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konstantin123 [22]
3 years ago
12

Can sum1 pls give me the answer to number 3

Physics
1 answer:
inna [77]3 years ago
7 0

maybe if you look it up it will come up

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Which statement regarding sound traveling in air is correct?
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Answer: D. A wave with a shorter wavelength is always faster than one with a longer wavelength

Explanation: "Imagine two sets of waves that have the same speed. <u><em>If one set has a longer wavelength, it will have a lower frequency (more time between waves). If the other set has a shorter wavelength, it will have a higher frequency</em></u> (less time between waves). Light moves even faster AND has shorter wavelengths."

Why it's not C: "The number of complete wavelengths in a given unit of time is called frequency (f). <em><u>As a wavelength increases in size, its frequency and energy (E) decrease</u></em>. From these equations you may realize that as the frequency increases, the wavelength gets shorter. As the frequency decreases, the wavelength gets longer."

Why it's not B: "The frequency does not change as the sound wave moves from one medium to another. Since the speed changes and the frequency does not, the wavelength must change."

Why it's not A: "Do loud sounds travel faster than soft sounds? No. Both travel at the same speed The speed depends on the medium it passes through. Louder sounds are simply sound waves with higher amplitude traveling at the same speed."

6 0
3 years ago
Two stones are launched from the top of a tall building. One stoneis thrown in a direction 30.0^\circ above the horizontal with
Butoxors [25]

Answer:

Part A)

t(1) > t(2), the stone thrown 30 above the horizontal spends more time in the air.

Part B)

x(f1) > x(f2), the first stone will land farther away from the building.

Explanation:

<u>Part A)</u>

Let's use the parabolic motion equation to solve it. Let's define the variables:

  • y(i) is the initial height, it is a constant.
  • y(f) is the final height, in our case is 0
  • v(i) is the initial velocity (v(i)=16 m/s)
  • θ1 is the first angle, 30°
  • θ2 is the first angle, -30°

For the first stone

y_{f1}=y_{i1}+v*sin(\theta_{1})t_{1}-0.5gt_{1}^{2}              

0=y_{i1}+16*sin(30)t_{1}-0.5*9.81*t_{1}^{2}

0=y_{i1}+8t_{1}-4.905*t_{1}^{2} (1)  

For the second stone  

0=y_{i2}+16*sin(-30)t_{2}-4.905t_{2}^{2}    

0=y_{i2}-8t_{2}-4.905t_{2}^{2} (2)            

 

If we solve the equation (1) we will have:

t_{1}=\frac{-8\pm \sqrt{64+19.62*y_{i}}}{-9.81}  

We can do the same procedure for the equation (2)

t_{1}=\frac{8\pm \sqrt{64+19.62*y_{i}}}{-9.81}

We can analyze each solution to see which one spends more time in the air.

It is easy to see that the value inside the square root of each equation is always greater than 8, assuming that the height of the building is > 0. Now, to get positive values of t(1) and t(2) we need to take the negative option of the square root.

Therefore, t(1) > t(2), it means that the stone thrown 30 above the horizontal spends more time in the air.

<u>Part B)</u>

We can use the equation of the horizontal position here.

<u>First stone</u>

x_{f1}=x_{i1}+vcos(30)t_{1}

x_{f1}=0+13.86*t_{1}

x_{f1}=13.86*t_{1}

<u>Second stone</u>

x_{2}=x_{i2}+vcos(-30)t_{2}

x_{1}=0+13.86*t_{1}

x_{1}=13.86*t_{2}

Knowing that t(1) > t(2) then x(f1) > x(f2)

Therefore, the first stone will land farther away from the building.

They land at different points at different times.

I hope it helps you!

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