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joja [24]
3 years ago
13

A boy yells in a narrow valley. The sound reflects off of a boulder that is 80 m away. How long will it take to hear the echo if

Physics
2 answers:
soldier1979 [14.2K]3 years ago
4 0

Answer:4 lubeconds (speed=distance over time)

Dima020 [189]3 years ago
3 0

Answer:

Time taken, t = 0.5 seconds

Explanation:

It is given that,

Speed of sound wave, v = 320 m/s

The sound reflects off of a boulder that is 80 m away.

Let t is the time taken to hear the echo. The distance covered by the sound to hear the echo is, 2d. So,

v=\dfrac{2d}{t}

t=\dfrac{2d}{v}  

t=\dfrac{2\times 80\ m}{320\ m/s}

t = 0.5 seconds

So, the time taken to hear the echo is 0.5 seconds. Hence, this is the required solution.

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Which of the following is an action-at-a-distance force? friction tension gravity air resistance
tatuchka [14]

Answer:

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Explanation:

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3 years ago
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What is the most common kind of element in the solar wind?
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3 years ago
When monochromatic light shines perpendicularly on a soap film (n = 1.33) with air on each side, the second smallest nonzero fil
Anika [276]

Let us start from considering monochromatic light as an incidence on the film of a thickness t whose material has an index of refraction n determined by their respective properties.

From this point of view part of the light will be reflated and the other will be transmitted to the thin film. That additional distance traveled by the ray that was reflected from the bottom will be twice the thickness of the thin film at the point where the light strikes. Therefore, this relation of phase differences and additional distance can be expressed mathematically as

2t + \frac{1}{2} \lambda_{film} = (m+\frac{1}{2})\lambda_{film}

We are given the second smallest nonzero thickness at which destructive interference occurs.

This corresponds to, m = 2, therefore

2t = 2\lambda_{film}

t = \lambda_{film}

The index of refraction of soap is given, then

\lambda_{film} = \frac{\lambda_{vacuum}}{n}

Combining the results of all steps we get

t = \frac{\lambda_{vacuum}}{n}

Rearranging, we find

\lambda_{vacuum} = tn

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4 0
3 years ago
During the spin cycle of a washing machine, the clothes stick to the outer wall of the barrel as it spins at a rate as high as 1
Darya [45]

To answer the two questions, we need to know two important equations involving centripetal movement:

v = ωr (ω represents angular velocity <u>in radians</u>)

a = \frac{v^{2}}{r}

Let's apply the first equation to question a:

v = ωr

v = ((1800*2π) / 60) * 0.26

Wait. 2π? 0.26? 60? Let's break down why these numbers are written differently. In order to use the equation v = ωr, it is important that the units of ω is in radians. Since one revolution is equivalent to 2π radians, we can easily do the conversion from revolutions to radians by multiplying it by 2π. As for 0.26, note that the question asks for the units to be m/s. Since we need meters, we simply convert 26 cm, our radius, into meters. The revolutions is also given in revs/min, and we need to convert it into revs/sec so that we can get our final units correct. As a result, we divide the rate by 60 to convert minutes into seconds.

Back to the equation:

v = ((1800*2π)/60) * 0.26

v = (1800*2(3.14)/60) * 0.26

v = (11304/60) * 0.26

v = 188.4 * 0.26

v = 48.984

v = 49 (m/s)

Now that we know the linear velocity, we can find the centripetal acceleration:

a = \frac{v^{2}}{r}

a = \frac{49^{2}}{0.26}

a = 9234.6 (m/s^{2})

Wow! That's fast!

<u>We now have our answers for a and b:</u>

a. 49 (m/s)

b. 9.2 * 10^{3} (m/s^{2})

If you have any questions on how I got to these answers, just ask!

- breezyツ

5 0
3 years ago
022 (part 1 of 4) 10.0 points A ball is thrown vertically upward with a speed of 24.5 m/s. How high does it rise? The accelerati
svetoff [14.1K]

1)

Answer:

Part 1)

H = 30.6 m

Part 2)

t = 2.5 s

Part 3)

t = 2.5 s

Part 4)

v_f = 24.5 m/s

Explanation:

Part 1)

initial speed of the ball upwards

v_i = 24.5 m/s

so maximum height of the ball is given by

H = \frac{v_i^2}{2g}

H = \frac{24.5^2}{2(9.80)}

H = 30.6 m

Part 2)

As we know that final speed will be zero at maximum height

so we will have

v_f - v_i = at

0 - 24.5 = (-9.8)t

t = 2.5 s

Part 3)

Since the time of ascent of ball is same as time of decent of the ball

so here ball will same time to hit the ground back

so here it is given as

t = 2.5 s

Part 4)

since the acceleration due to earth will be same during its return path as well as the time of the motion is also same

so here its final speed will be same as that of initial speed

so we have

v_f = 24.5 m/s

2)

Answer:

a = 9.76 m/s/s

Explanation:

As we know that the object is released from rest

so the displacement of the object in vertical direction is given as

y = \frac{1}{2}at^2

4.88 = \frac{1}{2}a(1^2)

a = 9.76 m/s^2

3)

Answer:

v = 29.7 m/s

Explanation:

acceleration of the rocket is given as

a = 90 m/s^2

time taken by the rocket

t = 0.33 min

final speed of the rocket is given as

v_f = v_i + at

v_f = 0 + (90)(0.33)

v_f = 29.7 m/s

4)

Answer:

Part 1)

y = 25.95 m

Part 2)

d = 6.72 m

Explanation:

Part 1)

As it took t = 2.3 s to hit the water surface

so here we will have

y = \frac{1}{2}gt^2

y = \frac{1}{2}(9.81)(2.3^2)

y = 25.95 m

Part 2)

Distance traveled by it in horizontal direction is given as

d = v_x t

d = 2.92 \times 2.3

d = 6.72 m

6 0
3 years ago
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