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yarga [219]
2 years ago
6

Two girls stand 500m away from a tall wall

Physics
1 answer:
Mama L [17]2 years ago
4 0

From the calculations, the speed of sound in this case is 16.9 m/s.

<h3>What is an echo?</h3>

The term echo has to do with the reflection of sound waves. Sound is a mechanical wave.

we know that the speed of sound is obtained from;

V = 2x/t

x = distance covered

t = time taken

V = 2(500)/59

v = 16.9 m/s

The error in the experiment could come from;

  • Lack of precise time measurement
  • Error can also arise from the environment of the experiment

Learn more about echo:brainly.com/question/9527413

#SPJ1

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charle [14.2K]
The answer is B

As seen on the graph, the bus maintains a 9m/s speed for a majority of the trip to school.
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Which of the following would be most likely to contribute to molecules
lana66690 [7]

Answer:

B. Containing charged regions

Explanation:

The term i.e. intermolecular forces would be used to explain the attraction forces. Here the interaction would be done between molecules etc that acts between the acts & the other types of particles i.e. neighboring like atoms or ions

So in the given case, the option b would be contributed to the molecules that have intermolecular forces

hence, the option b is correct

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Differentiate scalar &amp; vector quantity?
Keith_Richards [23]
\textbf{Hello Friend}


Scalar Quantity :-

→ These are the quantities with magnitude only . These quantities doesn't have to be mentioned with direction

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3 years ago
TRUE OR FALSE? earth revolves around the sun tilted on its axis
Alex777 [14]

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6 0
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Read 2 more answers
Help with physics projectile motion​
BlackZzzverrR [31]

Answer:

10.4 m/s

Explanation:

First, find the time it takes for the projectile to fall 6 m.

Given:

y₀ = 6 m

y = 0 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: t

y = y₀ + v₀ t + ½ at²

(0 m) = (6 m) + (0 m/s) t + ½ (-9.8 m/s²) t²

t = 1.11 s

Now find the horizontal position of the target after that time:

Given:

x₀ = 6 m

v₀ = 5 m/s

a = 0 m/s²

t = 1.11 s

Find: x

x = x₀ + v₀ t + ½ at²

x = (6 m) + (5 m/s) (1.11 s) + ½ (0 m/s²) (1.11 s)²

x = 11.5 m

Finally, find the launch velocity needed to travel that distance in that time.

Given:

x₀ = 0 m

x = 11.5 m

t = 1.11 s

a = 0 m/s²

Find: v₀

(11.5 m) = (0 m) + v₀ (1.11 s) + ½ (0 m/s²) (1.11 s)²

v₀ = 10.4 m/s

3 0
3 years ago
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