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3 years ago

7

Which sound wave-object interaction is used by animals in echolocation?

2 answers:

iris [78.8K]3 years ago

8 0

Answer:

Explanation:

Animals who use echolocation send a special kind of sound, when this sound wave collides with the object it reflect back and animal detect the location, shape and size of object.Example of such animal is bats, whales, dolphins and oilbird. Bats uses echolocation to reach their pray and for their movement at night.

tekilochka [14]3 years ago

5 0

The answer is reflection

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State the formula for calculating power

Alex

Where power<span> P is in watts, voltage V is in volts and current I is in amperes (DC).</span>Power Formula<span> 2 – Mechanical </span>power equation<span>: </span>Power<span> P = E ⁄ t where </span>power<span> P is in watts, </span>Power<span> P = work / time (W ⁄ t). Energy E is in joules, and time t is in seconds.</span>

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2 years ago

During the first 6 years of its operation, the Hubble Space Telescope circled the Earth 37,000 times, for a total of 1,280,000,0

madreJ [45]

Answer:

$Kilometers\ in\ 1\ Orbit=\frac{1.28*10^9 Km}{3.7*10^4orbits}$

$Kilometers\ in\ 1\ Orbit=3.46*10^4 Km/Orbit$

Explanation:

Given Data:

Numbers of times Telescope cycled around the earth in 6 years=37,000 times

Total Distance traveled in 6 years by the Hubble Space Telescope=1,280,000,000 Km

Find:

Kilometers in one Orbit=?

Solution:

Kilometers in 37,000 Orbits=1,280,000,000 Km

Kilometers in 1 Orbit=1,280,000,000/37,000

In Scientific Notation:

$Kilometers\ in\ 3.7*10^4\ Orbits=1.28*10^9 Km$

$Kilometers\ in\ 1\ Orbit=\frac{1.28*10^9 Km}{3.7*10^4 orbits}$

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Kilometers in 1 Orbit in Scientific notation:

$Kilometers\ in\ 1\ Orbit=3.46*10^4 Km$

8 0

3 years ago

A particle with mass 1.81*10^-3kg and a charge of 1.22*10^-8C has, at a given instant, a velocity v= (3.0*10^4 m/s)j.

charle [14.2K]

Answer:

a = -0.33 m/s² k^

Direction: negative

Explanation:

From Newton's law of motion, we know that;

F = ma

Now, from magnetic fields, we know that;. F = qVB

Thus;

ma = qVB

Where;

m is mass

a is acceleration

q is charge

V is velocity

B is magnetic field

We are given;

m = 1.81 × 10^(−3) kg

q = 1.22 × 10 ^(−8) C

V = (3.00 × 10⁴ m/s) ȷ^.

B = (1.63T) ı^ + (0.980T) ȷ^

Thus, since we are looking for acceleration, from, ma = qVB; let's make a the subject;

a = qVB/m

a = [(1.22 × 10 ^(−8)) × (3.00 × 10⁴)ȷ^ × ((1.63T) ı^ + (0.980T) ȷ^)]/(1.81 × 10^(−3))

From vector multiplication, ȷ^ × ȷ^ = 0 and ȷ^ × i^ = -k^

Thus;

a = -0.33 m/s² k^

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2 years ago

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seropon [69]

since centripetal acceleration is always towards the center of the circle

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so the coordinates of the center will be

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$(x, y) = (3.20, 4.04)$

so the coordinate is (3.20 m, 4.04 m)

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3 years ago

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Verdich [7]

I believe that the answer is C. Hope this Helps:)))

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3 years ago