Answer:
The frequency increases.
Explanation:
When the Musician draws the slide in the length of the horn gets shorter, which causes a decrease in the wavelength. A decrease in the wave length results in an increase in frequency.
Note:
The diameter of the horn has an effect on frequency, so a wider horn is effectively a long horn - open end correction ( distance between the the antinode and the open end of a pipe).
Frequency also depends on how hard the musician blows the trombone. The musician can change the frequency with the lip pressure being applied.
Answer:
(a) 0.613 m
(b) 0.385 m
(c) vₓ = 1.10 m/s, vᵧ = 3.50 m/s
v = 3.68 m/s², θ = 72.6° below the horizontal
Explanation:
(a) Take down to be positive.
Given in the y direction:
v₀ = 0 m/s
a = 10 m/s²
t = 0.350 s
Find: Δy
Δy = v₀ t + ½ at²
Δy = (0 m/s) (0.350 s) + ½ (10 m/s²) (0.350 s)²
Δy = 0.613 m
(b) Given in the x direction:
v₀ = 1.10 m/s
a = 0 m/s²
t = 0.350 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (1.10 m/s) (0.350 s) + ½ (0 m/s²) (0.350 s)²
Δx = 0.385 m
(c) Find: vₓ and vᵧ
vₓ = aₓt + v₀ₓ
vₓ = (0 m/s²) (0.350 s) + 1.10 m/s
vₓ = 1.10 m/s
vᵧ = aᵧt + v₀ᵧ
vᵧ = (10 m/s²) (0.350 s) + 0 m/s
vᵧ = 3.50 m/s
The magnitude is:
v² = vₓ² + vᵧ²
v = 3.68 m/s²
The direction is:
θ = atan(vᵧ / vₓ)
θ = 72.6° below the horizontal
Answer:
4. B and D
Explanation:
Two points along a transverse wave (such as the one in the figure) are said to be in phase when:
- the vertical position of the two points is the same
- The oscillation of the wave is going in the same way for both points
Basically, we say that two points are in phase when they are separated by a complete cycle (one complete oscillation) of the wave.
For this wave, we see that point B and C have same displacement, but they are not in phase since in B the oscillation is going down while in C is going up.
Instead, B and D are in phase, because they are separated by one complete cycle: both points have same displacement and the oscillation is going in the same way for both of them.
Answer:
= 2630.6 N.m
Explanation:
(FR)x = ΣFx = -F4 = -407 N
(FR)y = ΣFy =-F1-F2 -F3 = -510 - 306 - 501 = -1317 N
(MR)B =ΣM + Σ(±Fd)
= MA + F1(d1 +d2) + F2d2 - F4d3
= 1504 + 510(0.880+1.11) +306(1.11) - 407(0.560)
= 2630.64 N.m (counterclockwise)
