Answer:
121.43 m
Explanation:
Solution
Since standing waves are set up, the expression for the first mode of frequency is f₁ = nv/4L. The next mode of frequency is f₂ = (n + 2)v/4L where L is the length of the tunnel and v the speed of sound. f₁ = 5.0 Hz, f₂ = 6.4 Hz, v = 340 m/s. We now subtract f₂ - f₁ = (n + 2)v/4L - nv/4L = v/2L.
So f₂ - f₁ = v/2L. and L = v/2(f₂ - f₁) = 340/[2× (6.4-5.0)] = 340/2×1.4 = 340/2.8 = 121.43 m
So, it is 121.43 m far to the end of the tunnel.
If I remember correctly, the correct answer is either:
A. Electrostatic
or
B. Gravitational
<span>51 degrees.
Since we're ignoring friction, we have to have a banking angle such that the normal force is exactly perpendicular to the banked curve. Since this problem says "ignore friction", if the bank angle is too shallow, the bobsled would slide outwards if the banking angle is too shallow and would fall inwards if the banking angle is too steep. So we have to exactly match the calculated centripetal acceleration.
The equation for centripetal acceleration is:
F = mv^2/r
I'll assume a mass of 1 kg to keep the math simple. Any mass could be used and the direction vectors would be the same except their magnitude would differ. So
F = 1 kg * (35 m/s)^2/100 m
F = 1225 kg*m^2/s^2 / 100 m
F = 12.25 kg*m/s^2
The local gravitational acceleration is 9.8 m/s^2, so the sum of those vectors will have a length of sqrt(12.25^2 + 9.8^2) and an angle of atan(9.8/12.25) below the horizon. The magnitude of the vector doesn't matter, merely the angle which is:
atan(9.8/12.25) = atan(0.8) = 38.65980825 degrees.
The banking angle needs to be perpendicular to the force vectors. So
90 - 38.65980825 = 51.34019175 degrees.
Rounding to 2 significant figures gives a bank angle of 51 degrees.</span>
The main difference is direction. C.
Speed is a scalar quantity, while Velocity is a vector quantity.
Speed is measured without taking into fact direction, but Velocity takes into fact the direction.
