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hjlf
2 years ago
13

Imagine that treble sounds could travel at 400 m/s, but bass sounds traveled at only 300 m/s. If sound actually behaved that way

, and you were listening to a brass band from a distance of 120m, how much delay would you hear between trumpet and tuba notes when they begin a chord together?
Physics
1 answer:
kirza4 [7]2 years ago
4 0

Answer:

\Delta t = 0.1s

Explanation:

We need to calculate the difference between the times the waves from the bass sounds (t_b) and treble sounds (t_t) take to reach us, that is, \Delta t = t_b-t_t.

Since v=\frac{d}{t}, then t=\frac{d}{v} and we can write:

\Delta t = t_b-t_t=\frac{d}{v_b} - \frac{d}{v_t}=d(\frac{1}{v_b} - \frac{1}{v_t})=(120m)(\frac{1}{300m/s} - \frac{1}{400m/s})=0.1s

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A ball is dropped out of a window and hits the ground at 14.5 m/s. How long did it take to fall to the ground?
Lerok [7]

Answer:

Explanation:

Use the one-dimensional equation:

v_f=v_0+at which says that the final velocity of a falling object is equal to its initial velocity times the acceleration of gravity times the time it takes to fall. We have the final velocity, -14.5 (negative because its direction is down and down is negative), initial velocity is 0 (because it was held still by someone before it was dropped), and acceleration is -9.8 (negative again, because direction is down while acceleration increases). Filling in:

-14.5 = 0 - 9.8t and

-14.5 = -9.8t so

t = 1.5 seconds

3 0
3 years ago
Redondear a dos decimales 12,4552​
NikAS [45]

Answer:

12,46

Explanation:

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3 years ago
How is a magnetic field generated?
alekssr [168]
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2 years ago
How does mass and distance affect gravitational force?
loris [4]

Answer:

It shows that the gravitational force is directly proportional to the product of masses of two bodies and inversely proportional to the square of distance between them, So, as the mass of bodies increases, the gravitational force increases and with the increase of distance between them, the gravitational force decreases.

Explanation:

4 0
3 years ago
ASK YOUR TEACHER A 2.0-kg mass swings at the end of a light string with the length of 3.0 m. Its speed at the lowest point on it
Nadya [2.5K]

Answer:

  K_b = 78 J

Explanation:

For this exercise we can use the conservation of energy relations

starting point. Lowest of the trajectory

        Em₀ = K = ½ mv²

final point. When it is at tea = 50º

        Em_f = K + U

        Em_f = ½ m v_b² + m g h

where h is the height from the lowest point

        h = L - L cos 50

        Em_f = ½ m v_b² + mg L (1 - cos50)

energy be conserve

        Em₀ = Em_f

         ½ mv² = ½ m v_b² + mg L (1 - cos50)

         K_b = ½ m v_b² + mg L (1 - cos50)

let's calculate

          K_b = ½ 2.0 6.0² + 2.0 9.8 6.0 (1 - cos50)

          K_b = 36 +42.0

          K_b = 78 J

4 0
3 years ago
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