<span>Data:
Initial velocity upward: Vo = 5.00 m/s ,
Initial position: h = 40.0 m above the ground
Type of motion: free fall.
A) Compute the position of the sandbag at a time 1.05 s after its release.
Equation: y = h + Vo*t - g*(t^2) / 2
y = 40.0 m + 5.00 m/s * 1.05s - (9.8 m/s^2) * (1.05 s)^2 / 2 = 39.8 m
B)Compute the velocity of the sandbag at a time 1.05 s after its release.
Equation: Vf = Vo - g*t
=> Vf = 5.00 m/s - (9.8m/s^2) * (1.05 s) = - 5.29 m/s
Negative sign means that the sandbag is going down.
c) How many seconds after its release will the bag strike the ground?
Equation:
y = yo + Vo*t - g*(t^2) / 2
0 = 40.0 + 5.00t - 4.9 t^2
=> 4.9 t^2 - 5t - 40 = 0
Use the quadratic formula and you get: t = 3.41 s
</span>
Answer:
A pipe open at both ends would have an antinode at each end and its length would be λ/ 2
The next such points would be λ and 3 λ / 2
The ratio of 522 / 348 is 1.5 so the harmonic at 348 is one wavelength and the next harmonic is 3 λ / 2 at 1 1/2 wavelengths
348 hz would occur at one wavelength
f λ = v = f L where the length of the pipe is one wavelength
if we use 331 as the speed of sound then
L = 331 / 348 = .95 m for the length of the pipe
Weight = 45 * 9.8 = 441 N
Answer:
A and C
Explanation:
drag (the area of lower air pressure behind the car when moving) and mostly air resistance (the work to push the air in front of us away to move through - the faster we go, the stronger the air resists to move aside).
