Answer:
87.5 m/s
Explanation:
The speed of a wave is given by
![v=\lambda f](https://tex.z-dn.net/?f=v%3D%5Clambda%20f)
where
v is the wave speed
is the wavelength
f is the frequency
In this problem, we have
is the frequency
is the wavelength
Substituting into the equation, we find
![v=(0.35 m)(250 Hz)=87.5 m/s](https://tex.z-dn.net/?f=v%3D%280.35%20m%29%28250%20Hz%29%3D87.5%20m%2Fs)
Answer:
(a) A = 1 mm
(b) ![V_{max}=0.77872 m/s](https://tex.z-dn.net/?f=V_%7Bmax%7D%3D0.77872%20m%2Fs)
(c) ![a_{max}=606.4 m/s^{2}/tex]Explanation:Distance moved back and forth = 2 mm Frequency, f = 124 HzSo, amplitude is the half of the distance traveled back and forth. (a) So, amplitude, A = 1 mm(b) Angular frequency, ω = 2 π f = 2 x 3.14 x 124 = 778.72 rad/s The formula for the maximum speed is given by [tex]V_{max}=\omega \times A](https://tex.z-dn.net/?f=a_%7Bmax%7D%3D606.4%20m%2Fs%5E%7B2%7D%2Ftex%5D%3C%2Fp%3E%3Cp%3E%3Cstrong%3EExplanation%3A%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3EDistance%20moved%20back%20and%20forth%20%3D%202%20mm%20%3C%2Fp%3E%3Cp%3EFrequency%2C%20f%20%3D%20124%20Hz%3C%2Fp%3E%3Cp%3ESo%2C%20amplitude%20is%20the%20half%20of%20the%20distance%20traveled%20back%20and%20forth.%20%3C%2Fp%3E%3Cp%3E%28a%29%20So%2C%3Cstrong%3E%20amplitude%2C%20A%20%3D%201%20mm%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3E%28b%29%20Angular%20frequency%2C%20%CF%89%20%3D%202%20%CF%80%20f%20%3D%202%20x%203.14%20x%20124%20%3D%20778.72%20rad%2Fs%20%3C%2Fp%3E%3Cp%3EThe%20formula%20for%20the%20maximum%20speed%20is%20given%20by%20%3C%2Fp%3E%3Cp%3E%5Btex%5DV_%7Bmax%7D%3D%5Comega%20%5Ctimes%20A)
![V_{max}=778.72 \times 0.001](https://tex.z-dn.net/?f=V_%7Bmax%7D%3D778.72%20%5Ctimes%200.001)
![V_{max}=0.77872 m/s](https://tex.z-dn.net/?f=V_%7Bmax%7D%3D0.77872%20m%2Fs)
(c) The formula for the maximum acceleration is given by
![a_{max}=\omega ^{2}A](https://tex.z-dn.net/?f=a_%7Bmax%7D%3D%5Comega%20%5E%7B2%7DA)
![a_{max}=778.72 ^{2}\times 0.001](https://tex.z-dn.net/?f=a_%7Bmax%7D%3D778.72%20%5E%7B2%7D%5Ctimes%200.001)
[tex]a_{max}=606.4 m/s^{2}/tex]
Answer:
4 km/hr
Explanation:
The computation of the actual velocity is shown below:
Because the path of its paddles is opposed to the current direction, the real velocity can be determined by deducting the current velocity to its velocity while paddling
So, the actual velocity is
= Upstream - downstream
= 19 km/hr - 15 km/hr
= 4 km/hr
As we can see it is in positive, so it is an upstream direction
The bullet travels a horizontal distance of 276.5 m
The bullet is shot forward with a horizontal velocity
. It takes a time <em>t</em> to fall a vertical distance <em>y</em> and at the same time travels a horizontal distance <em>x. </em>
The bullet's horizontal velocity remains constant since no force acts on the bullet in the horizontal direction.
The initial velocity of the bullet has no component in the vertical direction. As it falls through the vertical distance, it is accelerated due to the force of gravity.
Calculate the time taken for the bullet to fall through a vertical distance <em>y </em>using the equation,
![y=u_yt+\frac{1}{2} gt^2](https://tex.z-dn.net/?f=y%3Du_yt%2B%5Cfrac%7B1%7D%7B2%7D%20gt%5E2)
Substitute 0 m/s for
, 9.81 m/s²for <em>g</em> and 1.5 m for <em>y</em>.
![y=u_yt+\frac{1}{2} gt^2\\ 1.5 m=(0m/s)t+\frac{1}{2} (9.81m/s^2)t^2\\ t=\sqrt{\frac{2(1.5m)}{9.81m/s^2} } =0.5530s](https://tex.z-dn.net/?f=y%3Du_yt%2B%5Cfrac%7B1%7D%7B2%7D%20gt%5E2%5C%5C%201.5%20m%3D%280m%2Fs%29t%2B%5Cfrac%7B1%7D%7B2%7D%20%289.81m%2Fs%5E2%29t%5E2%5C%5C%20t%3D%5Csqrt%7B%5Cfrac%7B2%281.5m%29%7D%7B9.81m%2Fs%5E2%7D%20%7D%20%3D0.5530s)
The horizontal distance traveled by the bullet is given by,
![x=u_xt](https://tex.z-dn.net/?f=x%3Du_xt)
Substitute 500 m/s for
and 0.5530s for t.
![x=u_xt\\ =(500m/s)(0.5530s)\\ =276.5m](https://tex.z-dn.net/?f=x%3Du_xt%5C%5C%20%3D%28500m%2Fs%29%280.5530s%29%5C%5C%20%3D276.5m)
The bullet travels a distance of 276.5 m.