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3 years ago

Which statement describes how echolocation and ultrasound imaging are the same?

Physics

2 answers:

Zielflug [23.3K]3 years ago

7 0

They both use the reflection of sound waves.<span>
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Mandarinka [93]3 years ago

6 0

Answer: The correct option is C.

Explanation:

The sound wave is reflected back after striking from the barrier. This is called the reflection of the sound.

When the sound wave incidents on the barrier then it is called incident wave.

When the sound wave gets reflected back from the barrier then it is called reflected wave.

Ultrasound imaging is a medical technique which is used to take the image of the internal parts of organ of the body.

Echolocation is the process which is used by several species. It is used by bats to locate its prey. It is the biological SONAR.

Therefore, echolocation and ultrasound imaging are same as they both use the reflection of sound waves.

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Using 1000 j of work, a toy elevator is raised from the ground floor to the second floor in 20 s. how much power does the elevat

Anastasy [175]

Power (watts) = work done (joules) ÷ time (seconds)

power (watts) = 1000J ÷ 20s

power (watts) = 50w

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3 years ago

A positron is an elementary particle identical to an electron except that its charge is . An electron and a positron can rotate

denis23 [38]

Explanation:

According to the energy conservation,

$F_{centripetal} = F_{electric}$

$\frac{mv^{2}}{r} = \frac{kq^{2}}{d^{2}}$

$v^{2} = \frac{kq^{2}r}{d^{2}m}$

= $\frac{9 \times 10^{9} N.m^{2}/C^{2} \times 1.6 \times 10^{-19} C \times 0.75 \times 10^{-9} m}{(1.50 \times 10^{-9}m)^{2} \times 9.11 \times 10^{-31} kg}$

= $8.430 \times 10^{10} m^{2}/s^{2}$

v = $\sqrt{8.430 \times 10^{10} m^{2}/s^{2}}$

= $2.903 \times 10^{5} m/s$

Formula for distance from the orbit is as follows.

S = $2 \pi r$

= $2 \times 3.14 \times 0.75 \times 10^{-9} m$

= $4.71 \times 10^{-9} m$

Now, relation between time and distance is as follows.

T = $\frac{S}{v}$

$\frac{1}{f} = \frac{S}{v}$

or, f = $\frac{v}{S}$

= $\frac{2.903 \times 10^{5} m/s}{4.71 \times 10^{-9} m}$

= $6.164 \times 10^{13} Hz$

Thus, we can conclude that the orbital frequency for an electron and a positron that is 1.50 apart is $6.164 \times 10^{13} Hz$ .

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3 years ago

A sealed syringe contains 10 × 10−6 m3 of air at 1 × 105 Pa. The plunger is pushed until the volume of trapped air is 4 × 10−6 m

Scilla [17]

Answer: $P_2=2.5\times 10^5\ Pa$

Explanation:

Given

Initial volume $V_1=10\times 10^{-6}\ m^3$

Initial Pressure $P_1=10^5\ Pa$

Final trapped air $V_2=4\times 10^{-6}\ m^3$

If there is no change in temperature then We can write

$P_1V_1=P_2V_2\quad \text{(From ideal gas equation)}$

$P_2=\dfrac{P_1V_1}{V_2}$

$P_2=10^5\times \frac{10\times 10^{-6}}{4\times 10^{-6}}$

$P_2=2.5\times 10^5\ Pa$

6 0

3 years ago

Absolute magnitude and apparent magnitude are the same thing. True False

Furkat [3]

False because absolute magnitude is the real magnitude but apparent is not

6 0

3 years ago

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Which of the following statements is false?

Irina18 [472]

d. is wrong...the temperature cannot be measured due to physical changes

7 0

2 years ago