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3 years ago

What is SONAR? How does it Work?

Physics

2 answers:

Advocard [28]3 years ago

7 0

Answer:

Sonar is simply making use of an echo. When an animal or machine makes a noise, it sends sound waves into the environment around it. Those waves bounce off nearby objects, and some of them reflect back to the object that made the noise.

murzikaleks [220]3 years ago

6 0

Answer: Sonar is simply making use of an echo. When an animal or machine makes a noise, it sends sound waves into the environment around it. Those waves bounce off nearby objects, and some of them reflect back to the object that made the noise.

Explanation:

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Drag the tiles to the boxes to form correct pairs. Match each hypothesis for how the Moon formed with the statement that best de

Roman55 [17]

Answer:

giant impact theory

Explanation:

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3 years ago

Which is an example of potential energy? Question 9 options: an electric current lighting a light bulb releasing the strings on

Feliz [49]

The pulled back string on an archer’s bow.

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3 years ago

An archer shoots an arrow at a 74.0 m distant target, the bull's-eye of which is at same height as the release height of the arr

GuDViN [60]

A. The angle at which the arrow must be released to hit the bull's-eye is 20.7 °

B. The arrow will go over the branch.

<h3>A. How to determine the angle</h3>

Range (R) = 74 m
Initial velocity (u) = 33 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Angle (θ) = ?

R = u²Sine(2θ) / g

74 = 33² × Sine (2θ) / 9.8

Cross multiply

74 × 9.8 = 33² × Sine (2θ)

725.2 = 1098 × Sine (2θ)

Divide both sides by 1098

Sine (2θ) = 725.2 / 1098

Sine (2θ) = 0.6605

Take the inverse of sine

2θ = Sine⁻¹ 0.6605

2θ = 41.3

Divide both sides by 2

θ = 41.3 / 2

θ = 20.7 °

<h3>B. How to determine if the arrow will go over or under the branch</h3>

To determine if the arrow will go over or under the branch situated mid way, we shall determine the maximum height attained by the arrow. This can be obtained as follow:

Initial velocity (u) = 33 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Angle (θ) = 20.7 °
Maximum height (H) = ?

H = u²Sine²θ / 2g

H = [33² × (Sine 20.7)²] / (2 ×9.8)

H = 6.94 m

Thus, the maximum height attained by the arrow is 6.94 m which is greater than the height of the branch (i.e 3.50 m).

Therefore, we can conclude that the arrow will go over the branch

Learn more about projectile motion:

brainly.com/question/20326485

#SPJ1

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2 years ago

A man walke towards the bus stop 100 meters away at a constant velocity. He reaches the bus stop in 200 seconds. Once at the bus

serg [7]

First:

d = 100 m
t = 200 sec

v = 100/200 = 0.5 m/s

Displacement is zero since he returned to his start point.

t2 = d/v2 = 100/2 = 50 sec

total time = 50 + 200 + 500 = 750 sec

3 0

4 years ago

A 0.560 kg snowball is fired from a cliff 14.2 m high with an initial velocity of 13.3 m/s, directed 26.0° above the horizontal.

enot [183]

Answer:

a) v = 21.34 m/s

b) v = 21.34 m/s

c) v = 21.34 m/s

Explanation:

Mass of the snowball, m = 0.560 kg

Height of the cliff, h = 14.2 m

Initial velocity of the ball, u = 13.3 m/s

θ = 26°

The speed of the slow ball as it reaches the ground, v = ?

The initial Kinetic energy of the snow ball, $KE_{0} = 0.5 mu^{2}$

Potential energy of the snow ball at the given height, PE = mgh

Final Kinetic energy of the ball as it reaches the ground, $KE_{f} = 0.5mv^{2}$

a) Using the principle of energy conservation,

$KE_{0} + PE = KE_{f} \\0.5mu^{2} + mgh = 0.5mv^{2}\\v^{2} =2( 0.5u^{2} + gh)\\v^{2} =u^{2} + 2gh\\v = \sqrt{u^{2} + 2gh} \\v = \sqrt{13.3^{2} + 2*9.8*14.2}\\v = 21.34 m/s$

b) The speed remains v = 21.34 m/s since it is not a function of the angle of launch

c)The principle of energy conservation used cancels out the mass of the object, therefore the speed is not dependent on mass

v = 21. 34 m/s

7 0

4 years ago