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Maksim231197 [3]
2 years ago
14

Sound waves travel at the rate of 343 m/s at 20°C. If a man standing 450 meters away from the wall of a canyon yells, “Hello,” h

ow long will it take for him to hear his own voice echo off of the wall?
Physics
2 answers:
Sophie [7]2 years ago
8 0

Answer:

\Delta t = 2.624\,s

Explanation:

Sound is a form of mechanic wave, which needs a medium to propagate itself and has a constant speed. The time needed to hear his echo is:

\Delta t = \frac{900\,m}{343\,\frac{m}{s} }

\Delta t = 2.624\,s

nekit [7.7K]2 years ago
5 0

Answer:

2.62seconds

Explanation:

Speed is defined as the ratio of the distance covered by a body with respect to time.

Speed v = Distance (s)/Time (t)

For a traveling sound wave, the distance between the source of a sound and the reflector is '2S'.

Speed v = 2 × distance (S)/Time (T)

V = 2S/t

2S = vt

Given speed of the wave = 342m/s

Distance covered = 450m

t = 2S/v

t = (2×450)/343

t = 900/343

t = 2.62seconds

It will take him 2.62seconds for him to hear his own voice echo off of the wall.

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ElenaW [278]

Answer:

I believe the answer is D

Explanation: The doppler affect isan increase or decrease in the frequency of sound, light, or other waves as the source and observer move toward or away from each other.

6 0
3 years ago
Projectile <br> SHOW WORK<br> WILL MARK BRANLIEST <br> (Draw Picture and Label)
m_a_m_a [10]

a) The horizontal distance covered by the projectile is 600 m

b) The projectile reaches its maximum height after 3.00 s

c) The altitude of the highest point is 44.1 m

Explanation:

a)

The motion of the projectile consists of two independent motions:  

- A uniform motion (constant velocity) along the horizontal direction  

- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction  

In this part A, we just need to analyze the horizontal motion. We know that:

  • The projectile travels horizontally with a constant velocity ov v_x = 100 m/s
  • The total time of flight of the projectile is t=6.00 s

Therefore, the horizontal distance covered by the projectile is given by

x=v_x t

And substituting, we find

x=(100)(6.0) = 600 m

b)

For this part, we need to analyze the vertical motion of the projectile.

First, we want to find the initial vertical velocity. We can do it by using the equation for the vertical displacement:

s=u_y t + \frac{1}{2}at^2

where:

u_y is the initial vertical velocity

a=g=-9.8 m/s^2 is the acceleration of gravity (negative because it is downward)

t is the time

s is the vertical displacement

We know that the total time of flight is t = 6.00 s: this means that when t=6, the projectile returns to its initial vertical position, so s = 0. Substituting and solving for u_y, we get

u_y = - \frac{1}{2}at=-\frac{1}{2}(-9.8)(6)=29.4 m/s

The vertical velocity then as a function of t is given by

v_y = u_y + at

And at the maximum height, it becomes zero: v_y = 0. Substituting and solving for t, we find the time at which the projectile reaches the maximum height:

t=-\frac{u_y}{a}=-\frac{29.4}{-9.8}=3.00 s

c)

To find the altitude of the highest point in the path, we use again the equation:

s=u_y t + \frac{1}{2}at^2

where

u_y = 29.4 m/s is the initial vertical velocity

t = 3.00 s is the time at which the projectile reaches the highest point

a=g=-9.8 m/s^2 is the acceleration of gravity

Substituting the values, we find

s=(29.4)(3.00)+\frac{1}{2}(-9.8)(3.00)^2=44.1 m

So, the highest point is at 44.1 m above the ground.

Learn more about projectiles:

brainly.com/question/8751410

#LearnwithBrainly

8 0
3 years ago
Froghopper insects have a typical mass of around 12.5 mg and can jump to a height of 42.3 cm. The takeoff velocity is achieved a
Maksim231197 [3]

Answer:

2065.005 m/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{v^2-0^2}{2\times h}\\\Rightarrow v^2=2ah\ m/s

Now H-h = 42.3 - 0.2 = 42.1 cm = 0.421 m

The final velocity will be the initial velocity

v^2-u^2=2as\\\Rightarrow 0^2-u^2=2gs\\\Rightarrow -2ah=2\times g(H-h)\\\Rightarrow -2a0.002=2\times g0.421\\\Rightarrow a=-\frac{0.421\times -9.81}{0.002}\\\Rightarrow a=2065.005\ m/s^2

Acceleration of the frog is 2065.005 m/s²

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What are the two parts of a force pair?
iragen [17]

These two forces are called action and reaction forces and are the subject of Newton's third law of motion.

<em>Have a luvely day!</em>

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2 years ago
Define the relationship between force acceleration and mass
goblinko [34]
Force is mass times acceleration. This means an object with a larger mass needs a stronger force to be moved along at the same acceleration as an object with a small mass
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