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snow_tiger [21]
2 years ago
8

During an ultrasound, sound waves are sent by a transducer through muscle tissue at a speed of 1300 m/s. Some of the sound waves

are reflected from a metal fragment in the muscle and are detected by the transducer 1.5 x 10−4 seconds after they were first emitted. What is the depth of the metal fragment in the muscle tissue?
Physics
1 answer:
Simora [160]2 years ago
4 0

Answer:

0.195 m

Explanation:

Speed is distance moved per unit time, expressed as s=d/t and making d the subject of the formula then d=st

Where d is distance/depth moved, s is rhe speed of waves and t is time in seconds.

Substituting s with 1300 m/s and t with 0.00015 s then the depth of metal segment will be

D=1300*0.00015=0.195 m

Therefore, the depth is equivalent to 0.195 m

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It is simply called Entropy.

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3 years ago
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The block in the figure below has a mass of 5.1 kg and it rests on an incline of angle . You pull on the rope with a force F = 3
viktelen [127]

42.9°

Explanation:

Let's assume that the x-axis is aligned with the incline and the positive direction is up the incline. We can then apply Newton's 2nd law as follows:

x:\;\;\;\;F - mg\sin{\theta} = 0\;\;\;\;

\Rightarrow mg\sin{\theta} = F

Note that the net force is zero because the block is moving with a constant speed when the angle of the incline is set at \theta. Solving for the angle, we get

\sin{\theta} = \dfrac{F}{mg}

or

\theta = \sin^{-1}\left(\dfrac{F}{mg}\right)

\;\;\;=  \sin^{-1}\left[\dfrac{34\:\text{N}}{(5.1\:\text{kg})(9.8\:\text{m/s}^2)}\right]

\;\;\;=42.9°

6 0
2 years ago
A 2,700-kg truck runs into the rear of a 1,000-kg car that was stationary. The truck and car are locked together after the colli
Leto [7]

Answer:

6200 J

Explanation:

Momentum is conserved.

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

The car is initially stationary.  The truck and car stick together after the collision, so they have the same final velocity.  Therefore:

m₁ u₁ = (m₁ + m₂) v

Solving for the truck's initial velocity:

(2700 kg) u = (2700 kg + 1000 kg) (3 m/s)

u = 4.11 m/s

The change in kinetic energy is therefore:

ΔKE = ½ (m₁ + m₂) v² − ½ m₁ u²

ΔKE = ½ (2700 kg + 1000 kg) (3 m/s)² − ½ (2700 kg) (4.11 m/s)²

ΔKE = -6200 J

6200 J of kinetic energy is "lost".

3 0
3 years ago
Starting from rest, a 6.79 kg block slides 2.82 m down a rough 20.7 ◦ incline. The coefficient of kinetic friction between the b
Veronika [31]

Answer:

23.52092 J

Explanation:

m = Mass of block = 6.79 kg

s = Sliding distance = 2.82 m

\theta = Angle of slide = 20.7°

\mu = Coefficient of kinetic friction = 0.425

g = Acceleration due to gravity = 9.8 m/s²

Work done by the force of gravity is given by

W=mgsin\theta\\\Rightarrow W=6.79\times 9.8\times sin20.7\\\Rightarrow W=23.52092\ J

The work done by the force of gravity is 23.52092 J

8 0
2 years ago
A steel cable has a cross-sectional area 2.54 10-3 m2 and is kept under a tension of 1.01 104 N. The density of steel is 7860 kg
ElenaW [278]

Answer:

The speed is equals to 22.49 m/s

Explanation:

Given Data:

Area = A=2.54*10^-^3m^2\\Force = F = 1.01*10^4N\\density = p = 7860 kg/m^3

Required:

Speed of Traverse wave = c =?

Solution:

As we know that

p=m/V\\\\ p=m/(L*A)\\p*A=m/L

Now the equation for speed of traverse wave is calculated through:

\sqrt \frac{F*L}{m}\\

=\sqrt\frac{F}{m/L} \\\sqrt{} \frac{F}{p*A}

Substituting the values

\sqrt\frac{1.01*10^4}{7860*2.54*10^-^3}  \\

=22.49 m/s

4 0
3 years ago
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