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denis-greek [22]
2 years ago
12

What does wave speed have to do with all SONAR AND RADAR technologies?

Physics
2 answers:
posledela2 years ago
8 0

Answer:

1) Radar uses radio waves, which are a type of electromagnetic energy. Sonar uses the echo principle by sending out sound waves underwater or through the human body to locate objects. Sound waves are a type of acoustic energy. Because of the different type of energy used in radar and sonar, each has its own applications.

2)Radar systems operate using radio waves primarily in air, while sonar systems operate using sound waves primarily in water (Minkoff, 1991). Despite the difference in medium, similarities in the principles of radar and sonar can frequently result in technological convergence.

Read more on Brainly.com - brainly.com/question/16593294#readmore

I can help if you want more

Explanation:

Elena L [17]2 years ago
4 0

Explanation:

1) Radar uses radio waves, which are a type of electromagnetic energy. Sonar uses the echo principle by sending out sound waves underwater or through the human body to locate objects. Sound waves are a type of acoustic energy. Because of the different type of energy used in radar and sonar, each has its own applications.

2)Radar systems operate using radio waves primarily in air, while sonar systems operate using sound waves primarily in water (Minkoff, 1991). Despite the difference in medium, similarities in the principles of radar and sonar can frequently result in technological convergence.

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You are given a parallel plate capacitor that has plates of area 29 cm2 which are separated by 0.0100 mm of nylon (dielectric co
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Answer:

20.60 kV

Explanation:

Capacitance of parallel plates without dielectric between them is:

C=\frac{\varepsilon_{0}A}{d}

with d the distance between the plates, A the area of the plates and ε₀ the constant 8.85419\times10^{-12}\frac{C^{2}}{Nm^{2}}, so :

C_0=\frac{(8.85419\times10^{-12})(0.0029)}{0.0100\times10^{-3}}=2.57\times10^{-9} F

But the dielectric constant is defined as:

k=\frac{C}{C_{0}}

With C the effective capacitance (with the dielectric) and Co the original capacitance (without the dielectric). So, the new capacitance is:

C=kC_0

But capacitance is related with voltage by:

C=\frac{Q}{V}

with Q the charge and V the voltage, using the new capacitance and solving for V:

kC_0=\frac{Q}{V}

V=\frac{Q}{kC_0}=\frac{0.18\times10{-3}}{(3.4)(2.57\times10^{-9})}=20599.68V=20.60 kV

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Answer:

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Explanation:

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Answer with Explanation:

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