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emmasim [6.3K]
3 years ago
5

An echo is sound that returns to you after being reflected from a distant surface (e.g., the side of a cliff). Assuming that the

distances involved are the same, an echo under water and an echo in air return to you _____________________
a. at different times, the echo under water returning more slowly.
b. at different times, the echo under water returning more quickly.
c. at the same time
Physics
2 answers:
Nuetrik [128]3 years ago
8 0

Answer:

b. at different times, the echo under water returning more quickly.

hoa [83]3 years ago
4 0

Answer:

The correct answer is B

An echo underwater and an echo in the air will return at different times. The echo underwater will return more quickly than the echo in the air.

Explanation:

The physics of this is simple.

Water and air are both made up of particles. The particles for water are more closely or densely arranged that those of the air molecules. Hence sound travels faster in water than in air. When measured, the speed actually differs by as much as 5 times with water being the fastest medium.

Think of it like this. Assume you have two stacks of dominoes, one closely packed than the other but exactly the same amount of dominos, you'd notice that the stack that is more tightly arranged will be the first to topple over because it takes less time for the kinetic energy from the first domino to reach the next and on and on like that until the last domino.

Cheers

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A) x and y components of the electric field  (Ep) at the origin.

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Epy = -1620.5 N/C

Ep = (1620.5(-i)+1620.5(-j)) N/C

B) Magnitude of the electric field  (Ep) at the origin.

   Ep= 2291.7 N/C

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Conceptual analysis

The electric field at a point P due to a point charge is calculated as follows:

E = k*q/d²

E: Electric field in N/C

q: charge in Newtons (N)

k: electric constant in N*m²/C²

d: distance from charge q to point P in meters (m)

Equivalence  

1nC= 10⁻⁹C

Data

K= 9x10⁹N*m²/C²

q₁ = q₂= +6.5nC=+6.5 *10⁻⁹C

d₁=d₂=0.190m

Graphic attached

The attached graph shows the field due to the charges:

Ep₁ : Electric Field at point P  (x=0, y=0) due to charge q₁. As the charge q₁ is positive (q₁+) ,the field leaves the charge.

Ep₂: Electric Field at point  P (x=0, y=0) due to charge q₂. As the charge q₂  is positive (q₂+) ,the field leaves the charge

Ep: Total field at point P due to charges q₁ and q₂.

Because q₁ = q₂ and d₁ = d₂, then, the magnitude of Ep₁ is equal to the magnitude of Ep₂

Ep₁ = Ep₂ = k*q/d² = 9*10⁹*6.5*10⁻⁹/0.190m² = 1620.5 N/C

Look at the attached graphic :

Epx = Ep₁= -1620.5 N/C

Epy = Ep₂= -1620.5 N/C

A) x and y components of the electric field  (Ep) at the origin.

Ep = (1620.5(-i)+1620.5(-j)) N/C

B) Magnitude of the electric field  (Ep) at the origin.

E_{p} =\sqrt{1620.5^{2}+1620.5^{2} } = 2291.7 \frac{N}{C}

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3 years ago
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