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BartSMP [9]
3 years ago
8

When a bat uses echolocation to determine the distance to an insect, it sends out a sound wave and waits to see how long the sou

nd takes to echo back. Suppose now that a bat hears its echo 0.06 LaTeX: ss after it emitted the sound. If the speed of sound is 344.9 LaTeX: \frac{m}{s}m s, how far away is the insect in meters?
Physics
1 answer:
ZanzabumX [31]3 years ago
3 0

Answer:

Explanation:

Distance travelled by sound in going to target( insect )  and returning back = 2d ,d is distance of target .

time t = .06 s

speed = distance / time

344.9 = 2d / .06

d = 10.35 m

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LenKa [72]

Answer:

The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C, but the direction is still to the right.

Explanation:

From coulomb's law, F = Eq

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F = E₁q₁

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Then

E₂q₂ = E₁q₁

E_2 = \frac{E_1q_1}{q_2}

where;

E₂ is the external electric field due to second test charge = ?

E₁ is the external electric field due to first test charge = 4 x 10⁶ N/C

q₁ is the first test charge = 13 mC

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Substitute in these values in the equation above and calculate E₂.

E_2 = \frac{4*10^6*13}{23} = 2.26 *10^6 \ N/C

The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C when 13 mC test charge is replaced with another test charge of 23 mC.

However, the direction of the external field is still to the right.

8 0
3 years ago
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harkovskaia [24]
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BlackZzzverrR [31]

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Hope this helped!

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Answer:

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