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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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You have a two-wheel trailer that you pull behind your ATV. Two children with a combined mass of 77.2 kg hop on board for a ride
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To solve this problem it is necessary to apply the kinematic equations of motion and Hook's law.

By Hook's law we know that force is defined as,

F= kx

Where,

k = spring constant

x = Displacement change

PART A) For the case of the spring constant we can use the above equation and clear k so that

k= \frac{F}{x}

k = \frac{mg}{x}

k= \frac{77.2*9.8}{0.0637}

k = 11876.92N/m

Therefore the spring constant for each one is 11876.92/2 = 5933.46N/m

PART B) In the case of speed we can obtain it through the period, which is given by

T = \frac{2\pi}{\omega}

Re-arrange to find \omega,

\omega = \frac{2\pi}{T}

\omega = \frac{2\pi}{2.14}

\omega = 2.93rad/s

Then through angular kinematic equations where angular velocity is given as a function of mass and spring constant we have to

\omega^2 = \frac{k}{m}

m = \frac{k}{\omega^2}

m = \frac{ 11876.92}{2.93}

m = 4093.55Kg

Therefore the mass of the trailer is 4093.55Kg

PART C) The frequency by definition is inversely to the period therefore

f = \frac{1}{T}

f = \frac{1}{2.14}

f = 0.4672 Hz

Therefore the frequency of the oscillation is 0.4672 Hz

PART D) The time it takes to make the route 10 times would be 10 times the period, that is

t_T = 10*T

t_T = 10 *2.14s

t_T = 21.4s

Therefore the total time it takes for the trailer to bounce up and down 10 times is 21.4s

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In an atom is the number of protons greater, less than or equal to the number of electrons?
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Answer:

<em>The number of electrons and protons are always equal in a neutral atom. The total number of protons in the nucleus of an atom gives us the atomic number of that atom.</em>

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What factors contribute to global winds identify areas where winds are weak
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A ball is thrown directly downward with an initial speed of 8.75 m/s, from a height of 29.4 m. After what time interval does it
steposvetlana [31]

After time 1.74 seconds ball will strike the ground.

A ball is thrown directly downward and information we have-

Initial velocity of ball (u) = 8.75m/s

distance/height (s) = 29.4m

Ball is thrown downward means there are gravitation force also worked and the acceleration of ball is-

acceleration (a) = g = 9.8 m/s²

When ball strike the ground then it's velocity should be 0 means-

Final velocity of ball (v) = 0 m/s²

For getting time period we can use 2nd equation of motion as-

s = ut + (1/2)at²

Now, 29.4 = (8.75 × t) + (1/2× 9.8 × t²)

29.4 = 8.75t + 4.9t²

Now we have a quadratic equation as -

4.9t² + 8.75t - 29.4 = 0

After solving it we get two values of t -

time = 1.74 and time = -3.5

But time can not be negative so we will reject the negative values.

so t = 1.74 seconds

So to conclude that after applying the second equation of motion the time taken by the ball to reach ground is 1.74 seconds.

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2 years ago
Se usa una bombona de He para inflar globos, de volumen 15L con una presión de 200 atm. Si los globos tienen un volumen de 1.5L
Bingel [31]

Answer:

La cantidad de globos llenos es de aproximadamente 1.974 globos

Explanation:

Los parámetros dados son;

El gas utilizado para inflar el globo = Él

El volumen del gas He, V₁ = 15 L

La presión del gas He, P₁ = 200 atm = 20,265,000 Pa

El volumen de cada globo = 1,5 L

La presión del He dentro del globo, P₂ = 770 mmHg = 102,641 Pa

Sea 'V₂', que represente el volumen total que el gas ocupa en todos los globos, y 'x', que represente el número de globos;

Por lo tanto, tenemos;

1,5 · x = V₂

Según la ley de Boyle, tenemos;

P₁ · V₁ = P₂ · V₂

∴ V₂ = P₁ · V₁ / (P₂)

Al introducir los valores conocidos, obtenemos;

V₂ = 20,265,000 Pa × 15 L / (102,641 Pa) = 2,961.53548 L

1,5 L × x = V₂

∴ 1,5 L × x = 2,961,53548 L

x = 2,961.53548 L / (1.5 L) = 1,974.35699 redondeamos hacia abajo para obtener un número entero de globos de la siguiente manera;

El número de globos, x ≈ 1,974 globos.

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3 years ago
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