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Veseljchak [2.6K]
4 years ago
12

What is the velocity of a wave with a frequency of 760 Hz and a wavelength of 0.45 m?

Physics
1 answer:
ad-work [718]4 years ago
3 0
<span>342 m/s This should be correct, but please tell me if it is incorrect!</span>
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The sensor in the torso of a crash test dummy records the magnitude and direction of the net force acting on the dummy.If the du
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Here in crash test the two forces are acting on the dummy in two different directions

As we know that force is a vector quantity so we need to use vector addition laws in order to find the resultant force on it.

So here two forces are given in perpendicular direction with each other so as per vector addition law we need to use Pythagoras theorem to find the resultant of two vectors

so we can say

F_{net} = \sqrt{F_1^2 + F_2^2}

here given that

F_1 = 130.0 N

F_2 = 4500.0 N

now we will plug in all data in the above equation

F_{net} = \sqrt{4500^2 + 130^}

F_{net} = 4501.9 N

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4 0
3 years ago
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A car horn emits a frequency of 400 Hz. A car traveling at 20.0 m/s sounds the horn as it approaches a stationary pedestrian. Wh
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Answer:

The observed frequency by the pedestrian is 424 Hz.

Explanation:

Given;

frequency of the source, Fs = 400 Hz

speed of the car as it approaches the stationary observer, Vs = 20 m/s

Based on Doppler effect, as the car the approaches the stationary observer, the observed frequency will be higher than the transmitted (source) frequency because of decrease in distance between the car and the observer.

The observed frequency is calculated as;

F_s = F_o [\frac{v}{v_s + v} ] \\\\

where;

F₀ is the observed frequency

v is the speed of sound in air = 340 m/s

F_s = F_o [\frac{v}{v_s + v} ] \\\\400 = F_o [\frac{340}{20 + 340} ] \\\\400 = F_o (0.9444) \\\\F_o = \frac{400}{0.9444} \\\\F_o = 423.55 \ Hz \\

F₀ ≅ 424 Hz.

Therefore, the observed frequency by the pedestrian is 424 Hz.

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3 years ago
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Hope this helps you!

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3 years ago
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jeka94

Answer:

A. alkali metals

Explanation:

im not sure about that

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In a building 30m up, the pressure will be regulated, as you are in a building. The floor will be strong enough to support the weight of the body, and the body will not recoil into itself.
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