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garik1379 [7]
2 years ago
7

 An ultrasound pulse used in medical imaging has a frequency of 5 MHz and a pulse width of 0.5 us. Ap- proximately how many osci

llations of the sound wave occur in the pulse? The number of oscillations is sometimes called the quality, Q, of the pulse. A pulse with little damping has Q 1, whereas a heavily damped pulse has e 1. Is the ultrasound pulse heavily damped?
Physics
1 answer:
podryga [215]2 years ago
4 0

At the frequency of 5 MHz, the period of the oscillations is 1/5meg. That's a period of 1/5 microsecond.

There are 5 full cycles in one full microsecond, and there are 2.5 full cycles in a 0.5 us pulse.

You'll have to decide for yourself how damped a pulse of 2.5 cycles is, because the parameters of the definition are corrupted in the question.

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True or false:slow movements of mantle rock called radiation transfer heat in the mantle.
RUDIKE [14]
The best answer between the two options would be the second choice B) FALSE.
8 0
3 years ago
The figure shows a 100-kg block being released from rest from a height of 1.0 m. It then takes it 0.90 s to reach the floor. Wha
Anna007 [38]

Answer:

The mass of the another block is 60 kg.

Explanation:

Given that,

Mass of block M= 100 kg

Height = 1.0 m

Time = 0.90 s

Let the mass of the other block is m.

We need to calculate the acceleration of each block

Using equation of motion

s=ut+\dfrac{1}{2}at^2

Put the value into the formula

1.0=0+\dfrac{1}{2}\times a\times(0.90)^2

a=\dfrac{2\times1.0}{(0.90)^2}

a=2.46\ m/s^2

We need to calculate the mass of the other block

Using newton's second law

The net force of the block M

Ma=Mg-T

T=Mg-Ma....(I)

The net force of the block m

ma=T-mg

Put the value of T from equation (I)

ma=Mg-Ma-mg

m(a+g)=M(g-a)

m=\dfrac{M(g-a)}{(a+g)}

Put the value into the formula

m=\dfrac{100(9.8-2.46)}{2.46+9.8}

m=59.8\ \approx60\ kg

Hence, The mass of the another block is 60 kg.

8 0
2 years ago
The altitude of the International Space Station ttt minutes after its perigee (closest point), in kilometers, is given by \qquad
Dmitriy789 [7]

Answer:

T = 92.8 min

Explanation:

Given:

The altitude of the International Space Station t minutes after its perigee (closest point), in kilometers, is given by:

                               A(t) = 415 - sin(\frac{2*\pi (t+23.2)}{92.8})

Find:

- How long does the International Space Station take to orbit the earth? Give an exact answer.

Solution:

- Using the the expression given we can extract the angular speed of the International Space Station orbit:

                                 A(t) = 415 - sin({\frac{2*\pi*t }{92.8} + \frac{23.2*2*\pi }{92.8} )

- Where the coefficient of t is angular speed of orbit w = 2*p / 92.8

- We know that the relation between angular speed w and time period T of an orbit is related by:

                                T = 2*p / w

                                T = 2*p / (2*p / 92.8)

Hence,                     T = 92.8 min

7 0
2 years ago
Calculate the speed of a proton after it accelerates from rest through a potential difference of 350 V.
AVprozaik [17]

The speed of a proton after it accelerates from rest through a potential difference of 350 V is 25.86 \times 10^4 ~m/s.

Initial velocity of the proton u = 0

Given potential difference \Delta V = 350V

let's assume that the speed of the proton is v,

Since the proton is accelerating through a potential difference, proton's potential energy will change with time. The potential energy of a particle of charge q when accelerated with a potential difference \Delta V is,

    U = q \Delta V

Due to Work-Energy Theorem and Conservation of Energy - <em>If there is no non-conservative force acting on a particle then loss in Potential energy  P.E must be equal to gain in Kinetic Energy K.E</em> i.e

\Delta K = \Delta V

If the initial and final velocity of the proton is u and v respectively then,

change in Kinetic Energy  \implies  \Delta K = \frac{1}{2}mv^2 -\frac{1}{2}mu^2 = \frac{1}{2}mv^2 - 0

change in Potential Energy \implies \Delta U = q\Delta V

from conservation of energy,

             v= \sqrt{\frac{2q\Delta V}{m}}

so,         v = \sqrt{\frac{2\times 350 \times 1.6\times 10^{-19}}{1.67 \times 10^{-27}}

                = 25.86 \times 10^4 ~m/s

To read more about the conservation of energy, please go to brainly.com/question/14668053

7 0
1 year ago
. Two identical vehicles traveling at the same speed are made to collide with barriers in an insurance company collision test. T
Serggg [28]

Answer:

F₁ / F₂ = 10

therefore the first out is 10 times greater than the second barrier

Explanation:

For this exercise let's use the relationship between momentum and momentum.

         I = F t = Δp

in this case the final velocity is zero

        F t = 0 -m v₀

        F = m v₀ / t

in order to answer the question we must assume that the two vehicles have the same mass and speed

concrete barrier

        F₁ = -p₀ / 0.1

        F₁ = - 10 p₀

barrier collapses

         F₂ = -p₀ / 1

let's look for the relationship of the forces

        F₁ / F₂ = 10

therefore the first out is 10 times greater than the second barrier

5 0
2 years ago
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