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denis-greek [22]
3 years ago
12

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

Physics
2 answers:
posledela3 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]3 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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A 392 N wheel comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at 24
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N=m*g\\m=\frac{N}{g}\\m=\frac{392N}{9.8\frac{m}{s^{2}}}

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M_{I}=0.8*40.0kg*(0.6m} )^{2}\\M_{I}=11.5 kg*m^{2}

Kinetic energy of the rotation motion

K_{r}=\frac{1}{2}*I*W^{2}\\K_{r}=\frac{1}{2}*11.52kg*m^{2}*(24\frac{rad}{s})^{2}\\K_{r}=3317.76J

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K_{t}=\frac{1}{2}*m*v^{2}\\v=w*r\\v=24rad/s*0.6m=14.4 \frac{m}{s}\\K_{t}=\frac{1}{2}*40kg*(14.4\frac{m}{s})^{2}\\K_{t}=4147.2J

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K=3317.79J+4147.2J\\K=7464.99J

Now the work done by the friction is acting at the motion so the kinetic energy and the work of motion give the potential work so there we can find height

K-W=E_{p}\\7464.99-2600J=m*g*h\\4864.99J=m*g*h\\h=\frac{4864.99J}{m*g}\\h=\frac{4864.99J}{392N}\\h=12.41m

6 0
3 years ago
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