The first one, directly proportional to the length that the spring is stretched.
If the canyon below is 100.0 m deep, the rocket travels from the edge of
the cliff 226m in the horizontal direction.
<span>When </span>air resistance<span> acts, acceleration during a fall will be less
than g because </span>air resistance affects<span> the motion of the falling objects by slowing it
down. </span>Air resistance<span> depends on two important factors -
the speed of the object and its urface area.
Increasing the surface area<span> of
an object decreases its speed.</span></span>
The correct answer between all
the choices given is the first choice or letter A. I am hoping that this answer
has satisfied your query and it will be able to help you in your endeavor, and
if you would like, feel free to ask another question.
T is the time for a whole round.
centripetal acceleration = V^2/R,
20 = 40^2 / R, find R = 40^2/20 = 40*40/20 = 80 m, right?
Now, one round is L = 2*pi*R = 2*pi*80 = 160*pi
And T = L/v (distance/speed) = 160*pi/40 = 4*pi seconds, or ~ 12.57 s
Answer:
The frequency increases with a shorter horn <em>(Option B)</em>.
Explanation:
The length of the horn determines the distance along which the wave travels; simply called the wavelength. Therefore, a short horn tube will produce a short wavelength and vice versa.
Sound waves have various characteristics that define pitches in musical instruments and these characteristics are interdependent on each other.
in this case, the frequency and the frequency and the wavelength are related.
The relationship between the wavelength and its frequency is given as:
<em> </em><em>c = f λ </em><em> </em>
<em>where 'c' is the speed of sound through the instrument; 'f ' is the frequency and 'λ' is the wavelength.</em>
Let's assume that the speed at which the musician blows air into the mouthpiece remains constant, an increase in wavelength will cause a decrease in frequency. Conversely, as the tube of the horn becomes shorter the frequency increases.
