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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?

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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A pulsar is a rapidly rotating neutron star that emits a radio beam the way a lighthouse emits a light beam. We receive a radio

Angelina_Jolie [31]

a) -0.259 rad/s/y

b) 1732.8 years

c) 0.0069698 s

Explanation:

a)

The angular acceleration of a rotating object is equal to the rate of change of angular velocity of the object.

Mathematically, it is given by

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

where

$\Delta \omega$ is the change in angular velocity

$\Delta t$ is the time elapsed

The angular velocity can be written as

$\omega=\frac{2\pi}{T}$

where T is the period of rotation of the object.

Therefore, the change in angular velocity can be written as

$\Delta \omega = \frac{2\pi}{T'}-\frac{2\pi}{T}=2\pi (\frac{1}{T'}-\frac{1}{T})$

In this problem:

T = 0.0140 s is the initial period of the pulsar

The period increases at a rate of 8.09 x 10-6 s/y, so after 1 year, the new period is

$T'=T+8.09\cdot 10^{-6} =0.01400809 s$

Therefore, the change in angular velocity after 1 year is

$\Delta \omega =2\pi (\frac{1}{0.01400809}-\frac{1}{0.0140})=-0.259 rad/s$

So, the angular acceleration of the pulsar is

$\alpha = \frac{-0.259 rad/s}{1 y}=-0.259 rad/s/y$

b)

To solve this part, we can use the following equation of motion:

$\omega'=\omega + \alpha t$

where

$\omega'$ is the final angular velocity

$\omega$ is the initial angular velocity

$\alpha$ is the angular acceleration

t is the time

For the pulsar in this problem:

$\omega=\frac{2\pi}{T}=\frac{2\pi}{0.0140}=448.8 rad/s$ is the initial angular velocity

$\omega'=0$ , since we want to find the time t after which the pulsar stops rotating

$\alpha = -0.259 rad/s/y$ is the angular acceleration

Therefore solving for t, we find the time after which the pulsar stops rotating:

$t'=-\frac{\omega}{\alpha}=-\frac{448.8}{-0.259}=1732.8 y$

c)

As we said in the previous part of the problem, the rate of change of the period of the pulsar is

$\frac{\Delta T}{\Delta t}=8.09\cdot 10^{-6} s/y$

which means that the period of the pulsar increases by

$\Delta T=8.09\cdot 10^{-6} s$

For every year:

$\Delta t=1 y$

From part A), we also know that the current period of the pulsar is

T = 0.0140 s

The current period is related to the initial period of the supernova by

$T=T_0+\frac{\Delta T}{\Delta t}\Delta t$

where $T_0$ is the original period and

$\Delta t=869 y$

is the time that has passed; solving for T0,

$T_0=T-\frac{\Delta T}{\Delta t}\Delta t=0.0140 - (8.09\cdot 10^{-6})(869)=0.0069698 s$

6 0

3 years ago

What physical property of Earth gives rise to the seasons?

Elza [17]

Answer:

Earth's tilted axis causes the seasons. Throughout the year, different parts of Earth receive the Sun's most direct rays. So, when the North Pole tilts toward the Sun, it's summer in the Northern Hemisphere. And when the South Pole tilts toward the Sun, it's winter in the Northern Hemisphere.

Explanation:

8 0

2 years ago

Which astronomer discovered that the planets orbit the sun in oval-shaped paths called ellipses?

Leviafan [203]

The answer is Kepler.

7 0

3 years ago

Read 2 more answers

Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.032.03 times a second. A tack is stuck in the ti

Luda [366]

Answer:

4.55 m/s

Explanation:

The frequency is defined as the number of rotations in one second

So, f = 2.03 Hz

r = 0.357 m

The tangential speed is

v = r ω = r x 2 x 3.14 x f = 0.357 x 2 x 3.14 x 2.03 = 4.55 m/s

6 0

3 years ago

An eagle is flying above a mountain of height 2,700 m above sea level. Calculate the pressure on the eagle if the average densit

nadya68 [22]

Answer:

The correct option is;

(A) 25.8 cmHg

Explanation:

The given parameters are;

The height of the mountain over which the eagle is flying = 2700 m

The average density of air, ρₐ = 1.3 kg/m³

The atmospheric pressure at sea level = 76 cmHg = 0.76 mHg

The density of mercury, $\rho _m$ = 13,600 kg/m³

The formula for pressure, p, is p = ρ·g·h

The pressure in the atmosphere = 13,600 × 0.76 × g = 10,336·g

The equivalent height of air is found as follows;

1.3 × g × h = 10,336·g

h = 10,336/1.3 = 7,950.769 meters

Therefore;

The pressure of the atmosphere at 2,700 meters, p₂₇₀₀ = (2,700/7,950.769) × 76 cm Hg

Which gives;

p₂₇₀₀ = 25.808 cmHg ≈ 25.8 cmHg.

3 0

3 years ago