The frequency of note C3 is 131 
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<u>Explanation:</u>
Frequency is the measure of repetition of same thing a certain number of times. So frequency is inversely proportional to the wavelength. As wavelength is distance between two successive crests or troughs in a sound wave.
And frequency is the completion of number of cycles in a given time in sound waves. The frequency and wavelength are inversely proportional to each other with velocity of sound being the proportionality constant.
Thus, here the speed of sound is given as 343 m/s, the wavelength of the note is also given as 2.62 m, then frequency will be as follows:
Thus,
So the frequency of note C3 is 131 .
Answer:
velocity = 1527.52 ft/s
Acceleration = 80.13 ft/s²
Explanation:
We are given;
Radius of rotation; r = 32,700 ft
Radial acceleration; a_r = r¨ = 85 ft/s²
Angular velocity; ω = θ˙˙ = 0.019 rad/s
Also, angle θ reaches 66°
So, velocity of the rocket for the given position will be;
v = rθ˙˙/cos θ
so, v = 32700 × 0.019/ cos 66
v = 1527.52 ft/s
Acceleration is given by the formula ;
a = a_r/sinθ
For the given position,
a_r = r¨ - r(θ˙˙)²
Thus,
a = (r¨ - r(θ˙˙)²)/sinθ
Plugging in the relevant values, we obtain;
a = (85 - 32700(0.019)²)/sin66
a = (85 - 11.8047)/0.9135
a = 80.13 ft/s²
Answer:
1/3 the distance from the fulcrum
Explanation:
On a balanced seesaw, the torques around the fulcrum calculated on one side and on another side must be equal. This means that:
where
W1 is the weight of the boy
d1 is its distance from the fulcrum
W2 is the weight of his partner
d2 is the distance of the partner from the fulcrum
In this problem, we know that the boy is three times as heavy as his partner, so
If we substitute this into the equation, we find:
and by simplifying:
which means that the boy sits at 1/3 the distance from the fulcrum.
The x-acis of a trajectory represents its C
