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morpeh [17]
4 years ago
7

The Doppler effect using ultrasonic waves of frequency 2.25 × 106 Hz is used to monitor the heartbeat of a fetus. A (maximum) be

at frequency of 240 Hz is observed. Assuming that the speed of sound in tissue is 1540 m/s, calculate the maximum velocity of the surface of the beating heart.
Physics
1 answer:
miss Akunina [59]4 years ago
4 0

Answer:

v = 2.88 \times 10^7 m/s

Explanation:

As per Doppler's effect we know that the frequency of the sound that is observed by the detector is the reflected sound

This reflected sound is given as

f' = f (\frac{v - v_h}{v + v_h})

f' = 2.25\times 10^6(\frac{v - 1540}{v + 1540})

so we know that the beat frequency is

\Delta f = 240 Hz

so we will have

f - f' = \Delta f

2.25 \times 10^6 - 2.25\times 10^6(\frac{v - 1540}{v + 1540}) = 240

1 - (\frac{v - 1540}{v + 1540}) = 1.07 \times 10^{-4}

so we have

0.99989 = (\frac{v - 1540}{v + 1540})

1.99989\times 1540 = 1.067 \times 10^{-4} v

v = 2.88 \times 10^7 m/s

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Ocean bulges on Earth would be bigger if the Moon had twice as much mass and yet orbited the planet at the same distance. Option B is correct.

<h3>What is ocean bludge?</h3>

The fluid and moveable ocean water are drawn towards the moon by the gravitational attraction between the moon and the Earth.

The ocean nearest to the moon experiences a bulge as a result, and as the Earth rotates, the affected seas' locations shift.

The Moon's bulges in the oceans would be larger if it had twice the mass and orbited Earth at the same distance.

Hence option B is corect.

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2 years ago
3. Christina is on her college softball team, and she is practicing swinging the bat. Her coach wants her to work on the speed a
inessss [21]

Answer:

The average angular acceleration is  \alpha =125.487 rad /s^2

Explanation:

From the question we are told that

  From the question we are told that

        The length of the bat is l = 0.85m  \

         The initial linear velocity is  u = 0 m/s

         The time is  t = 0.15s

         The velocity at t is  v = 16 m/s

  Generally average  angular acceleration is mathematically represented as

                \alpha  = \frac{w_f - w_o}{t}

        Where w_f is the finial angular velocity which is mathematically evaluated as  

            w_f = \frac{v}{l}

                  w_f = \frac{16}{0.85}

                        = 18.823 rad/s

 and w_o is the initial angular velocity which is zero since initial linear velocity is zero

               So

                         \alpha  = \frac{18.823 - 0}{0.15}

                               \alpha =125.487 rad /s^2

5 0
4 years ago
Which of the following statements is part of the theory?
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4 years ago
Use the Bohr model to address this question. When a hydrogen atom makes a transition from the 5 th energy level to the 2nd, coun
iris [78.8K]

Answer:

A. 2.82 eV

B. 439nm

C. 59.5 angstroms

Explanation:

A. To calculate the energy of the photon emitted you use the following formula:

E_{n1,n2}=-13.4(\frac{1}{n_2^2}-\frac{1}{n_1^2})     (1)

n1: final state = 5

n2: initial state = 2

Where the energy is electron volts. You replace the values of n1 and n2 in the equation (1):

E_{5,2}=-13.6(\frac{1}{5^2}-\frac{1}{2^2})=2.82eV

B. The energy of the emitted photon is given by the following formula:

E=h\frac{c}{\lambda}   (2)

h: Planck's constant = 6.62*10^{-34} kgm^2/s

c: speed of light = 3*10^8 m/s

λ: wavelength of the photon

You first convert the energy from eV to J:

2.82eV*\frac{1J}{6.242*10^{18}eV}=4.517*10^{-19}J

Next, you use the equation (2) and solve for λ:

\lambda=\frac{hc}{E}=\frac{(6.62*10^{-34} kg m^2/s)(3*10^8m/s)}{4.517*10^{-19}J}=4.39*10^{-7}m=439*10^{-9}m=439nm

C. The radius of the orbit is given by:

r_n=n^2a_o   (3)

where ao is the Bohr's radius = 2.380 Angstroms

You use the equation (3) with n=5:

r_5=5^2(2.380)=59.5

hence, the radius of the atom in its 5-th state is 59.5 anstrongs

8 0
3 years ago
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