Answer: when the wave encounters something, it can bounce (reflection) or be bent (refraction). In fact, you can "trap" waves by making them bounce back and forth between two or more surfaces. Musical instruments take advantage of this; they produce pitches by trapping sound waves.

Explanation: Any bunch of sound waves will produce some sort of noise. But to be a tone - a sound with a particular pitch - a group of sound waves has to be very regular, all exactly the same distance apart. That's why we can talk about the frequency and wavelength of tones.

8 0

3 years ago

A snowball is launched horizontally from the top of a building at v = 16.9 m/s. If it lands d = 44 meters from the bottom, how h

Aloiza [94]

$v_x=16,9\frac{m}{s}\\
s=44m\\
g=10\frac{m}{s^2}\\\\
v_x=\frac{s}{t}\Rightarrow t=\frac{s}{v_x}\\\\
t=\frac{44}{16,9}\\\\
t=2,6s\\\\
h=\frac{gt^2}{2}\\\\
h=\frac{10*2,6^2}{2}\\\\
h=5*6,76\\\\
h=33,8m$

6 0

4 years ago

A 1300 kg car moving at 20 m/s and a 900 kg car moving at 15 m/s in precisely oppositedirections participate in a head-on crash.

miskamm [114]

Given

Car 1

m1 = 1300 kg

v1 = 20 m/s

m2 = 900 kg

v2 = -15 m/s

(Negative sign shows that direction of car 2 is opposite to car 1)

Procedure

As per the conservation of linear momentum, "The total momentum of the system before the collision must be equal to the total momentum after the collision". And this applies to the perfectly inelastic collision as well. Then the expression is,

$\begin{gathered} m_1v_1+m_2v_2=(m_1+m_2)v \\ v=\frac{m_1v_1+m_2v_2}{m_1+m_2} \\ v=\frac{1300\operatorname{kg}\cdot20m/s-900\operatorname{kg}\cdot15m/s}{1300\operatorname{kg}+900\operatorname{kg}} \\ v=5.681m/s \end{gathered}$

Thus, we can conclude that the speed and direction of the cars after the impact is 5.68 m/s towards the first car.

5 0

1 year ago

Nvm i got the answer but <br> free points

faust18 [17]

Answer:

thank you 谢谢

Explanation:

8 0

3 years ago

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