medium between transducer and skin. The ultrasonic waves easily transmit from the probe to the tissues because of gel. A tight bond is created between the probe and skin layer and the gel acts as a coupling agent. The density of the gel is similar to the skin layer. This reduces the attenuation of the waves. A thin layer of gel is applied which fills the air gaps and helps in transmission of waves to the tissues. Hence, the technician apply ultrasound gel to the patient before beginning the examination

Answer 2

Answer:

The gel has a density similar to that of skin, so very little of the incident ultrasonic wave is lost by reflection.

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An object of mass m moves to the right with a speed v. It collides head-on with an object of mass 3m moving with speed v/3 in the opposite direction. If the two objects stick together, what is the speed of the combined object, of mass 4m, after the collision

Answers

The speed of the combined object after collision is 0 m/s.

Total momentum before collision = total momentum after collision

m₁u₁ + m₂u₂ = (m₁ + m₂)a

m₁ = object 1 mass = m, u₁ = velocity of object 1 before collision = v, m₂ = mass of object 2 = 3m, u₂ = velocity of object 2 before collision = -v/3, a = velocity after collision

mv + 3m(-v/3) = (m + 3m)a

mv - mv = 4ma

0 = 4ma

a = 0 m/s

The speed of the combined object after collision is 0 m/s.

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Answer:

the answer is 0 m/s

Explanation:

This question is describing the law of conservation of momentum

First object has mass =m

velocity of first object = v

second object = 3m

velocity of second object = v/3

the law of conservation of momentum is expressed as

m1V1 - m2V2 = (m1+ m2) V

substituting the parameters given;

making V as the subject of formular

V = $(m_(1 ) V_(1) -m_(2)V_(2) )/(m_(1)+m_(2) )$

V =

$(mV - 3mv/3)/(m+ 3m)$

V = $(0)/(4m)$

= 0 m/s

Which correctly describes how the energy of a wave on the electromagnetic spectrum depends on wavelength and frequency?A.
Energy decreases with decreasing wavelength and decreasing frequency.
B.
Energy increases with decreasing wavelength and increasing frequency.
C.
Energy increases with decreasing wavelength and decreasing frequency.
D.
Energy decreases with increasing wavelength and increasing frequency.

Answers

Answer:

B. Energy increases with decreasing wavelength and increasing frequency.

Explanation:

A 2.13-kg object on a frictionless horizontal track is attached to the end of a horizontal spring whose force constant is 5.00 N/m. The object is displaced 3.54 m to the right from its equilibrium position and then released, initiating simple harmonic motion. (a) What is the force (magnitude and direction) acting on the object 3.50 s after it is released

Answers

Answer:

17.54N in -x direction.

Explanation:

Amplitude (A) = 3.54m

Force constant (k) = 5N/m

Mass (m) = 2.13kg

Angular frequency ω = √(k/m)

ω = √(5/2.13)

ω = 1.53 rad/s

The force acting on the object F(t) = ?

F(t) = -mAω²cos(ωt)

F(t) = -2.13 * 3.54 * (1.53)² * cos (1.53 * 3.50)

F(t) = -17.65 * cos (5.355)

F(t) = -17.57N

The force is 17.57 in -x direction

A syringe containing 12.0 mL of dry air at 25 C is placed in a sterilizer and heated to 100.0 C. The syringe is sealed, but the plunger can move and the volume can change. What is the volume of the air in the syringe at 100.0 C, assuming no change in pressure?

Answers

Answer:

15.01 Liters

Explanation:

T₁ = Initial temperature = 25°C = 298.15 K

T₂ = Final temperature = 100°C = 373.15 K

V₁ = Initial volume = 12 mL

Here, pressure is constant so we apply Charles Law

$(V_1)/(T_1)=(V_2)/(T_2)\n\Rightarrow {V_2}=(V_1)/(T_1)* T_2\n\Rightarrow {V_2}=(12)/(298.15)* 373.15\n\Rightarrow {V_2}=15.01 L$

∴ Final volume at 100°C is 15.01 Liters.

A 60.0-kg boy is surfing and catches a wave which gives him an initial speed of 1.60 m/s. He then drops through a height of 1.57 m, and ends with a speed of 8.50 m/s. How much nonconservative work was done on the boy

Answers

Answer:

Work = 1167.54 J

Explanation:

The amount of non-conservative work here can be given by the difference in kinetic energy and the potential energy. From Law of conservation of energy, we can write that:

Gain in K.E = Loss in P.E + Work

(0.5)(m)(Vf² - Vi²) - mgh = Work

where,

m = mass of boy = 60 kg

Vf = Final Speed = 8.5 m/s

Vi = Initial Speed = 1.6 m/s

g = 9.8 m/s²

h = height drop = 1.57 m

Therefore,

(0.5)(60 kg)[(8.5 m/s)² - (1.6 m/s)²] - (60 kg)(9.8 m/s²)(1.57 m) = Work

Work = 2090.7 J - 923.16 J

Work = 1167.54 J

Two parallel wires carry currents in the same direction. If the currents in the wires are 1A and 4A and the wires are 5 m apart. Calculate the force between the 2 wires. Is it attractive or repulsive? Calculate the magnetic field midway between the wires.